text:
The Leon Show
The Leon Show, Guest Dr. Chris Thurman "Building a Great Marriage"
[cc]
-LEON FONTAINE: Today, the ten most common lies couples believe
about their marriage.
- Dr. Chris Thurman knew he wanted to be a psychologist from
an early age. With a superior knowledge of human emotions and
mental health, Thurman credits his Christian faith as the lens
through which he treats all patients and strives to see
healing, wholeness and freedom. On top of consulting with
churches and running a very successful private practice, Dr.
Thurman finds time to write best-selling books exploring
common emotional, relational and spiritual struggles. Dr. Thurman
and his wonderful wife Hope have been married for over 30 years
and call Austin, Texas home. Here to offer wisdom and
insight, please welcome Dr. Chris Thurman. [Applause] Dr.
Thurman, welcome! It's good to have you here!
-DR. CHRIS THURMAN: Thank you, Leon. Great to be here!
-LEON: That's quite the title for a book, 'Lies. 'The Lies
Couples Believe.' But if they did believe a lie, that's
completely destructive.
-DR. THURMAN: Very destructive and it's not an easy title to
sell, because who wants to admit to believing a lie, right?
-LEON: That's very true. Now, you work in Austin, Texas. Tell
me a little bit about what you do there.
-DR. THURMAN: Well, I'm a psychologist in private
practice. I went to the University of Texas, which is in
Austin, as an undergraduate and that's where I got my doctorate
from as well. And I got my wife and kids back to Austin as soon
as I could after spending about ten years up in Dallas.
-LEON: Okay. So now, how many years have you been working with
marriages?
-DR. THURMAN: I have been in the field of counseling for 40
years, so I would say 35 of those I've worked with couples.
-LEON: Couples, so this book is coming from a whole lot of
-DR. THURMAN: It is. It is, yes.
-LEON: Well, I thought one of the fun things, like people? We
cannot get through the whole book in the shows that we're
going to do, but I think that if we can touch on some of the
things that you have noticed about couples that I think it
would help people immensely, because there's a lot of married
people out there and the truth is, all of us want to work and
have better marriages. If someone doesn't, then there's a
lie there somewhere already. Where would you start with the
lies couples believe?
-DR. THURMAN: Well, let me start with the one that's pretty
common and incredibly destructive, which is that the
purpose of marriage is to be happy, and that is a tough sell
in this day and age to say that it's not the purpose of
marriage. So that one is, what I'm trying to go after there is
the idea that the main focus in your marriage needs to be your
happiness, and I think that gets a lot of couples in trouble
because inevitably in marriage you're going to wound each
other, you're going to be selfish at times, you're going
to be irresponsible. And therefore, if your happiness is
dependent on how you're being treated, then you're going to be
unhappy and then you start to think about the grass has got to
be greener somewhere else. I'll go find somebody that I'll be
happier with. So what I try to drive home in that chapter is,
the real focus in marriage is on healthy, intimate companionship
and growth. If you can make it about growth, then you've got
the right mindset because God is using that relationship to help
you mature into more loving, healthy Christ-like people.
-LEON: So if I'm hearing you right, then if you focus on your
happiness you're pretty self-centered, then?
-DR. THURMAN: You are very self-centered, you are being
what we would call narcissistic because you're making it all
about you, and you're going to end up unhappy which is the
paradox of it. The more you seek your happiness directly, the
more unhappy you get over time. -LEON: You know, it's
interesting. We use the word 'happy' now a lot. Like years
and years ago, even parents, if you'd ask them, "What do you
want for your kids?" They would say something like, "Great
purpose, living meaningful lives." Today, I hear this all
the time as a Pastor, "I just want them to be happy." And now
you're saying, "If you pursue happiness, you're guaranteed
unhappiness."
-DR. THURMAN: It is the way the god of darkness and the father
of lies has set it up. If you can get to believing what he
wants you to believe, which is your whole issue needs to be:
are you happy? You're done! You are going to get creamed.
-LEON: So when it comes to looking at our marriages, we
pursue happiness; we're going to find unhappiness. But when it
comes to growth, I heard someone say somewhere that the thing
that makes depression work the most in our lives is when we
stop growing. Is that true?
-DR. THURMAN: I think depression is a symptom of: we
have quit growing, that we've metaphorically curled up in a
fetal position you know, and we've just kind of tossed in the
towel, and we have gotten hopeless about life ever being
enjoyable or meaningful or happy. But I think again, the
depression can be a symptom that somebody wasn't focused on
growth, and therefore they just kept pursuing happiness. And
when a couple does that, I think they have a relatively depressed
marriage. It's real down, it's in the ditch, it's dark. There's
not much joy or happiness in the marriage again, when that's
their number one reason for being together.
-LEON: When you go out for lunch, and you notice young
couples that are newly in their relationship, it's animated
talking. Then when you notice like, a lot, not all but a lot
of older couples, it's like not a word the entire meal. It just
stands out to me sometimes. And so, it doesn't matter how old
you are though, you're saying there's lots to grow in. You
-DR. THURMAN: You never stop if you're thinking about the right
focus. Young couples? I'm trying not to sound like I'm down on
people being happy because I'm sure not.
-LEON: But still, it's a lie you're saying that is hurting
people. Let's fix it!
-DR. THURMAN: It's a lie and they key with young couples is,
they fall in love and there's nothing that'll make you happier
than being in love but 'in love' is not love, okay? In love is
more of a psychosis, you know? You're walking around, you're
nine feet off the ground. You don't know what in the world
you're even thinking or feeling half the time. You sure don't
know the person yet, because you're in love with the fantasy
of who you think they are.
-LEON: That's true. You're in love with the dream you have of
-DR. THURMAN: That's right, and you walk down the aisle and it
doesn't take too long before the reality that you've married a
fallen human being who can be what I call the unholy trinity;
selfish, lazy and immature, and you're on the way to some
wounds. And can you weather those and can you come together
as a couple, because of your vow and because of your commitment
and really roll up your sleeves about: we need to grow, we need
to come from being children who thought and reasoned like
children, to full-fledged adults who no longer think childish
-LEON: Now, most couples are pretty aware of where their
spouse needs to grow. So, what? You're going to cause a big
fight here if we all go home and go, "Okay, I just heard a thing
on growth and you need to grow." So how do you resolve that?
-DR. THURMAN: Well, when I'm working with couples, I really
get on them if they're in the other person's backyard, so I do
not allow that in my office. It's like, "You are not going to
work on them. You are not going to play Holy Spirit. You are not
going to look down your nose like they're the one that needs
to grow and you don't need to have anything worked on in you."
So I lovingly get on them pretty good if that's the name of the
game in a session that we're doing.
-LEON: All right, so I think there's a couple things you
said. So the first? The things to focus on weren't: am I happy?
It was growth. What was the other one you mentioned?
-DR. THURMAN: Well, companionship obviously. It's
not good for man to be alone, so when God created marriage He was
looking at Adam's situation. And even though Adam had a perfectly
intimate relationship with God, he didn't have another human
being to be with. So I think we have to keep our eye on the
purpose of a marriage is healthy, intimate bonding,
healthy attachment. But it's also about children as well, we
are to fill the earth and we are to also I think, change is the
final issue. It's a marriage is not going to do anything but die
if it's not growing. Marriages do not stay in neutral. They're
either getting better or they're getting worse.
-LEON: What are some of the signs of a growing marriage, or
of a changing marriage? Is there anything that you've kind of
zoned in one?
-DR. THURMAN: I think a growing marriage is marked by a couple
of things, for sure. Service is one. A couple that you can see
is really interested in serving each other, that's a marriage
that's going to grow because that's what Christ came to do.
He came to serve, not to be served. So if you see two people
who are looking out for the other, who are looking for that
other person's needs to matter as much as their own do, that's
a marriage that's going to grow. It can't not grow. I think a
humble marriage is a marriage that's going to grow. When you
have two people who are not full of themselves and who are
willing to say, "I am a human being. I'm fallen, I have my
flaws." That's a marriage that's going to really take flight and
be, stand out like a sore thumb to the rest of the world.
-LEON: So would you say then, as we come to the close of this
first lie, that if you will focus on change, if you'll focus
on growth that happiness seems to follow that?
-DR. THURMAN: I believe it does. I think that's how God has
wired it. I think He's saying, "If you will put your eye on the
right thing, you get this additional wonderful thing added
to it, but don't go after that directly. Go after the real
deal, and happiness comes and it lands on your shoulder."
-LEON: Let's take a break right here and then when we come back,
let's unpack some more lies. My guest today is Dr. Chris Thurman
with this book, 'The Lies that Couples Believe.' You need to
get a copy! We'll be right back.
-DR. THURMAN: Yeah, the bar is just set way too high. Again,
nobody can meet all of your needs, and if you think that
they can and should then you end up resentful, you end up bitter.
You can't help but feel that.
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-LEON: Welcome back. My guest today is
Dr. Christ Thurman and he's written this book, 'The
Lies Couples Believe.' And we're just kind of unpacking some of
these lies. So, the first one is: don't pursue your own
happiness. You're going to end up in the ditch.
-DR. THURMAN: Don't make it the focus of the marriage as far as
the main purpose.
-LEON: All right, what's another one we can cover?
-DR. THURMAN: Well, another one that's very common in marriage
is that my spouse can completely meet all of my needs, and I see
that one really being destructive in a marriage,
because the simple truth is, nobody can completely meet all
your needs. So if you go up to a spouse and you drop that on
their doorstep and you say, "That's your job, and if you
don't meet every need I have in total, then you're not doing
your job." It leads to a lot of resentment, a lot of bitterness,
a lot of anger. So what I try to teach my couples is: there is
not one single person on the planet who can completely meet
your needs. That's God's job and He works through a variety of
people and activities and hobbies and passions, and His
Word and the Holy Spirit to meet your needs. Don't do that to a
person. Don't do that to them. That's just too much to ask of
any person, and they'll never feel like they are enough.
Nobody wants to walk around feeling like, I just can never
be enough for you.
-LEON: Yeah, that would. You know, I've met people just in
pastoral counseling. We don't do a lot of marriage counseling but
you know, just getting together a couple times with couples and
then moving them somewhere, that some people are so bitter and so
angry when you first talk to them, and then you meet their
spouse and you kind of go, "There's no real problem here."
But that, you said it, that bitterness comes from their
expectations and their wrong expectations.
You can't help but feel that. So the expectation that they will
meet all of my needs, otherwise why did we marry? That's kind of
an attitude that a lot of people have about marriage is, you
marry the person who's going to meet your complete package of
needs, completely. And again, it's just not realistic. Nobody
can do that. On our best day, we can probably meet a part of your
needs to some degree if we're trying to die to self enough.
But this whole thing about the movie with Tom Cruise and you
know, you complete? It's like, no! Nobody completes you, okay?
That's just worldly thinking, it's Hollywood and that's not
the way to look for a spouse. You don't go out there looking
for somebody that just completely fulfills everything
you are. It's just a delusion to think that way.
-LEON: Even, I think even when people are dating. I love your
thought on that. They find someone that they think is going
to do that and people seem to like that, "I complete you! Oh,
I can really help you. I'm the answer." But then, once you get
married for a couple of years, I think you really resent. Is that
true? Like having to meet that person's needs all the time and
be the reason they're not? Like, any time you're the reason
they're not happy, that's got to be one of the most destructive
-DR. THURMAN: Right, right. I think the resentment builds up
in the other person too, because they feel like, "Well, I'm just
basically here for your use. I'm just here for your disposal."
And I think unconsciously if not consciously, we start to be
bitter about, "Is that all I am to you? I'm just here for your
needs and you're going to get on me if I don't meet them
completely, which I can't do?" It's just a real sick, unhealthy
pattern for a couple to get into.
-LEON: I was talking earlier about being in a restaurant and
watching these young couples animatedly talking, and then an
older couple hardly talking at all. But then, on the other
hand, there's something pretty cool about being so at home in
each other's presence. You could read a book. You don't have to
impress them. You're just so comfortable being together as
friends, I think is probably a pretty cool side as well.
-DR. THURMAN: That's a very cool side, but I will say this
about the older couple that's sitting there in silence. When
they're in my office, I get on them pretty good too because
it's like, "Well, let me tell you. You've gone the other
extreme, okay? You have quit meeting each other's needs and
you guys do not need to be sitting there in silence at a
restaurant, okay?" So Satan to me is always into extremes. He
either wants you to think you should meet all of them, or I
don't have to meet any of them. He's the god of extremes, if you
will. So I lovingly push my older couples. You were meant to
be happy in each other's arms, you were meant to be lively, you
were meant to be passionate, you were meant to be visceral around
each other. So if you're sitting there in dead silence for an
hour meal at a restaurant, something has gone south on you
and maybe it's you're not meeting each other's needs much
at all. So I push them pretty hard on that.
-LEON: That's very cool. So all right, so lies, and if couples
have, like you're going through ten of them. You've probably
found more than ten, but any one of these could probably destroy
a marriage.
-DR. THURMAN: Any one of them by themselves could destroy and
does destroy marriage. I think we have to really be mindful of
the enemy's schemes and that, "We war not against flesh and
blood, but against spiritual forces of darkness." So as the
father of lies, what other way to torpedo a marriage than to
get people thinking things that are unbiblical and carry that
into the marriage every day, and thus the anger ratchets up, the
resentment and bitterness grows stronger. People start
withdrawing from each other or attacking each other, and the
marriage is really under a great deal of duress and it's going to
take a lot of work to get it out of that.
-LEON: It's so true. I think that people need to recognize
the enemy, he really shoots darts which are lies. Every
thought that seems to come into your head isn't true, and he
just seems to make us ask the wrong questions. It seems like
wisdom is in the questions, and we start asking questions that
are so self-centered and so negative about our partner, and
then that lie just takes us down. And the Bible say, "The
truth sets us free." So almost all truth is confrontational.
-DR. THURMAN: I think it is, and I think that's another
reason why we are not all that open to it being pointed out
that we're not thinking right. I think the truth has a way of
being pretty confrontational, pretty in your face, so I know
when I'm working with my couples, it's I'll challenge
them on some of their thinking, but you can see the resistance.
It's like, "I may not make an appointment with you for next
week, Dr. Thurman, and that's what you get for telling me the
truth today because I'm not really super open to it." But
whether you like it or not, you're to speak the truth in
love and we are all to try to let God have His way with us
about when He wants to hold up a mirror and tell us that we're
not thinking correctly.
-LEON: Well, I think some people, once you even believe
one lie. Like, let's pick on the first one, that you're to make
me happy. So you begin to develop an entire system of
thinking. Every flaw, every failure, everything you have or
haven't done, you begin to build this line upon line. But your
entire thoughts of your marriage, your past, your future
are all around this one lie. So if you can really bring that up
and show them, you can save a marriage. If not, they'll
continue to build layer after layer. Is that true?
-DR. THURMAN: Yes, and that's where you run into whether or
not the client is truly open to saving their marriage. If I can
hold that truth up for them and if they are willing to humble
themselves and say, "You know what? I think I need to admit
that I've been thinking a wrong way for decades."
-LEON: That's hard to do!
-DR. THURMAN: It's hard to do; I mean, any of us don't want to.
-LEON: Like, talk to me about humility because you said this
two or three times now that people who are proud, you almost
can't help. But Proverbs says that, "Pride goeth before
destruction," is it? "And a haughty spirit before a fall."
So once a person is locked into their belief, some don't want a
change in there.
-DR. THURMAN: And that's why as a Christian, it's especially
powerful to be able to say, "Let me, not in a Bible-thumping way,
but let me show you where this is in the Word of God, because
you're not arguing with me, okay? You're arguing with God,
and if you want to take Him on you can, but you're going to
lose. So it's not just me telling you that the purpose of
marriage is to grow. Let me show you in scripture where it says
that, and if you want to take Him on and argue with Him? Okay,
but you're going to lose, because He is truth and He
doesn't mess around and He's not going to negotiate with you
about how to think in a marriage in order for it to be a loving
one."
-LEON: I think one of the best things that married couples can
do, you know pastoring over the years, is continually together
read a marriage book every month or two and it gives you stuff to
talk about. And then just rehearse it, just go through the
same one together and talk about what you agree with, what you
don't agree with. But it just plants seeds of truth and even
if you don't want to go there today, it just seems to begin to
grow. Holy Spirit has this amazing ability to take that
truth and it begins to grow in you, and then you start dealing
with yourself because it's like-- it's not like a
mechanical paradigm where, "Find broken person. Fix broken
person, you know? Find the part." It's like an agricultural
where it just grows, and so do you find that frustrating
sometimes that you just, you can't fix things right away but
you just teach them truth and they go back home?
-DR. THURMAN: I do, because I am more of a fixer by nature. I
like to really get into something and fix it, so it is
frustrating for me just because I'm not all that patient. So
with my clients, we have to all learn how to be patient and let
it be a process, and let it be a lengthy one because it is, it is
a long time of hard work.
-LEON: It's worth it.
-DR. THURMAN: But it is worth it.
-LEON: Man, our time is up! Thank you so much for being with
-DR. THURMAN: My pleasure. Thank you!
-LEON: We've been talking today with Dr. Chris Thurman about
this book, 'The Lies Couples Believe.' We got through two,
barely! And only scratched the surface. You need to get a hold
of this book. Go to work on your marriage. We'll be right back.
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text:
Features » Process Controls & Plant » Boilers Burners & Controls
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text:
\section{generalities}

If $G$ is a group then $G',$ and $ \zeta{G}$ shall denote the commutator subgroup,
and centre of $G$, respectively.
If $J$ is a subgroup of $G$ then $C_G(J)$ is the centralizer of $J$ in $G$.
Let $G^{(0)}=G$ and $G^{(n+1)}={G^{(n)}}'$ be the terms of the derived series for $G$,
and let $\gamma_1G=G$ and $\gamma_{n+1}G=[G,\gamma_nG]$ be the terms
of the lower central series for $G$.
(This notation is from \cite{Ro}.)
Let $I(G)$ be the {\it isolator\/} of $G'$ in $G$,
so that $G/I(G)$ is the maximal torsion-free abelian quotient of $G$.
Let $\sqrt{G}$ denote the Hirsch-Plotkin radical of $G$.
If $G$ is virtually polycyclic then $\sqrt{G}$ is the unique maximal nilpotent
normal subgroup of $G$, and is also known as the Fitting subgroup of $G$.
We shall use ``simple"  to mean ``non-abelian and simple" throughout.

A virtually solvable group $S$ has a solvable radical $\widetilde{S}$ of finite index.
If $S$ is not solvable then the lowest term of a composition series
for $S/\widetilde{S}$ is a finite simple group,
and so contains a minimal simple group \cite{BW}.
We shall say that $S$ is {\it minimal TFNS\/}
if it is finitely generated,  torsion-free,
$S/\widetilde{S}$ is a minimal simple group and $S$
has minimal Hirsch length for such groups.
Thus $h(S)\leq15$.  (Cf. \cite[Definition 2.1]{LuS}.)

An {\it outer action\/} of a group $H$ on a group $G$
is a homomorphism $\theta:H\to{Out(G)}$.
Such a homomorphism induces homomorphisms from $H$ to $Aut(G/G')$
and $Aut(\zeta{G})$.
We shall say that the action is {\it effective\/} if it is a monomorphism.

A {\it crystallographic\/} group $G$ is a group which is an extension $\xi$
of a finite group $H$ by a finitely generated free abelian group $A$,
such that the action $\theta:H\to{Aut(A)}$ induced by conjugation
in $G$ is effective.
The {\it holonomy group\/} of $G$ is the quotient $G/A$.
Such extensions are classified by $\theta$ and a cohomology class
$c(\xi)$ in $H^2(H;A)$, where $H$ acts on $A$ via $\theta$.

A {\it Bieberbach\/} group is a torsion-free crystallographic group.
A crystallographic group corresponding to an extension $\xi$
is a Bieberbach group if and only if  $c(\xi)$ has non-zero restriction
in $H^2(C;A)$ for every cyclic subgroup $C<H$ of prime order $>1$
\cite[Theorem 3.1]{Sz}.

The condition that $S$ be torsion-free is largely irrelevant for
many of our arguments limiting the possible finite perfect quotients $S/K$
that can act effectively on a normal subgroup $K$ of finite index in $S$.

\section{some useful lemmas}

We have adapted the first two lemmas in this section for our needs.

\begin{lemma}
\label{crys}
\cite[Theorem 2.2]{Sz}
Let $G$ be a finitely generated, virtually abelian group.
Then $G$ is a crystallographic group if and only if it has no non-trivial finite normal subgroup.
In that case $\sqrt{G}$ is  free abelian and is
the maximal normal abelian subgroup of $G$,
and $G$ has holonomy $H=G/\sqrt{G}$.
If $H$ is not solvable then $h(G)\geq4$.
\end{lemma}

\begin{proof}
We may assume that $G$ is an extension of a finite group $H$
by a finitely generated free abelian normal subgroup $A$.
Then $A$ has finite index in $\sqrt{G}$,
and so $\sqrt{G}$ is nilpotent and has finite torsion subgroup.

Let $C=C_G(A)$ be the centralizer of $A$ in $G$.
Then $G/C$ acts effectively on $A$.
Since $[C:A]$ is finite, $C'$ is finite,
by a theorem of Schur \cite[10.1.4]{Ro}.
Hence if $G$ has no non-trivial finite normal subgroup then
$C'=1$ and $C$ is free abelian.
By the same reasoning, $\sqrt{G}$ is abelian and so $C=\sqrt{G}$
is the maximal abelian normal subgroup of $G$.
Since $G/C$ acts effectively on $A$ it acts effectively on $C$.
Thus $G$ is a crystallographic group with holonomy $H_1=G/C$.

Conversely,  suppose that $G$ is a crystallographic group with holonomy $H$.
if $F$ is a finite normal subgroup of $G$ then $A\cap{F}=1$,
since $A$ is torsion-free, and so $F$ projects injectively to $H$.
Moreover, since $A$ and $F$ are each normal,
$[A,F]\leq{A}\cap{F}=1$ and so $AF\cong{A}\times{F}$.
Since the action of $H$ on $A$ is effective
we have $F=1$.

The final observation holds since finite subgroups of
$GL(3,\mathbb{Q})$ are solvable.
\end{proof}

\begin{lemma}
\label{HS?}
\cite[Prop.  4.1]{Sz}
Let $G$ be a crystallographic group with holonomy $H$.
If $H^1(G;\mathbb{Q})=0$ then ${Hom(\sqrt{G},\mathbb{Q})}^H=1$.
Hence the $\mathbb{Q}[G]$-module $\mathbb{Q}\otimes\sqrt{G}$
has no summand $\mathbb{Q}$ with trivial $H$-action.
\end{lemma}

\begin{proof}
This follows from the exact sequence of low degree in the
LHS spectral sequence for $G$ as an extension of $H$ by $\sqrt{G}$,
since $H^1(G;\mathbb{Q})=H^2(H;\mathbb{Q})=0$.
The second assertion is then clear.
\end{proof}

Let $N$ be a nilpotent group,
and let $tN$ be its torsion subgroup.
Then $N/tN$ is torsion-free and has a  central series
with torsion-free abelian subquotients \cite[5.2.7 and 5.2.20]{Ro}.
Let
\[
\tilde\gamma_1N=N>\dots>\tilde\gamma_cN=tN
\]
be the preimage in $N$ of the most rapidly descending such central series.
Then $\tilde\gamma_2N=I(N)$.
Let $\mathbb{Q}N^{ab}=\mathbb{Q}\otimes{N/N'}=
\mathbb{Q}\otimes{N/\tilde\gamma_2N}$.

\begin{lemma}
\label{vnil}
Let $G$ be a torsion-free virtually solvable group of finite Hirsch length
and let $N=\sqrt{G}$.
Suppose that $G/N$ is a torsion group.
Then the  homomorphism from $G/N$ to $Aut(\mathbb{Q}N^{ab})$
induced by conjugation in $G$ is a monomorphism and  $G/N$ is finite.
If $G$ is finitely generated then $G/I(N)$ is crystallographic.
\end{lemma}

\begin{proof}
Since $G/N$ is a torsion group $h(N)=h(G)<\infty$,
and so $r=h(N^{ab})$ is finite.
Let $C$ be the kernel of the homomorphism from $G$ to
$Aut(\mathbb{Q}N^{ab})\cong{GL(r,\mathbb{Q})}$ induced by conjugation in $G$.
Then $N\leq{C}$ and $C/I(N)$ contains
$N/I(N)$ as a central subgroup of finite index.
Conjugation by elements of $C$ also induces the identity on
the subquotients $\tilde\gamma_iN/\tilde\gamma_{i+1}N$, for all $i>1$.
Hence $C$ is nilpotent,
by Baer's extension of a theorem of Schur \cite[14.5.1]{Ro}
and the fact that $G$ is torsion-free.
Hence $C=N$, by the maximality of $N$,
and so $G/N$ embeds in $Aut(\mathbb{Q}N^{ab})$.
Since $G/N$ is a torsion subgroup of $GL(r,\mathbb{Q})$
 it is finite \cite[8.1.11]{Ro}.

This monomorphism factors through $Aut(N/I(N))$,
and so $G/N$ acts effectively on $N/I(N)$.
If $G$ is finitely generated then $N/I(N)\cong\mathbb{Z}^r$,
and so $G/I(N)$ is a crystallographic group.
\end{proof}

\begin{lemma}
\label{HPnilp}
Let $S$ be a torsion-free solvable group such that $h(S)$ is finite.
Then $\sqrt{S}$ is nilpotent, and $C_S(\sqrt{S})=\zeta\sqrt{S}$.
\end{lemma}

\begin{proof}
If $N$ is a  finitely generated subgroup of $\sqrt{S}$ then $h(N)\leq{h}=h(S)$,
and so $\gamma_{h+1}N=1$, since $S$ is torsion-free.
It follows immediately that $\gamma_{h+1}\sqrt{S}=1$,
and so $\sqrt{S}$ is nilpotent.

Let $C=C_S(\sqrt{S})$,
and suppose that $\sqrt{S}$ is a proper subgroup of $C.\sqrt{S}$.
Since $S$ is solvable it has a subgroup $D$ containing $C.\sqrt{S}$
and such that $D/\sqrt{S}$ is abelian.
But then $D$ is nilpotent,
contradicting the maximality of the Hirsch-Plotkin radical.
Hence $C=C_S(\sqrt{S})\leq\sqrt{S}$ and so $C=\zeta\sqrt{S}$.
\end{proof}

In the virtually polycyclic case $S/\sqrt{S}$ is virtually abelian \cite[15.1.6]{Ro}.
However, this not so for all finitely presentable solvable groups \cite{RS}.

If $G$ is virtually polycyclic and all of its abelian subnormal subgroups
have rank $\leq{n}$ then $G/\sqrt{G}$ is virtually abelian of rank $<n$
\cite[Theorem 2]{Wi}.
Hence $h(G/\sqrt{G})<h(\sqrt{G})$,
since $\sqrt{G}$ contains all abelian subnormal subgroups.
We may push this inequality a little further.

\begin{lemma}
\label{Wilson}
Let $G$ be a virtually polycyclic group, and let $V$ be a
normal subgroup of finite index which contains $\sqrt{G}$
and such that $V/\sqrt{G}$ is abelian.
Let $C$ be the preimage in $G$ of the centre of $V/\sqrt{G}'$.
Then $h(G/\sqrt{G})<h(\sqrt{G}/\sqrt{G}'.V'\cap{C})$.
\end{lemma}

\begin{proof}
We note first that $\sqrt{G}=\sqrt{V}$,
since $V$ is a normal subgroup and $\sqrt{G}\leq{V}$.
Hence we may assume that $G=V$, since $h(G)=h(V)$,
and so $G'\leq\sqrt{G}$.
The preimage of $\sqrt{G/\sqrt{G}'}$ in $G$ is $\sqrt{G}$
\cite[5.2.10]{Ro},
and so we may pass to the quotient $G/\sqrt{G}'$.
Hence we may also assume that $\sqrt{G}$ is abelian.
We then have $C=\zeta{G}\leq\sqrt{G}$ and
$\sqrt{G/(G'\cap{C})}=\sqrt{G}/(G'\cap{C})$,
since $G'\cap{C}$ is central.
Applying \cite[Theorem 2]{Wi} to $G/G'\cap{C}$ gives the result.
\end{proof}

This bound is sharp.
For example, if $K$ is a totally real number field of degree 5,
then the group of integral units $\mathcal{O}_K^\times$ has rank 4,
and acts effectively on $\mathcal{O}_K$.
Hence $G=\mathcal{O}_K\rtimes\mathcal{O}_K^\times$ is
virtually torsion-free poly-$\mathbb{Z}$.
The abelian normal subgroup $\mathcal{O}_K$ is its own centralizer in $G$,
and so $\sqrt{G}=\mathcal{O}_K$ and $\zeta{G}=1$.
Hence  $h(G/\sqrt{G})=4<h(\sqrt{G})=5$.

\section{minimal TFNS groups}

If a virtually solvable group $S$ is minimal TFNS then $S^{ab}$ is finite.
For otherwise there would be an epimorphism $\phi:S\to\mathbb{Z}$,
and $K=\mathrm{Ker}(\phi)$ would not be solvable.
Since $K$ is virtually solvable,
it has finitely generated subgroups which are non-solvable,
but have Hirsch length $<h(S)$,
contradicting the minimality of $S$.
We can improve on this result.

\begin{lemma}
\label{perf}
Let $S$ be a virtually solvable group such that $H=S/\widetilde{S}$ is perfect.
If $\widetilde{S}^{(n)}=1$ then $S^{(n)}$ is perfect,
and if $H$ is simple then $S^{(n)}/\widetilde{S^{(n)}}\cong{H}$.
\end{lemma}

\begin{proof}
Let $f:S\to{T}=S/S^{(n+1)}$ be the natural epimorphism.
Then $f$ induces an epimorphism from $H$ onto $T/f(\widetilde{S})$,
since $f(\widetilde{S})$ is normal in $T$.
Since $H$ is perfect and $T$ is solvable it follows that $f(\widetilde{S})=T$,
and so $f$ induces epimorphisms $f^{(k)}:S^{(k)}\to{T^{(k)}}$,
for all $k\geq1$.
Hence $T^{(n)}=1$ and so $S^{(n)}=S^{(n+1)}$.
Thus $S^{(n)}$ is perfect.
The final assertion is clear,
since $H$ is the only non-solvable quotient of $S$.
\end{proof}

In particular,  if $S$ is minimal TFNS then it has a perfect subgroup of finite index
which is also minimal TFNS.
We may assume that $S$ is finitely generated,
since some finitely generated subgroup of $S$ must map onto $H$.
We may also assume that $S$ is perfect, by Lemma \ref{perf}.

Specializing further, we see that if $G$ is a crystallographic group
with perfect holonomy $H$ then $G'$ is a perfect crystallographic group,
and if $H$ is simple then $G'$ also has holonomy $H$.

\begin{lemma}
\label{overline}
Let $S$ be a finitely generated, infinite group which is torsion-free,
virtually solvable and of finite Hirsch length.
Then $S$ has normal subgroups $V\leq{U}\leq\widetilde{S}$ such that
$S/U$ is crystallographic,  $U/V$ is finite and $V/I(V)$ has positive rank.
\end{lemma}

\begin{proof}
Since $S$ is finitely generated, infinite and virtually solvable it has a
a normal subgroup $T\leq\widetilde{S}$ of finite index
which is solvable and has infinite abelianization.
The quotient $S/T'$ is finitely generated and virtually abelian,
and so has a maximal finite normal subgroup.
Let $U$ be the preimage in $S$ of this subgroup and let $\overline{S}=S/U$.
Then $A=\sqrt{\overline{S}}$ has finite index in $\overline{S}$
and so is finitely generated.
The torsion subgroup $tA$ is trivial,
since it is a finite normal subgroup of $\overline{S}$.
Since $A=\sqrt{\overline{S}}$ and is torsion-free and virtually abelian,
it is abelian.
It is also its own centralizer in $\overline{S}$,
and so $\overline{S}/A$ embeds in $Aut(A)$.
Hence $\overline{S}$ is a crystallographic group.

Since $U\leq\widetilde{S}$ it is solvable.
Let $U^{(i)}$ be the largest member of the derived series for $U$ such that
$U^{(i)}/I(U^{(i)})$ has positive rank,
and let $V$ be the preimage in $U$ of the centralizer of
$U^{(i)}/I(U^{(i)})$ in $U/I(U^{(i)})$.
Then $U^{(i)}\leq{V}\leq{U}$, and so $U/V$ is a torsion group.
Since it acts effectively on $U^{(i)}/I(U^{(i)})$, which is a torsion-free abelian group
of finite rank, it is finite  \cite[8.1.11]{Ro}.
Since $U^{(i)}$ is central in $V$ and $V/U^{(i)}$ is a torsion group,
the commutator subgroup $V'I(U^{(i)})/I(U^{(i)})$ is also a torsion group.
Thus $V/I(V)$ is torsion-free, of positive rank, and $S/V$ is virtually abelian.
\end{proof}

Note that if $U$ is nilpotent then $V=U^{(i)}=U$,
and $\sqrt{S/U}$ is the unique free abelian normal subgroup of minimal index in $S/U$.
Moreover, $S$ is then virtually polycyclic if and only if $U$ is finitely generated.

\begin{cor}
If $S$ is minimal TFNS then $h(\overline{S})\geq4$
and we may assume that $\overline{S}$ is perfect.
\end{cor}

\begin{proof}
If $S$ is minimal TFNS then $\overline{S}$ is not solvable,
since $U$ is solvable.
Hence $h(A)\geq4$, by Lemma \ref{crys}.
After replacing $S$ by $S'$, if necessary,
we may assume that $\overline{S}$ is perfect, by Lemma \ref{perf}.
\end{proof}

\section{the relevant minimal perfect groups}

A finite perfect group is {\it minimal\/} if all of its proper subgroups are solvable.
The minimal simple groups are $PSL(2,q)$ for $q=2^p$ with $p$ a prime,
$q=3^\ell$ with $\ell$ odd, $q=p$ a prime such that $p^2\equiv-1$ mod (5),
a Suzuki group $Sz(2^p)$ with $p$ an odd prime or $PSL(3,3)$ \cite{Th}.
Every finite simple group contains a minimal simple group \cite{BW}.
(This is not entirely obvious from the definitions!)

There appears to be no corresponding determination of minimal perfect groups.
Every such group $H$ is an extension of a minimal simple group $J$ by
a solvable normal subgroup $K$.
Since the extension does not split,
the prime divisors of the order $|K|$ divide $|J|$ \cite[9.1.2]{Ro}.

We are interested in minimal perfect groups $H$ which have non-trivial
homomorphisms to $GL(n,\mathbb{Z})$, for some $n\leq14$.
We may in fact work with coefficients $\mathbb{Q}$,
as every finite subgroup of $GL(n,\mathbb{Q})$ is conjugate into
$GL(n,\mathbb{Z})$.
Moreover,  the rational group ring $\mathbb{Q}[H]$ is semisimple,
which simplifies our analysis.
The image of $H$ in $GL(n,\mathbb{Q})$ is again a minimal perfect group,
but may be a proper quotient of $H$.
(When $S$ is virtually nilpotent and $N=\sqrt{S}$,
we may assume that $H=S/N$ embeds in $Aut(\mathbb{Q}N^{ab})$,
by Lemma \ref{vnil}.)

There is a further simplification.
The groups of interest to us which are not subgroups of $GL(10,\mathbb{Q})$
are subgroups of the symplectic group $Sp(12,\mathbb{Q})$,
by Lemma \ref{Sp12} below.

Quite a lot can  be done by hand.
If $p$ is a prime then the maximal power of $p$ dividing the order
of a finite subgroup $G<GL(n,\mathbb{Q})$ is
$e_n(p)=\Sigma_{j\geq0}\lfloor{\frac{n}{p^j(p-1)}}\rfloor$.
In particular,  $e_n(p)=0$ if $p>n+1$,
while if $n<12$ then $e_n(7)\leq1$ and $e_n(5)\leq2$,
and if $12\leq{n}\leq14$ then $e_n(11)=e_n(13)=1$,
$e_n(7)=2$ and $e_n(5)=3$.
Moreover, if $G$ is cyclic and of prime power order $p^k\geq3$ then
$n\geq{p^{k-1}(p-1)}$,
while if $G$ is cyclic of composite order not congruent to 2 mod (4)
then $n$ is bounded below by the corresponding sum.
In the remaining case the minimal value of $n$ is 1 less than this sum \cite{KP}.
Thus the only prime powers of interest to us are $2^k$ with $k\leq4$,
3, 5, 7, 9, 11 and 13.

The projective linear groups $PSL(2,q)$ contain cyclic subgroups of order
$q-1$, and $q+1$,
and so we may eliminate such groups with $q=2^k$ or $3^k$ and $k\geq5$.
The Suzuki group $Sz(2^p)$ has a cyclic subgroup of order $2^p+2^s+1$,
where $2s=p+1$, and so we may eliminate such groups with $p>3$.
This leaves only $A_5=PSL(2,4)=PSL(2,5)$, $PSL(2,7)=SL(3,2)$,  $PSL(2,8)$,
$PSL(2,13)$,  $PSL(2, 3^3)$,  $Sz(8)$ and $PSL(3,3)$.
The last four groups each have elements of order 13.

However we still need to consider extensions of such groups by
solvable normal subgroups.
At this point we shall simplify our task by using the findings of
Lutowski and Szczep\'anski,
who show that the  minimal perfect groups with irreducible embeddings in
$GL(n,\mathbb{Q})$ (for $n\leq10$) are:
$A_5$, $PSL(2,7)$,  $PSL(2,8)$,
the universal central extensions $SL(2,5)$ and $SL(2,7)$,
and $L_3(2)N2^3$, the non-split extension of $PSL(2,7)$ by $\mathbb{F}_2^3$ \cite{LuS}.

If $H$ is simple then all non-trivial representations are faithful.
We shall label the non-trivial $\mathbb{Q}$-irreducible $\mathbb{Q}$-rational
characters of the groups of most interest to us by their degree.
We shall also use the same symbols to denote the associated $\mathbb{Q}[H]$-modules.

$A_5$ has three: $\rho_4,\rho_5,\rho_6$.

$PSL(2,7)$ has four: $\tau_{6a},\tau_{6b},\tau_7,\tau_8$.

$SL(2,8)$ also has four: $\psi_7,\psi_8,\psi_{21},\psi_{27}$.

The other three groups are not simple,
but have faithful representations in dimensions 7 or 8.

$SL(2,5)$ has two: $\pi_{8a}$ and $\pi_{8b}$.

$SL(2,7)$ has one: $\xi_8$.

$L_3(2)N2^3$ has two: $\lambda_{7a}$ and $\lambda_{7b}$.

We shall also consider faithful representations which are reducible,
but have no trivial summands.
In particular, $SL(2,5)$ has two such representations in each of dimensions 12 and 13,
given by $\pi_{8i}\oplus\widehat\rho_j$, where $i=a$ or $b$ and $j=4$ or 5,
and $\widehat\rho_j$ is the representation which factors through $\rho_j$.
Similarly, $L_3(2)N2^3$ has four such representations in dimension $13$,
but $SL(2,7)$ has none in dimensions $<14$.

In the sections below on nilpotent groups we shall need to consider also
{\it symplectic\/} representations.
A representation $\rho$ into  $GL(2k,\mathbb{Q})$ is symplectic if it is conjugate
into the subgroup $Sp(2k,\mathbb{Q})$.

\begin{lemma}
\label{symplectic}
Let $\rho:F\to{GL(2k,\mathbb{Q})}$ be an irreducible representation of
a finite group $F$.
Then $\rho$ is symplectic if and only if the trivial representation $\mathbf{1}$ is a summand of the exterior square $\wedge_2\rho$.
\end{lemma}

\begin{proof}
If $\rho$ is symplectic then the associated skew-symmetric pairing defines
a non-zero $F$-linear homomorphism from $\wedge_2\mathbb{Q}^n$ to $\mathbb{Q}$.
Since $\mathbb{Q}[F]$ is semisimple,
$\mathbf{1}$ is a summand of $\wedge_2\rho$.

Conversely,  a projection from $\wedge_2\rho$ onto $\mathbf{1}$ gives a skew-symmetric pairing on $\wedge_2\mathbb{Q}^n$.
Since $\rho$ is irreducible the radical of this pairing is 0,
and so the pairing is non-singular.
Hence $\rho$ is symplectic.
\end{proof}

The finite subgroups of $Sp(2k,\mathbb{Q})$ are determined in \cite{Ki},
for $2k\leq12$.
In particular, all such groups with order divisible by 13 are solvable,
and the representation of $L_3(2)N2^3$ in $GL(8,\mathbb{Q})$ is not symplectic.

We shall need to know how the exterior squares of faithful representations
of degree $\leq10$ decompose as a sum of irreducible representations.
This is an easy exercise in comparing characters.
See \cite[Chapter 2]{Se}.

\medskip
\noindent$A_5$:\quad $\wedge_2\rho_4\cong\rho_6$,
\quad$\wedge_2\rho_5\cong\rho_4\oplus\rho_6$,
\quad$\wedge_2\rho_6\cong\rho_4\oplus\rho_5\oplus\rho_6$,

$\wedge_22\rho_4\cong\mathbf{1}\oplus\rho_4\oplus\rho_5\oplus3\rho_6$,\quad
$\wedge_22\rho_5\cong\mathbf{1}\oplus4\rho_4\oplus2\rho_5\oplus3\rho_6$,

$\wedge_22\rho_6\cong2\mathbf{1}\oplus4\rho_4\oplus6\rho_5\oplus3\rho_6$,\quad
$\wedge_2(\rho_4\oplus\rho_5)\cong5\rho_4\oplus\rho_5\oplus3\rho_6$,

$\wedge_2(\rho_4\oplus\rho_6)\cong3(\rho_4\oplus\rho_5\oplus\rho_6)$\quad
{and}\quad$\wedge_2(\rho_5\oplus\rho_6)\cong4\rho_4\oplus3\rho_5\oplus4\rho_6$.

\noindent$PSL(2,7)$:\quad
$\wedge_2\tau_{6a}\cong2\mathbf{1}\oplus\tau_{6a}\oplus\tau_7$,\quad
$\wedge_2\tau_{6b}\cong2\mathbf{1}\oplus\tau_{6b}\oplus\tau_7$,

$\wedge_2\tau_7\cong6\tau_{6a}\oplus\tau_7\oplus\tau_8$\quad
{and}\quad$\wedge_2\tau_8\cong6\tau_{6a}\oplus2\tau_7\oplus\tau_8$

\noindent$SL(2,5)$:\quad
$\wedge_2\pi_{8a}\cong3\mathbf{1}\oplus\widehat\rho_4\oplus3\widehat\rho_5\oplus\widehat\rho_6$,\quad{and}\quad
$\wedge_2\pi_{8b}\cong6\mathbf{1}\oplus4\widehat\rho_4\oplus\widehat\rho_6$.

\noindent$SL(2,7)$:\quad
$\wedge_2\xi_8\cong\mathbf{1}\oplus\widehat\tau_{6b}\oplus
2\widehat\tau_7\oplus\widehat\tau_8$.

\noindent$SL(2,8)$:\quad
$\wedge_2\psi_7\cong\psi_{21}$\quad{and}\quad
$\wedge_2\psi_8\cong\psi_7\oplus\psi_{21}$.

\noindent$L_3(2)N2^3$:\quad
$\wedge_2\lambda_{7a}\cong\lambda_{7a}\oplus\lambda_{14}$\quad{and}
\quad $\wedge_2\lambda_{7b}\cong\lambda_{7b}\oplus\lambda_{14}$.

\smallskip
The following information on torsion was provided by R. Lutowski.
Let $G$ be a crystallographic group with holonomy $H$.
If $H\cong{A_5}$ and $h(G)=4$ or 5 then $G$ has 5-torsion.
If $H\cong{PSL(2,7)}$ and $h(G)=6$ or 7,
or if  $H\cong{SL(2,7)}$ and $h(G)=8$,
or if $H\cong{L_3(2)N2^3}$ and $h(G)=7$ or 8 then $G$ has 7-torsion.
However there are examples with $H\cong{PSL(2,7)}$ or $SL(2,8)$ and $h(G)=8$
with no 7-torsion.

\section{ $S$ virtually solvable but not virtually nilpotent}

Let $S$ be a finitely generated, torsion-free group which is
virtually solvable and of finite Hirsch length.
Then $S$ has solvable normal subgroups $V\leq{U}$ such that
$S/U$ is crystallographic,  $U/V$ is finite and $V/I(V)$ has positive rank,
by Lemma \ref{overline}.
Since $S/V$ is virtually abelian,
it has a free abelian normal subgroup of finite index.
If $U$ is nilpotent then we may assume $V=U$, and then
$\sqrt{S/U}$ is the unique such subgroup of minimal index,
but otherwise $S/V$ may have a non-trivial torsion normal subgroup,
and there may several such subgroups.
Note also that in this section $H=S/\widetilde{S}$ is a minimal simple group.

Let $A$ be one such free abelian normal subgroup of minimal index in $S/V$.
Then $A\cong\mathbb{Z}^m$, where $m=h(S/V)$,
and $V^{ab}$ is a finitely generated $\mathbb{Z}[A]$-torsion module.
The $\mathbb{Q}[A]$-module $M=\mathbb{Q}\otimes{V}^{ab}$ is
a finite dimensional $\mathbb{Q}$-vector space,
and $\mathbb{Q}[A]/Ann(M)$ is an Artinian ring.
Let $Supp(M)$ be the set of maximal ideals $\mathfrak{m}$ in  $\mathbb{Q}[A]$
which contain $Ann(M)$, and let $M_\mathfrak{m}$ be the
submodule annihilated by a power of $\mathfrak{m}$.
Then $M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m}$
is a non-trivial vector space over the field
$\mathbb{K}_\mathfrak{m}=\mathbb{Q}[A]/{\mathfrak{m}}$,
for each $\mathfrak{m}\in{Supp(M)}$,
and  $M=\oplus{M_\mathfrak{m}}$,
where the summation is over $Supp(M)$.

The following lemma is a variation on Hall's criterion.

\begin{lemma}
\label{vHall}
Let $S$ be a finitely generated,  torsion-free group with subgroups $V\leq{U}$
as above.
Then $S$ is virtually nilpotent if and only if $U$ is virtually nilpotent
and $Supp(M)=\{\mathfrak{e}\}$.
\end{lemma}

\begin{proof}
Let $W$ be the preimage of $A$ in $S$, so $V<W$ and $W/V\cong{A}$.
If $W$ is virtually nilpotent then $W/I(V)$ is also nilpotent,
and so has an ascending central series.
The successive subquotients of the intersections of the terms of this central series
with $V/I(V)$ give rise to a filtration of $M$
with subquotients annihilated by $\mathfrak{e}$.

Conversely, if $Supp(M)=\{\mathfrak{e}\}$ then $M$ has such a filtration.
This determines an ascending series $\{C_i\mid0\leq{i}\leq{n}\}$ for
$V$ such that $C_0=I(V)$,
each subquotient $C_{i+1}/C_i$ is central in $W/C_i$ and $C_n=V$.
Hence $W/I(V)$ is nilpotent.
Since $W$ is torsion-free,
it follows that if $U$ is virtually nilpotent then
$W$ is virtually nilpotent, by a mild variation of Hall's criterion \cite[5.2.10]{Ro}.
\end{proof}

We shall assume henceforth that $S$ is perfect and is minimal TFNS.

\begin{lemma}
\label{sl2}
Let $G$ be a finitely generated, perfect, virtually solvable subgroup of $GL(d,\mathbb{Q})$.
If $d\leq5$ then the order of $G/\widetilde{G}$ divides $2^{10}3^25$, and if $d\leq3$ then $S=1$.
\end{lemma}

\begin{proof}
Since $G$ is perfect, it is a subgroup of $SL(d,\mathbb{Q})$.
Let $m$ be the lowest common denominator for
the entries of a generating set for $G$,
and let $R=\mathbb{Z}[\frac1m]$.
Then $G\leq{SL(d,R)}$.
Let $p>m$ be prime,
and let $G_k$ be the kernel of the projection of $G$ into $SL(d,R/p^kR)$.
Then $G/G_1$ is a perfect subgroup of $SL(d,p)$,
while $G_k/G_{k+1}$ is an elementary $p$-group for all $k\geq1$,
and $\cap_{k\geq1}G_k=1$.
In particular, either $G_1=1$, in which case $G$ is finite,
or $G_1$ maps onto $\mathbb{Z}/p\mathbb{Z}$,
and so the maximal finite simple quotient of $G$ has order dividing $|SL(d,p)|$.

Using the infinitude of primes in arithmetic progressions,
we find that if $d=5$ then the highest common factor of such orders
(taken over all $p>m$) is $2^{10}3^25$.
If $d=3$ the highest common factor is 48,
and so $G$ has no simple quotient.
Hence in this case $G=1$.
\end{proof}

The argument extends to show that
finitely generated perfect subgroups of $GL(n,\mathbb{K})$
have maximal simple quotients of order bounded as a function of $n$
and $\mathbb{K}$, for any algebraic number field $\mathbb{K}$.

If $G<GL(5,\mathbb{Q})$ and $G/\widetilde{G}$ is a minimal simple group then
$G/\widetilde{G}\cong{A_5}$ (which is a subgroup of $GL(4,\mathbb{Z})$).
On the other hand, $PSL(2,7)<GL(\mathbb{Q},6)$.
While $SL(2,8)<GL(7,\mathbb{Q})$,
the arguments of Lemma \ref{sl2} alone are not enough to decide
whether an extension of $SL(2,8)$ by a solvable normal subgroup may
embed in $GL(6,\mathbb{Q})$.

We shall sketch an argument.
If such a group $G$ with $G/\widetilde{G}\cong{SL(2,8)}$ has non-trivial image
in $SL(6,p)$ for some $p>7$ then it follows from the Aschbacher-Dynkin Theorem
on subgroups of $GL(n,q)$ that the image is a central extension of $SL(2,8)$.
Since $SL(2,8)$ is superperfect, central extensions split,
and so $SL(2,8)<GL(6,p)$.
If $p\equiv1~mod~e$, where $e=126$ is the exponent of $SL(2,8)$
then  representations of $SL(2,8)$ over $\mathbb{F}_p$
lift to representations over $\mathbb{Q}(\zeta_e)$ of the same degree.
(This uses Theorems 9.2.7 and 9.3.6 of \cite{We16}.)
Since the minimal non-trivial complex representation of $SL(2,8)$
has degree 7 this gives a contradiction.

\begin{lemma}
\label{mfixed}
If $\mathfrak{m}$ is fixed by $S$ and $d=[\mathbb{K}_\mathfrak{m}:\mathbb{Q}]\leq3$
then $\mathfrak{m}=\mathfrak{e}$.
\end{lemma}

\begin{proof}
If $\mathfrak{m}$ is fixed by $S$ then $S$ acts on $\mathbb{K}_\mathfrak{m}$.
The action is $\mathbb{Q}$-linear, and so factors through $GL(d,\mathbb{Q})$.
If $d\leq3$ this group has no non-trivial finitely generated perfect subgroups,
by Lemma \ref{sl2},
and so the action is trivial.
Hence $\mathfrak{m}=\mathfrak{e}$.
\end{proof}

Lemma \ref{sl2} gives a sharper result when $S/\widetilde{S}\cong{PSL(2,7)}$.

\begin{lemma}
\label{mnotfixed}
If $\mathfrak{m}$ is not fixed by $S$ then $h(S)\geq9$.
If $S/\widetilde{S}\cong{A_5}$ then $\dim_\mathbb{Q}M_\mathfrak{m}\leq2$
and $[\mathbb{K}_\mathfrak{m}:\mathbb{Q}]\leq2$.
If $S/\widetilde{S}\cong{PSL(2,7)}$ then $M_\mathfrak{m}\cong\mathbb{Q}$
and $\mathbb{K}_\mathfrak{m}=\mathbb{Q}$.
If $S/\widetilde{S}\cong{SL(2,8)}$ then $S$ acts trivially on $Supp(M)$.
\end{lemma}

\begin{proof}
If $S/\widetilde{S}\cong{A_5}$ then each non-trivial orbit $S\mathfrak{m}$
of the action of $S$ on $Supp(M)$ must have at least 5 members,
since $S_4$ has no nontrivial perfect subgroup.
Hence $h(S)\geq9$.
Since $h(S/U)\geq4$ we must have $|S\mathfrak{m}|\dim_\mathbb{Q}M_\mathfrak{m}\leq10$
and so $\dim_\mathbb{Q}M_\mathfrak{m}\leq2$.
The largest proper subgroup of $PSL(2,7)$ has index 7,
and so in this case each orbit has at least 7 members.
If $S/\widetilde{S}\cong{PSL(2,7)}$ then $h(S/U)\geq6$,
so $|S\mathfrak{m}|\dim_\mathbb{Q}M_\mathfrak{m}\leq8$
and $\dim_\mathbb{Q}M_\mathfrak{m}=1$.
Similarly,
the largest proper subgroup of $SL(2,8)$ has index 9,
but $h(S/U)\geq8$ and so there can be no such ideal.
\end{proof}

Together these lemmas give the following theorem.

\begin{theorem}
\label{vsolvthm}
Let $S$ be minimal TFNS but not virtually nilpotent, and let $H=S/\widetilde{S}$.
Then $h(S)\geq9$.
If $H\cong{PSL(2,7)}$ then $h(S)\geq13$, and if $H\cong{SL(2,8)}$ then $h(S)=15$.
\end{theorem}

\begin{proof}
Since $S$ is not virtually nilpotent there is a $\mathfrak{m}\not=\mathfrak{e}$ in
$Supp(M)$.
If $H\cong{A_5}$ then $h(S/U)\geq4$ and either $|S\mathfrak{m}|\geq5$ or
$[\mathbb{K}_\mathfrak{m}:\mathbb{Q}]\geq4$, and so $h(S)\geq8$.
If $h(S)=8$ then $h(S/U)=h(U)=4$, and so $S/U$ has an element of order 5,
and $V$ is abelian.
Since $U/V$ is finite,
there is an element $s\in{S}$ whose image in $S/V$ has order 5.
Since $S$ is torsion-free  and $V$ has rank 4 it follows that $s$ must centralize $V$.
But then the normal closure of $s$ in $S$ centralizes $V$ also, and $V$ is virtually nilpotent.
Thus we must have $h(S)\geq9$ if $H\cong{A_5}$.

Similarly,
if $H\cong{PSL(2,7)}$ then $h(S/U)\geq6$, and if $h(S/U)=6$ or 7 then $h(U)\geq7$.
In this case we find that $h(S)\geq13$.

If $H\cong{SL(2,8)}$ then $h(S/U)\geq8$ and either $|S\mathfrak{m}|\geq9$ or
$[\mathbb{K}_\mathfrak{m}:\mathbb{Q}]\geq7$,
by the argument following Lemma \ref{sl2}.
The torsion argument gives no extra leverage here.
Nevertheless, $h(S)=15$ is the only possibility.
\end{proof}

There is an alternative argument for polycyclic groups, with a slightly sharper result.

\begin{lemma}
Let $S$ be virtually polycyclic.
If $\mathbb{K}_\mathfrak{m}=\mathbb{Q}$ for all $\mathfrak{m}\in{Supp(M)}$ then
$S$ is virtually nilpotent.
\end{lemma}

\begin{proof}
In this case we may assume that $V=\sqrt{S}$.
Since $S$ is virtually polycyclic the eigenvalues of the action of $A$ on
$\mathbb{Q}\otimes{V^{ab}}$ are algebraic integers.
If $\mathbb{K}_\mathfrak{m}=\mathbb{Q}$ then the eigenvalues for the action on
$M_\mathfrak{m}$ lie in $\mathbb{Q}$, and so must be $\pm1$.
Hence the subgroup of $S$ generated by $\sqrt{S}$ and all squares of elements of $A$
is nilpotent.
\end{proof}

\begin{theorem}
\label{vpcnonvnil}
Let $S$ be a virtually polycyclic group which is minimal TFNS,
but not virtually nilpotent.
Then $h(S)\geq9$.
If $U$ is a normal subgroup which contains $\sqrt {S}$ and such that $S/U$
is crystallographic then $h(S/U)\leq6$ and so the holonomy of $S/U$
is $A_5$ or $PSL(2,7)$.
\end{theorem}

\begin{proof}
We may assume that $U$ contains $\sqrt{S}$ as a subgroup of finite index.
Then $h(\sqrt{S})>h(S/\sqrt{S})$,
by Lemma \ref {Wilson}, and so $h(S)>2h(S/U)$.
Since $S/U$ is perfect, $h(S/U)\geq4$ and so $h(S)\geq9$.
If the holonomy of $S/U$ is not $A_5$ or $PSL(2,7)$ then $h(S/U)\geq8$.
But then $h(S)\geq17$, and so $S$ is not minimal.
\end{proof}

If $S$ is as in Theorem \ref{vpcnonvnil}, $h(S)\leq14$ and $S/U$ has holonomy $PSL(2,7)$
then $h(S/\sqrt{S})=6$, $h(\sqrt{S}^{ab})=7$ or $8$ and $I(\sqrt{S})\leq\zeta{S}$.

\section{nilpotent groups}

We shall say that  a finitely generated nilpotent group $N$
 is {\it of type $[m,n]$} if
$N/\tilde\gamma_2N\cong\mathbb{Z}^m$ and
$\tilde\gamma_2N/\tilde\gamma_3N\cong\mathbb{Z}^n$.
(Note that $n\leq\binom{m}2$, since $N/\tilde\gamma_3N$
is a quotient of $F(m)$,  the free group of rank $m$.)

Let $\mathbb{Q}N_{i/i+1}=\mathbb{Q}\otimes\tilde\gamma_iN/\tilde\gamma_{i+1}N$,
for $i\geq1$.
Automorphisms of $N$ must preserve the rational {\it commutator pairing}
\[
[-,-]_\mathbb{Q}:\mathbb{Q}N^{ab}\wedge\mathbb{Q}N^{ab}\to
\mathbb{Q}N_{2/3}.
\]
This pairing has two related aspects.
It is a skew-symmetric pairing on $\mathbb{Q}N^{ab}$,
and also is an epimorphism of $\mathbb{Q}$-vector spaces.
In the latter context we shall use the term {\it commutator epimorphism\/}.
Let
\[
R(N)=
\{x\in\mathbb{Q}N^{ab}\mid[x,y]_\mathbb{Q}=0,~\forall~y\in\mathbb{Q}N^{ab} \}
\]
be the radical of the commutator pairing,
and let $\overline{\mathbb{Q}N^{ab}}=\mathbb{Q}N^{ab}/R(N)$.
Then $R(N)\cong\mathbb{Q}\otimes\zeta{N}/\tilde\gamma_3N$.

\begin{lemma}
\label{rad dim}
Let $N$ be a finitely generated nilpotent group such that
$\tilde\gamma_3N<\tilde\gamma_2N$.
Suppose that a nontrivial finite perfect group $H$ acts effectively on $N$
and fixes no nontrivial subspace of $\mathbb{Q}N^{ab}$.
Then
\begin{enumerate}
\item$\dim_\mathbb{Q}\overline{\mathbb{Q}N^{ab}}\ge4$ and either $R(N)=0$
or $\dim_\mathbb{Q}R(N)\geq4$;
\item{}if there is a $\mathbb{Q}[H]$-linear epimorphism $\lambda:
\mathbb{Q}\otimes\tilde\gamma_2N/\tilde\gamma_3N\to\mathbb{Q}$
then $\dim_\mathbb{Q}\overline{\mathbb{Q}N^{ab}}\geq6$;
\item$\mathbb{Q}N_{2/3}$ is a direct summand of
$\mathbb{Q}N^{ab}\wedge\mathbb{Q}N^{ab}$.
\end{enumerate}
\end{lemma}

\begin{proof}
The radical $R(N)$ is an $H$-invariant subspace of $\mathbb{Q}N^{ab}$.
Since $\mathbb{Q}[H]$ is a semisimple ring,
$R(N)$ has an $H$-invariant complement in $\mathbb{Q}N^{ab}$,
which projects isomorphically onto $\overline{\mathbb{Q}N^{ab}}$.
The complement is non-zero, since $N$ is not virtually abelian.

Since finite subgroups of $GL(3,\mathbb{Q})$ are solvable,
any $H$-invariant subspace of $\mathbb{Q}N^{ab}$ of dimension $\leq3$
is fixed pointwise.
Hence $\dim_\mathbb{Q}\overline{\mathbb{Q}N^{ab}}\ge4$
and either $R(N)=0$ or $\dim_\mathbb{Q}R(N)\geq4$.

Suppose that $\lambda:\mathbb{Q}\otimes\tilde\gamma_2N/\tilde\gamma_3N\to\mathbb{Q}$ is a $\mathbb{Q}[H]$-linear epimorphism.
The composite $\lambda\circ[-,-]_\mathbb{Q}$ is then a non-zero
skew-symmetric pairing.
Let $R(\lambda)$ be the radical of this pairing.
The induced pairing on $V=\mathbb{Q}N^{ab}/R(\lambda)$ is nonsingular,
and so has even dimension.
Since $\mathbb{Q}[H]$ is semisimple and $H$ fixes no
non-trivial subspace of $\mathbb{Q}N^{ab}$,
the action of $H$ on $V$ is nontrivial.
Hence $\dim_\mathbb{Q}V\geq6$,
and either $R(\lambda)=0$ or $\dim_\mathbb{Q}R(\lambda)\geq4$.
In particular,  $\dim_\mathbb{Q}\mathbb{Q}N^{ab}\geq6$.

The final assertion is clear, since $\mathbb{Q}[H]$ is semisimple and
$[-,-]_\mathbb{Q}$ is an epimorphism.
\end{proof}

\begin{lemma}
Let $N$ be a finitely generated nilpotent group such that
$\tilde\gamma_3N<\tilde\gamma_2N$.
Suppose that a nontrivial finite perfect group $H$ acts effectively on $N$
and fixes no nontrivial subspace of $\mathbb{Q}N^{ab}$.
If $h(N/\tilde\gamma_3N)\leq9$ then $[m,n]=
[5,4],[6,1],[6,2], [6,3]$ or $[8,1]$.
\end{lemma}

\begin{proof}
The kernel of the commutator epimorphism has dimension
$\binom{m}2-n$ and is $H$-invariant.
Hence if $N$ has type $[4,n]$ with $n=3,4$ or 5 then this kernel
has dimension $\leq3$,  and so $H$ acts trivially on it.
Let $\omega$ be a nonzero 2-form in this kernel.
Since 2-forms determine skew-symmetric pairings on the dual vector space,
we  may choose a basis $\{e_1,e_2,e_3,e_4\}$ for $\mathbb{Q}N^{ab}$,
so that $\omega$ is one of $e_1\wedge{e_2}$ or
$e_1\wedge{e_2}+e_3\wedge{e_4}$.
If an automorphism of $N$ fixes $e_1\wedge{e_2}$ then it fixes
the subspace of $\mathbb{Q}\otimes{N^{ab}}$ generated by
$e_1$ and $e_2$.
Hence we may assume that $\omega=e_1\wedge{e_2}+e_3\wedge{e_4}$,
and that $H$ does not fix any 2-dimensional subspace of
$\mathbb{Q}N^{ab}$.
In this case the skew-symmetric pairing determined by $\omega$
is non-degenerate and $H$ acts symplectically.

If $h(N/\tilde\gamma_3N)\leq7$ then
$\dim_\mathbb{Q}\overline{\mathbb{Q}N^{ab}}\geq4$,
and so $\dim_\mathbb{Q}
\mathbb{Q}\otimes\tilde\gamma_2N/\tilde\gamma_3N\leq3$.
Hence $H$ acts trivially on
$\mathbb{Q}\otimes\tilde\gamma_2N/\tilde\gamma_3N$,
and so $N$ is of type $[6,1]$, by (2).
Similar arguments show that if $N$ is of type $[m,n]$ with $n\leq3$
then $m\geq5+n$, and if $m>6$ then $m\geq8$.

Thus we are left with only types [5,4], [6,1], [6,2],  [6,3] and [8,1].
\end{proof}

In the next section we shall see that [6,3] is not possible.

\section{$S$ virtually nilpotent}

In this section we shall assume that $S$ is virtually nilpotent,
but not virtually abelian, and that $S$ is perfect,
$h(S)\leq14$ and $H=S/\sqrt{S}$ is a minimal perfect group.
(We do not assume here that $H$ is simple.)
Our goal is to limit the possibilities for $H$ and for
the type of $N=\sqrt{S}$.
The fact that $S$ is torsion-free is used only through Lemma \ref{vnil},
to show that $H$ embeds in $Aut(\mathbb{Q}N^{ab})$.

We shall play off the $\mathbb{Q}[H]$-module
structures of $\mathbb{Q}N^{ab}$ and
$\mathbb{Q}N_{2/3}=\mathbb{Q}\otimes\tilde\gamma_2N/\tilde\gamma_3N$
against each other.
The key conditions are
\begin{enumerate}
\item$\dim_\mathbb{Q}N^{ab}+\dim_\mathbb{Q}\mathbb{Q}N_{2/3}\leq{h(S)}\leq14$
and  $\dim_\mathbb{Q}N_{2/3}\geq1$;
\item$\mathbb{Q}N_{2/3}$ is a $\mathbb{Q}[H]$-summand of $\wedge_2\mathbb{Q}N^{ab}$;
\item{}$\mathbb{Q}$ is not a $\mathbb{Q}[H]$-summand of $\mathbb{Q}N^{ab}$,
by Lemmas \ref{HS?} and \ref{vnil};
\item{}if $\mathbb{Q}$ is a $\mathbb{Q}[H]$-summand of $\mathbb{Q}N_{2/3}$ then
$\mathbb{Q}N^{ab}$ has a symplectic summand, by Lemma \ref{symplectic}.
\end{enumerate}

Let $[m,n]$ be the type of $N$.
Then $m+n\leq{h(S)}\leq14$ and $n>0$, by (1).
If $ \mathbb{Q}$ is a $\mathbb{Q}[H]$-summand of $\mathbb{Q}N_{2/3}$
then $\mathbb{Q}N^{ab}\cong{V_0}\oplus{V_1}$,
where $V_1$ supports a nonsingular skew-symmetric pairing,
with even rank $r$.
Since finite subgroups of $Sp(4,\mathbb{Q})$ are solvable, $r\geq6$,
and since $H$ acts effectively on $V_0$,
either $V_0=0$ or $m-r=\dim_\mathbb{Q}V_0\geq4$.
In the latter case $m\geq10$.
There is always such a summand $\mathbb{Q}$ if $n\leq3$.

On the other hand,
the fact that $A_5$ acts effectively on $F(4)/\gamma_3(F(4))$ shows that
$\mathbb{Q}N^{ab}$ need not have a symplectic summand.

\begin{lemma}
\label{Sp12}
If $H=S/\sqrt{S}$ is not a subgroup of $GL(10,\mathbb{Q})$ then
it is a subgroup of $Sp(12,\mathbb{Q})$.
\end{lemma}

\begin{proof}
We must have $11\leq{m}\leq13$, and so $n\leq3$.
Hence  $\mathbb{Q}N_{2/3}$ is a trivial $H$-module.
Let $\lambda:\mathbb{Q}N_{2/3}\to\mathbb{Q}$ be an epimorphism.
Then $H$ preserves the non-zero skew-symmetric pairing
$\omega=\lambda\circ[-,-]_\mathbb{Q}$,
and $\mathbb{Q}N^{ab}\cong{R(\omega)\oplus{V}}$,
where the induced pairing on $V$ is non-singular,
and so has even dimension.
If $R(\omega)=0$ then $m=12$ and $H$ is a subgroup of $Sp(12,\mathbb{Q})$,
since $\mathbb{Q}N^{ab}$ is a faithful $H$-module.
Otherwise $r=\dim_\mathbb{Q}R(\omega)\geq4$ and $d=\dim_\mathbb{Q}V=6$ or 8,
since $\mathbb{Q}N^{ab}$ has no trivial summands.
Let $\pi_R$ and $\pi_V$ be the projections of $H$ into $GL(r,\mathbb{Q})$ and
$Sp(d,\mathbb{Q})$, respectively.
Then $K_R=\mathrm{Ker}(\pi_R)$ and $K_V=\mathrm{Ker}(\pi_V)=1$
are each subgroups of $\widetilde{H}$, the solvable radical of $H$,
and $K_R\cap{K_V}=1$.
Since $d\leq8$ and $\pi_V(H)$ is perfect,
$\pi_V(H)$ is either $A_4$, $PSL(2,7)$ or $SL(2,5)$ \cite{Ki}.
In the first two cases $K_V=\widetilde{H}$,
and so $K_R\leq{K_V}$.
Hence $\pi_R$ is injective.
Otherwise,
$\pi_V$ maps $K_R$ injectively to $\mathbb{Z}/2\mathbb{Z}$,
the solvable radical of $SL(2,5)$.
Hence $H\cong{SL(2,5)}$ and $\pi_V$ is injective.
Since $r<10$ and $d<12$,
this proves the lemma.
\end{proof}

The only groups with representations satisfying conditions (1)--(4) above are
$H\cong{A_5}$,  $PSL(2,7)$,  $SL(2,5)$, $SL(2,7)$ or $L_3(2)N2^3$.
In particular, this observation together with Theorem \ref{vsolvthm}
shows that there is no virtually solvable,  minimal TFNS group $S$ with
$S/\widetilde{S}\cong{SL(2,8)}$ and $h(S)\leq14$.

Consideration of the decompositions of $\mathbb{Q}N^{ab}$ and
$\mathbb{Q}N_{2/3}$ as $\mathbb{Q}[H]$-modules shows that the
remaining possibilities are:

$m=4$ and $n=6$. Only $H=A_5$.

$m=5$ and $n=4$ or 6. Only $A_5$.

$m=6$ and $n=1$ or 2. Only $PSL(2,7)$.

$m=6$ and $n=4$ or 5. Only  $A_5$.

$m=6$ and $n=6$.  $A_5$ or $PSL(2,7)$.

$m=6$ and $n=7$ or 8.  Only $PSL(2,7)$.

$m=7$ and $n=6$ or 7.  Only $PSL(2,7)$.

$m=8$ and $n=1$.  $A_5$, $SL(2,5)$ or $SL(2,7)$.

$m=8$ and $n=2$ or 3.  Only $SL(2,5)$.

$m=8$ and $n=4$ or 5.  $A_5$ or $SL(2,5)$.

$m=8$ and $n=6$. $A_5, PSL(2,7),SL(2,5)$ or $L_3(2)N2^3$.

$m=9$ and $n=4$ or 5.  Only $A_5$.

$m=10$ and $n=1$. Only $A_5$ and $\mathbb{Q}N^{ab}\cong\rho_5\oplus\rho_5$.

$m=10$ and $n=4$. Only $A_5$.

$m=12$ and $n=1$ or 2.  $A_5, PSL(2,7)$ or $SL(2,5)$.

$m=13$ and $n=1$.  $A_5$, $PSL(2,7)$ or $SL(2,5)$.

\smallskip
If $h(S)\leq9$ then $[m,n]=[5,4]$, [6,1] or [6,2],
and $\tilde\gamma_2N\cong\mathbb{Z}^n$.

If $n=1$ then $\tilde\gamma_2N$ is central in $S$, and $h(S)=m+1$.
If $h(\zeta{N})\leq3$ then $H$ acts trivially on $\zeta{N}$,
and $\zeta{N}$ is then central in $S$.
Hence there is at most one extension of $H$ by $N$
corresponding to each action on $N$,
since $H^2(H;\zeta{N})=Ext(H^{ab},\zeta{N})=0$.
On the other hand,  there are infinitely many pairwise non-isomorphic
but mutually commensurable candidates for $N$.

Parallel arguments apply further down the $\mathbb{Q}$-lower central series,
since there are natural epimorphisms from
$\mathbb{Q}N^{ab}\otimes\mathbb{Q}N_{i/i+1}$ to $\mathbb{Q}N_{i+1/i+2}$,
for all $i\geq1$ \cite[5.2.5]{Ro}.

\begin{theorem}
If $S$ is  minimal TFNS,  virtually nilpotent and $h(S)\leq14$
then $\tilde\gamma_4\sqrt{S}=1$, and so $S$ is virtually metabelian.
\end{theorem}

\begin{proof}
Let $N=\sqrt{S}$.
If $n=1$ then $\tilde\gamma_2N\cong\mathbb{Z}$, and so $\tilde\gamma_3N=1$.
Similarly, if $p=\dim_\mathbb{Q}\mathbb{Q}N_{3/4}=1$ then $\tilde\gamma_4N=1$.
Thus we may assume that $n\geq2$ and $p\geq2$, and so $m+n\leq11$ and $m\leq9$.
In each case $\mathbb{Q}N_{3/4}$ must be a summand of
$\mathbb{Q}N^{ab}\otimes\mathbb{Q}N_{2/3}$.
Checking the possibilities leaves only $A_5$, with $[m,n]=[5,4]$.
We then have $\mathbb{Q}N_{3/4}=\rho_4$,
since $\rho_4\otimes\rho_5\cong\rho_4\oplus2\rho_5\oplus\rho_6$.
But then $\mathbb{Q}N_{4/5}$ is a summand of $\rho_5\otimes\rho_4$
of dimension $\leq1$, and so $\mathbb{Q}N_{4/5}=1$.
Therefore $\tilde\gamma_4N=1$.
Hence $N$ is metabelian,
since  $G''\leq\gamma_4G$ for any group $G$.
\end{proof}

We have not yet made much use of the condition that $S$ be torsion-free
(except in Theorem \ref{vsolvthm}).

\begin{lemma}
Let $S$ be a finitely generated, perfect, torsion-free group with a normal subgroup $N$
such that $H=S/N$ is finite and $I(N)\cong\mathbb{Z}^r$.
If $S/I(N)$ has an element of prime order $p>r$
whose image in $H$ normally generates $H$ then $I(N)\leq\zeta{S}$.
\end{lemma}

\begin{proof}
Let $s\in{S}$ be an element whose image in $S/I(N)$ has order $p$.
Then $s$ normalizes $I(N)$.
Since the subgroup generated by $I(N)$ and $s$ is torsion free and $p>r$,
this is only possible if $s$ centralizes $I(N)$.
Since $I(N)$ is normal in $S$,
the normal closure of $s$ centralizes $I(N)$,
and so $I(N)\leq\zeta{S}$.
\end{proof}

With this lemma we may exclude the cases with $H=A_5$,
$m=5,8$ or 9 and $n=4$
(for which we may apply the lemma with $p=5$),
and those with $H\cong{PSL(2,7)}$ and $[m,n]=[6,6]$ or $[7,6]$
(for which we may apply the lemma with $p=7$).

More generally,  since $S$ is minimal TFNS,
$\overline{S}=S/\tilde\gamma_2N$ has torsion,  and since $S$ is perfect,
$H^2(\overline{S};\mathbb{Z})$ is free abelian.
If restriction from $\overline{S}$ to a finite cyclic subgroup induces
0 on $H^2$ then every central extension of $\overline{S}$ has torsion.
Can we apply this observation effectively?

We cannot exclude type $[4,6]$ by arguments involving just the lower central series.
Let $\{w,x,y,z\}$ be a basis for the free group $F(4)$,
and define endomorphisms $\sigma$ and $\tau$ by
$\sigma(w)=w^{-1}$, $\sigma(x)=wxy$, $\sigma(y)=y^{-1}$,
$\sigma (z)=yz$ and $\tau(w)=x^{-1}$, $\tau(x)=xyz$, $\tau(y)=z^{-1}$ and
$\tau(z)=(wxy)^{-1}$.
These are automorphisms,
since $\sigma^2=\tau^3=(\sigma\tau)^5=1$,
and so $A_5$ embeds in $Aut(F(4))$.
This is an integral version of $\rho_4$.
The induced action of $A_5$ on $N=F(4)/\gamma_3F(4)$
(the free class 2 nilpotent group on 4 generators)
satisfies the hypotheses of Lemma \ref{rad dim}.

The character $\rho_6$ is induced from the non-trivial
1-dimensional character $sg:D_{10}\to\mathbb{Z}^\times$.
Is $\gamma_2N\cong\mathbb{Z}[A_5]\otimes_{D_{10}}\mathbb{Z}^{sg}$?
If so, then $H^2(A_5;\gamma_2N)$
$\cong\mathbb{Z}/5\mathbb{Z}$
and extensions of $A_5$ by $N$
must have 2- and 3-torsion.

The arguments of this section can be used to further sharpen some of the results of
\S5.
Suppose that $S$ is virtually solvable but not virtually nilpotent,
and let $V\leq{U}<S$ be as in the first paragraph of \S5.
Let $W$ be a normal subgroup of finite index in $S$
which contains $V$ and is such that $A=W/V$ is free abelian
and of minimal index in $S/V$.
Let $M=\mathbb{Q}\otimes{V}^{ab}$,
and let  $m=h(S/U)$ and $n=\dim_\mathbb{Q}M$.
Let $\overline{S}=S/I(V)$ and $\overline{W}=W/I(V)$,
and let $\breve{S}=\overline{S}/\cap_{i\geq1}\tilde\gamma_i\overline{W}$.
Then $N=\overline{W}/\cap_{i\geq1}\tilde\gamma_i\overline{W}$ is nilpotent
and $\breve{S}/N\cong{S}/W$.
If $M$ has a summand annihilated by $\mathfrak{e}$ then
$\mathbb{Q}N_{2/3}$ has $\mathbf{1}$ as a summand.
But if $S/U$ has holonomy $A_5$ and $h(S/U)\leq6$ then there are no such summands.

\section{some questions}

1) Is there a minimal TFNS group which is an extension of  a crystallographic group
by $\mathbb{Z}$?

2) If $H$ is the holonomy group of an $n$-dimensional infranilmanifold
is it also the holonomy group of a flat $n$-manifold?
This is so if $n\leq4$, by inspection of the known groups.
In general, $H$ is the holonomy of a crystallographic group in dimension $\leq{n}$,
by Lemma \ref{vnil}.

(The converse fails for $n=4$, since $A_4$ is the holonomy group of a flat 4-manifold,
but not of any other 4-dimensional infranilmanifold.)

3) If $S$ is a minimal TFNS group must it be virtually polycyclic?

4) If $S$ is a minimal TFNS group must it be perfect?

\newpage

redpajama_set_name: RedPajamaArXiv


text:
Jarābulus (arabiska: جرابلس) är en distriktshuvudort i Syrien.   Den ligger i provinsen Aleppo, i den norra delen av landet,  kilometer norr om huvudstaden Damaskus. Jarābulus ligger  meter över havet och antalet invånare är .

Terrängen runt Jarābulus är platt. Den högsta punkten i närheten är  meter över havet,  kilometer väster om Jarābulus. Det finns inga andra samhällen i närheten.

Trakten runt Jarābulus består till största delen av jordbruksmark. Runt Jarābulus är det ganska tätbefolkat, med  invånare per kvadratkilometer.  Stäppklimat råder i trakten. Årsmedeltemperaturen i trakten är  °C. Den varmaste månaden är juli, då medeltemperaturen är  °C, och den kallaste är januari, med  °C. Genomsnittlig årsnederbörd är  millimeter. Den regnigaste månaden är december, med i genomsnitt  mm nederbörd, och den torraste är juli, med  mm nederbörd.

Se även
Karkemisch (en forntida stad på samma plats)
Operation Eufrats sköld

Kommentarer

Källor

Orter i Aleppo (provins)

redpajama_set_name: RedPajamaWikipedia


text:
New Cars 2020 Coming To Ireland
|In Features
What cars are new in Ireland in 2020? We've listed all the new models from the brands reaching Ireland in 2020 and those already here for the 202 registration!
The Alfa Romeo range in Ireland at the moment is based around the Giulietta hatchback, Giulia saloon and Stelvio SUV. We haven't seen any new models yet in 2020. But a new compact SUV called the Tonale is expected in 2021 and will slot in below the Stelvio in the range. It will be available as a plug-in hybrid and the brand's first electrified car. In Ireland, the brand is offering some limited edition Giulietta models, along with refreshed styling and interiors for the Giulia and Stelvio.
A revised Audi A4 and Audi Q7 arrived in dealers at the beginning of 2020. Both models feature changes in design, connectivity and a new engine line-up incorporating mild hybrid technology. Since January the range has also been bolstered by the arrival of the Audi A1 citycarver, Q3 Sportback and Audi Q5 TFSI e, the latter which has 367 hp and a pure electric range of over 40 km. This summer will see the arrival of the new generation of the Audi A3 Sportback, with the new A3 Saloon scheduled to arrive in the autumn. The new A3 will be available with petrol and diesel engines, and a digital cockpit will come as standard. Audi will also expand the electric range with the new Audi e-tron Sportback. This SUV coupé derivative of the e-tron will have a range up to 446 kilometres and will arrive later in summer 2020.
There are new plug-in hybrids in the BMW camp including the X5 xDrive45e, BMW X3 xDrive30e and X1 xDrive25e. In March the all-new BMW 2 Series Gran Coupé arrived in Ireland. BMW recently took the covers off the facelifted 5 Series range and the new 4 Series Coupé so we should see these models in Ireland before the end of the year. Mild hybrid technology is also being introduced across 51 models from this year including the 5 Series, 3 Series, 6 Series, 7 Series and X range, to improve dynamic performance and reduce fuel consumption in BMW combustion engines.
The new BMW 2 Series Gran Coupé, one of the cars heading to Ireland in 2020
The Citroen C5 Aircross is one of the brand's most popular models now in dealers and will be joined by a new C5 Aircross PHEV before the end of 2020. The plug-in hybrid combines a PureTech 180 hp petrol engine and an 80 kW electric motor to deliver an electric range of 50 km. The new C4 will debut in 2021 as a crossover, with petrol, diesel and electric powertrains.
The new CUPRA Formentor is expected in the second half of 2020 and will be the first model launched exclusively to the CUPRA brand. The high-performance plug-in hybrid will have 245 hp and an electric range of up to 50km. A CUPRA Leon is also expected in 2020.
The Dacia Duster gets a more efficient TCe 100 petrol engine. Air conditioning is now standard for the 2020 model year.
The new Fiat 500 Hybrid is available now. Expect to see a new Fiat 500e La Prima battery electric vehicle in the first few months of 2021.
There are also 2020 model year updates for the 500X, 500L and Tipo.
The new Ford Puma compact crossover arrived in Ireland in January and introduces mild hybrid technology. The new Kuga SUV has just arrived in dealers. It's available as a plug-in hybrid or with a 1.5-litre diesel engine. The new Mustang Mach-E is expected in Ireland before the end of the year. There are also mild hybrid variants of the Fiesta and Focus expected here.
The new Ford Mustang Mach-E!
The new Honda Jazz and Honda e are expected later in 2020. The new Jazz will be hybrid only. Buyers will also find the Civic and CR-V ranges in dealers, including the CR-V Hybrid. A subtly updated Civic range is expected from August.
New in dealers is the Hyundai Kona Hybrid, Ioniq Electric – which comes with an increased range of 312 km – and the new generation of the Hyundai i10 city car. Other interesting new models include the Tuscon now with mild hybrid technology and the Tucson N Line.
There's a new Jaguar F-TYPE in dealers with updated styling, equipment, engine and technical upgrades. Jaguar also has special 'Chequered Flag' editions of models like the E-PACE, F-PACE, XE and XF with more equipment and sportier styling.
The new Jaguar F-TYPE
The first Renegade plug-in hybrid badged '4xe' is expected by the end of the year. Elsewhere, Renegade, Compass and Wrangler have all received updates for the 2020 model year.
New models include a facelifted Kia Niro PHEV and e-Niro. There's also the new XCeed range, which has been bolstered by the recent arrival of the XCeed plug-in hybrid. The Kia Sportage is now available with a 1.6 diesel with mild hybrid technology. A new seven seater Kia Sorento is expected before the end of the year, as is a facelifted Rio range.
The new Land Rover Defender has just arrived in dealers. Plug-in hybrids of the Discovery Sport and Range Rover Evoque are also on the way, along with a new three-cylinder petrol Ingenium engine.
The new Land Rover Defender
And updated RX large SUV arrived in January. A new UX 300e all electric compact crossover is expected in 2021.
The latest arrival in the Mazda camp is the all-new CX-30 SUV. There's also a revised Mazda2 in dealers. The brand's first electric crossover, the Mazda MX-30, is expected in early 2021.
The latest arrivals in dealerships are the new GLB and GLA compact SUVs. The GLB is available with five or seven seats. Further SUV launches will be the GLS and the GLE Coupé. A facelifted E-Class range will arrive in July, followed by a new S-Class before the end of 2020. We will also see the second EQ model in the form of a fully electric powered EQA SUV at some point this year. New hybrids are joining also, with the first to arrive being the new E300de and the E300e saloon and estate models.
The new Mercedes-Benz GLB will arrive in Ireland in 2020!
A new MINI Electric based on the brand's iconic three door hatchback has arrived in Ireland with a range of 235 to 270 kilometres.
A revised Space Star arrived in January of this year. The 1.2-litre petrol model is available in both manual and automatic transmission. In the second half of 2020, a Mitsubishi Eclipse Cross PHEV will arrive.
The new Nissan Juke is the latest arrival in dealer showrooms this summer. It debuts with a new 1.0-litre petrol engine. For 2020, the Micra will be available with a new three cylinder 100 hp engine – the same as the one currently in the CVT (automatic). The SV Premium grade has also been added to the Micra range and includes privacy glass, front fog lights, automatic air con and iKey, rear parking sensors and extra personalisation options. The Qashqai SE replaces the Qashqai SV with a price reduction of about €1,200. ProPilot has been added as an option to the Qashqai and X-Trail range for €1,000. Pricing has come down across the X-Trail range for example there's €3,000 off the volume selling 7 seater SVE (top spec model).
The new Nissan Juke has just arrived in Ireland
A new generation of the Opel Corsa arrived in dealers at the beginning of the year. The range has now been joined by an all-new electric Corsa-e and the Grandland X plug-in hybrids (front wheel drive and all wheel drive). Arriving later in the year will be a facelifted Insignia range and the new electric Opel Zafira-e Life people carrier. Early 2021 will see the arrival of the new Mokka crossover, which will be available with combustion engines and as an electric vehicle. There will also be a new Crossland X.
The new Peugeot 208 arrived in Ireland earlier in 2020 with petrol, diesel and electric variants available. The new Peugeot 2008 is also on sale in dealers now. Plug-in hybrid versions of the 3008 and 508 are also new for Peugeot.
There's a new Renault Clio and Captur range in dealers with petrol and diesel engines available. Hybrid variants are expected soon also. A new electric Renault Zoe with 52 kWh battery and a range of up to 395 km has just arrived in dealers. There are also facelifted ranges of the Koleos, and the Mégane due soon.
The second generation of the popular Renault Captur heading this way in 2020
The new SEAT Leon has just arrived in dealers this summer with petrol and diesel engines, mild hybrid technology and a plug-in hybird to follow later in the year. The SEAT Tarraco FR is also expected in 2020. The FR trim will complete the Tarraco range with sportier styling. The SEAT el-Born is the brand's first fully electric car and is expected by the end of 2020 with a range of up to 420 km.
The new Kamiq crossover and facelifted Superb range are in dealers now. The Superb is available with petrol and diesel engines and as a new plug-in hybrid 'Superb iV'. A new Octavia is on the way in the summer also. Skoda's first electric SUV, the Enyaq, is expected in 2021.
The 2020 ŠKODA Octavia!
There's a new Korando SUV range and this summer sees the arrival of an updated Tivoli crossover range with a new 1.2-litre turbo petrol engine.
Subaru has introduced new e-BOXER mild hybrid technology to the Subaru XV Crossover and the all-new Forester. Both are on sale now.
There's a new facelifted Ignis range now in dealers, with hybrid now available. New 48-volt self-charging hybrid technology is also now available in the Vitara and S-Cross. These models are equipped with a new K14D Boosterjet turbocharged engine, an even higher standard specification than before (including Automated Emergency Braking now standard on all Vitara and S-Cross Hybrid variants) and directly replace the outgoing 1.4-litre Boosterjet models with manual transmission.
There's an updated Toyota C-HR range in dealers now with two hybrid powertrains available. A new 2.0-litre petrol electric hybrid joins the current 1.8-litre model. A new GR Sport trim level has been added to the Toyota Corolla range, along with 2.0-litre hybrid powertrain with more power. A new generation of the Toyota Yaris supermini is one the way before the end of 2020 and will be available as a 1.5-litre hybrid. There's also a RAV4 plug-in hybrid on the way. The new Yaris Cross small SUV and Higlander seven seat SUV are expected in Ireland in early 2021.
The 2020 Toyota Yaris
The eighth generation of the iconic Volkswagen Golf has just arrived in dealers, and the new all electric ID.3 hatchback will arrive in September. Buyers will also find an updated Passat and Passat GTE in showrooms. A new T-ROC R performance SUV also arrived in 2020. The Toureg R hybrid is also expected before the end of the year, as are Golf plug-in hybrid variants and new Golf GTI, GTD and GTE. An updated Arteon range will arrive from September with a 1/5 litre petrol and 2.0 litre diesel. For the first time the range will be joined by a new Arteon Shooting Brake. The new Arteon R and Arteon eHybrid are expected in November. The popular Tiguan range is also being updated and will arrive in Ireland with petrol and diesel engines in November. The new performance Tiguan R will follow in January 2021 and a new plug-in hybrid will join the range in the summer.
A new Volvo XC40 plug-in hybrid has joined the range. There are also recently added V60 and S60, T8 plug in hybrids and the V60 Cross Country. This summer, buyers can expect to see a refreshed Volvo S90 saloon and Volvo V90/Cross Country estate models in showrooms. The S90 T8 plug-in hybrid is available from the end of June, while the V90 is available as a D4 FWD Automatic and a T6 plug-in hybrid. The V90 Cross Country is available in D5 AWD Automatic.
A new Bowers & Wilkins audio system and Advanced Air Cleaner technology are now available on all 90 and 60 Series models, along with double USBC charging points in the rear, replacing the 12 volt outlets. The wireless charging functionality for smartphones, first introduced on the XC40 SUV, is now also available on most variants in the 90 and 60 Series. The exclusive tailored wool blend seats first introduced on the XC90 last year are now also available on all 90 and 60 models, while there is also a leather free interior option for the top trim levels.
An all electric XC40 Recharge is expected in 2021.
The new Volkswagen Golf expected in Ireland in 2020


The Boston Red Sox have offered third baseman Pablo Sandoval a five-year $95 million contract. The free agent is also being targeted by the San Diego Padres and his former team the San Francisco Giants. The Red Sox and Padres seem to be after everybody right now, with San Diego targeting some bats. Boston also offered Jon Lester earlier in the week and they could get him back to go with Yoenis Cespedes, who they got in return for Lester at the trade deadline. You add those two players, with Sandoval, I think you are looking at the favorite to win the American League East and most likely pennant.
Pablo Sandoval hit .279 last season with 16 homers and 73 RBIs, but I think that his performance in the playoffs is were you will get your monies worth. Sandoval has a career .344 batting average in 167 postseason plate appearances. He has been on three World Series winning teams. Sandoval just knows how to win and that's why I think he's the best player available this offseason.
Sandoval is just 28 years old and should be entering his prime, which is scary if you ask me. The Red Sox are getting a steal, if they get him for an average of $19 million a year.


The complainant, who worked for a transportation company, objected to injured fellow workers, temporarily employed in the company's office, handling confidential payroll information. He argued that only authorized personnel should have access to such sensitive information.
In its submission to this Office, the company explained that it provides a return to work program for injured bargaining unit workers, as required under the Canadian Human Rights Act and the Workplace Safety and Insurance Act, 1997. The program assists injured workers in their rehabilitation and return to work by allowing them to work temporarily in a company office before returning to their regular duties. These workers perform various duties on an "as needed" basis, including stuffing envelopes with employee pay stubs prior to delivery. The workers do not, however, have access to computer records or other forms of personal information.
At the time of the complaint, the workers were given no training regarding the sensitivity and confidentiality of the information they might be handling; nor were they required to sign any type of confidentiality agreement.
Jurisdiction: As of January 1, 2001, the Personal Information Protection and Electronic Documents Act (the Act) applies to any federal work, undertaking or business. The Assistant Privacy Commissioner had jurisdiction in this case because a transportation company such as this one is a federal work, undertaking or business as defined in the Act.
Application: Principle 4.7 states that personal information shall be protected by security safeguards appropriate to the sensitivity of the information. Principle 4.7.2 states that the nature of the safeguards will vary depending on the sensitivity of the information that has been collected, the amount, distribution, and format of the information, and the method of storage. More sensitive information should be safeguarded by a higher level of protection. Principle 4.7.3(b) indicates that methods of protection should include organizational measures, for example, limiting access on a "need-to-know" basis. Principle 4.7.4 states that organizations shall make their employees aware of the importance of maintaining the confidentiality of personal information.
Although the company's return to work program obviously benefited injured workers, the fact remained that, as part of their duties, these workers were handling highly sensitive information, namely, the payroll information of their fellow employees, without signing any agreements or receiving any training regarding confidentiality.
This practice posed a serious risk that the workers could have accessed sensitive personal information to which they should not have been privy.
In such a situation, a reasonable person would expect appropriate security safeguards to be in place to protect this information from all but a few authorized personnel.
At the time of the complaint, the company had no confidentiality agreement or other appropriate safeguards in place.
The Assistant Commissioner therefore found that the company was not in compliance with Principles 4.7, 4.7.2, 4.7.3(b) and 4.7.4 of the Act.
The Assistant Commissioner was pleased that, as a result of this complaint, the company established a procedure to safeguard the personal information of its employees by developing and instituting a confidentiality agreement. Employees required to sign this agreement will thus have an increased awareness of the importance of maintaining confidentiality. The complainant indicated that he was satisfied with the company's action.
In order to be fully compliant with the expectations in Principle 4.7.3(b), regarding organizational measures to limit access on a "need to know" basis, the company should refrain from giving its injured workers temporarily employed in the office access to such highly sensitive personal information as pay stubs. A more stringent control process, for example, making the handling of payroll information part of the permanent duties of a few authorized office personnel, should be established to appropriately safeguard such information.
With respect to Principle 4.7.4, which speaks to the requirement of making employees aware of the importance of confidentiality, employees who are required to sign the confidentiality agreement should receive training to ensure that they fully understand its meaning and implications.