text:
\section{Introduction and notation}

The map which takes every element of a group
to its inverse is an automorphism
if and only if the group is abelian. Following \cite{HL}, we say that a
group is N.I. (no inversions occur) if no automorphism of the group sends
any non-identity element to its inverse{\footnote{In \cite{MS}, the term
I.F.P. (inverse-point-free) is used instead.}}. Thus an N.I. group is
non-abelian. It is easy to see that a finite group $G$ is N.I. if and
only if Aut$(G)$ has odd order{\footnote{We do not know if there exists an
infinite group which has no automorphism of order $2$ but is not N.I.}}, and
that the order of such a group must be
odd. N.I. groups have attracted some attention, because of the
surprising difficulty in finding explicit
examples of them. Though there are a number of results indicating that
such groups are ubiquitous (for example, see \cite{HL},
\cite{HR} and \cite{M}), no brute force search among $\lq$small' groups
will yield any examples. More precisely,
MacHale and Sheehy \cite{MS} proved that there is no finite N.I.
group of order less than $3^6$. There
exist groups of order $3^6$ with automorphism group of order $3^7$ (see
the Small Groups Library \cite{BEB}). In this
paper we prove the following complement to MacHale and Sheehy's result :

\begin{thm}
If $G$ is a non-trivial finite group such that
$|{\hbox{Aut}}(G)|$ is odd, then \\ $|{\hbox{Aut}}(G)| \geq 3^7$.
\end{thm}

The proof will be presented in the next two sections. In Section 4 we
present some remaining open problems concerning small N.I. groups.
\\
\\
All groups considered in the remainder of this paper are finite. The
following notational conventions will be adhered to throughout.
\\
\\
We denote by $\Omega(n)$ the number of prime factors of the
positive integer $n$, counting repetitions, and by $\omega(n)$ the number of distinct prime factors of $n$. If $G$ is a group then $\pi(G)$ denotes
the set of distinct prime
factors of $|G|$. If $p \in \pi(G)$ then $G_p$ denotes a Sylow $p$-subgroup of $G$.
\par
The direct product of $n$ copies of a group $G$ with itself will be denoted $G^n$.
\par
The center of a group $G$ is denoted by
$Z(G)$. We will often just write $Z$ when it is clear
to which group we are referring.
\par
Inn$(G)$ and Cent$(G)$ denote, respectively, the groups of inner and central
automorphisms
of $G$. For $g \in G$, the inner automorphism $x \mapsto g^{-1}xg$ will be denoted
$I_g$. The commutator $x^{-1}g^{-1}xg$ will be denoted $[x,g]$.
\par
Let $A$ be a subgroup of $G$. Let $\phi
\in {\hbox{Aut}}(A)$ and $\phi^{*} \in {\hbox{Aut}}(G)$. We say
that $\phi^{*}$ is an {\em extension} of $\phi$ (or, equivalently, that $\phi$ {\em
extends} to $\phi^{*}$), if
$(A)\phi^{*} = A$ and $\phi^{*} |_{A} = \phi$.
\par
If $A$ is a normal subgroup of $G$, we say that $G$ {\em splits} over $A$ if there
is a subgroup $B \subseteq G$ such that $A \cap B = \{1\}$ and $G = AB$. Then $G$ is
said to be a {\em split extension} of $A$ by $B$ and we write $G = A \rtimes B$. The
subgroup $B$ is said to be a {\em complement} of $A$. It acts on $A$ by conjugation
and the induced subgroup of Aut$(A)$ is denoted by
$A^{B}$. This last piece of notation
will be used even in the case of a non-split extension.
\par If $q$ is a prime power, then $\mathbb{F}_q$ denotes the finite field with $q$
elements and
GL$(n,q)$ the group of invertible linear transformations on the
$n$-dimensional vector space over
$\mathbb{F}_q$. If $q$ is a prime, this group can be identified with Aut$(G)$ where
$G \cong C_{p}^{n}$, the
elementary abelian $p$-group of rank $n$.
\par

\setcounter{equation}{0}

\section{Toolbox}

All of the results in this section, except for Theorem 2.13, are taken from the existing literature. The first seven lemmas are, moreover, standard material.
For these the reader seeking
details beyond what we provide
is referred to \cite{Sco} in the absence of any alternative citations.

\begin{lem}
Cent$(G) \; \cap$ Inn$(G) = Z[{\hbox{Inn}}(G)] \cong Z[G/Z(G)]$.
\end{lem}

\begin{lem}
Let $G$ be a group without any abelian direct factor. Then
\be
|{\hbox{Cent}}(G)| = |{\hbox{Hom}}(G/G^{\prime},Z(G))|.
\ee
\end{lem}

\begin{proof}
The natural one-to-one correspondence is given by $\phi \rightarrow
(G^{\prime}g \mapsto g^{-1} g\phi)$. For a rigorous proof of the lemma, see
\cite{Sa}.
\end{proof}

\begin{lem}
Let $G$ be a group with center $Z$ and suppose $G/Z$ is a $p$-group for some prime
$p$.
\\
(i) If $G/Z$ is abelian, say
\be
G/Z \cong \prod_{i=1}^{r} C_{p^{\alpha_i}},
\ee
with $\alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_r$, then $r \geq 2$ and
$\alpha_1 = \alpha_2$.
\\
(ii) If $G/Z$ is not elementary abelian then it possesses a non-trivial central
characteristic subgroup $K$
of index at least $p^2$.
\end{lem}

\begin{proof}
(i) Clearly $r \geq 2$.
Let $Za$ be a generator of a highest order cyclic factor. Then for any $Zb$ in
any complementary factor we
have $b^{p^{\alpha_2}} \in Z$. Since $G/Z$ is abelian we have $1 =
[a,b^{p^{\alpha_2}}]= [a,b]^{p^{\alpha_2}} = [a^{p^{\alpha_2}},b]$.
Thus $a^{p^{\alpha_2}} \in Z$, so $\alpha_1 = \alpha_2$.
\\
(ii) If $G/Z$ is not abelian, take $K$ to be its center. Otherwise take $K$ to be
the subgroup generated by
the elements of order $p$. In both cases $K$ is non-trivial of index at least $p^2$,
by part (i).
\end{proof}

\begin{lem}
Let $G$ be a group, $N$ a normal subgroup of $G$. Let $\phi$ be an automorphism of
$G$ whose order is
relatively prime to that of $G$ and which acts trivially on both $N$ and $G/N$. Then
$\phi$ is the trivial
automorphism.
\end{lem}

\begin{proof}
Let $g \in G$. Since $\phi$ acts trivially on $G/N$, we have $g\phi = ng$ for some
$n \in N$. Since $\phi$ acts
trivially on $N$ it follows that for every $i$, $(g)\phi^{i} = n^{i}g$. In
particular, taking $i = |\phi|$ we find
that $n^{|\phi|}=1$. But $|\phi|$ is prime to $|G|$ so $n=1$, thus $g\phi = g$ as
desired.
\end{proof}

\begin{lem}
Let $G$ be a group, $p \in \pi(G)$ and $P$ a Sylow $p$-subgroup of $G$.
\\
(i) The number of conjugates of $P$ in $G$ is congruent to one modulo $p$.
\\
(ii) If $P \lhd G$ then $G$ splits over $P$.
\end{lem}

\begin{lem}
For any positive integer $n$ and prime power $q$, we have
\be
|{\hbox{GL}}(n,q)| = \prod_{i=0}^{n-1} (q^{n} - q^{i}).
\ee
\end{lem}

\begin{lem}
Let $G$ be a group of odd order. Then $G$ possesses an elementary abelian
characteristic $p$-subgroup, for
some $p \in \pi(G)$.
\end{lem}

\begin{proof}
A group is characteristically simple if and only if it is the internal direct
product of isomorphic simple
subgroups. Since every group of odd order is soluble, it follows that a
characteristically simple group
of odd order must be an elementary abelian $p$-group. This and the fact that
\be
B \; {\hbox{char}} \; A \; {\hbox{char}} \; G \; \Rightarrow \; B \; {\hbox{char}}
\; G,
\ee
allows us to prove the lemma by induction on the group order.
\end{proof}

\begin{lem}
If $\Omega(|{\hbox{Aut}}(G)|) \leq 4$ then $|{\hbox{Aut}}(G)|$ is even.
\end{lem}

\begin{proof}
This follows from the work of various authors. See the introduction to \cite{SZ}, where
all the necessary references are given.
\end{proof}

\begin{lem}
Let $p$ be an odd prime and $G$ a $p$-group such that Aut$(G)$ is also a $p$-group.
Then
$|{\hbox{Aut}}(G)| \geq 3^7$ if $p = 3$ and $|{\hbox{Aut}}(G)| \geq p^6$ if $p > 3$.
\end{lem}

\begin{proof}
See \cite{C1} and \cite{MS}.
\end{proof}

\begin{cor}
If $G$ is a group for which Aut$(G)$ is an odd order group of order strictly
less than $3^7$ then Aut$(G)$ is not a $p$-group.
\end{cor}

\begin{proof}
If Aut$(G)$ is a $p$-group then so is the central factor group $G/Z$, being
isomorphic to a normal subgroup of
the former. Thus $G$ is nilpotent, hence the internal direct product of its Sylow
subgroups, say $G = P_1
\times \cdots \times P_k$. But then also
\be
{\hbox{Aut}}(G) = \prod_{i=1}^{k} {\hbox{Aut}}(P_i).
\ee
Since the automorphism group of a $2$-group is either trivial or of even order, we
now obtain the desired
conclusion from Lemma 2.9.
\end{proof}

The next three results will be the main tools in our approach to proving Theorem
1.1. The first two already appear in several papers, so we omit proofs. See, for
example, Theorem 2.1 of \cite{C2}.

\begin{lem}
Suppose $G = AB$ for some pair of subgroups $A$ and $B$, where $A \lhd G$.
Let $\phi \in {\hbox{Aut}}(A)$. Then $\phi$
extends to an automorphism $\phi^{*}$ of $G$ such that $|\phi| = |\phi^{*}|$
if the following two conditions are satisfied :
\par (i) $\phi$ is trivial on $A \cap B$,
\par (ii) in the group Aut$(A)$, $[\phi,A^{B}] = 1$.
\end{lem}

This lemma will be used in the case when $\phi$ has order 2 and the subgroup $A$ is
either abelian or a $p$-group of class 2. In the former case we have

\begin{cor}
Suppose $G = A \rtimes B$ where $A$ is abelian of order greater than 2. Then $G$
possesses an automorphism of order 2. In particular, if
$G$ has a normal, abelian Sylow $p$-subgroup, for some odd prime $p$, then
$|{\hbox{Aut}}(G)|$ is even.
\end{cor}

In the case of non-abelian $A$ we have the following result. To the best of our
knowledge it is a new result, and we thus state it as a theorem in its own
right.

\begin{thm}
Let $G = AB$, where $A$ is a normal $p$-subgroup of class 2 with \\ $A/(Z \cap A)$
elementary abelian, $B/(Z \cap B)$ is an abelian $p^{\prime}$-group
and $A \cap B \subseteq Z$. Assume $B$ acts
non-trivially on $A$. Then $G$ possesses an
automorphism of order 2
.
\end{thm}

While our proof is lengthy, it is
crucial to our analysis in Section 3.

\begin{proof}
Since $B$ acts non-trivially on $A$, it does so even on $A/(Z \cap A)$, by Lemma 2.4.
Let $n$ be the rank of $A/(Z \cap A)$ as an elementary abelian $p$-group. The
group $A^{B}$ thus describes a non-trivial
representation of the abelian $p^{\prime}$-subgroup $B/(Z \cap B)$ in
GL$(n,p)$.
By Maschke's theorem, this representation is completely reducible, say
\be
A^B = \psi_1 \times \cdots \times \psi_r,
\ee
where rank$(\psi_i) = n_i$ say. Moreover,
Schur's lemma implies that $A^B/{\hbox{ker}} \psi_i$ is cyclic for each $i$.
We first prove
the theorem in two special cases, before establishing the general result.
\\
\\
{\sc Case 1} : Suppose that $A^B$ is irreducible.
\\
\\
Let $A^B = \; <I_b>$. Clearly, there is a choice of basis $Z\alpha_1,....,Z\alpha_n$
for $A/(Z \cap
A)$, an element $z \in Z$ and
non-negative integers $k_1,...,k_n$ such that the action of $b$ on $A$ is given by
\begin{eqnarray}
b^{-1} \alpha_i b = \alpha_{i+1}, \;\;\; i = 1,...,n-1, \\
b^{-1} \alpha_n b = z \alpha_{1}^{k_1} \cdots \alpha_{n}^{k_n}.
\end{eqnarray}
In other words, the matrix of $I_b$ with respect to this basis is
\be
M = M_b = \left( \begin{array}{ccccccc} 0 & 1 & 0 & \cdot & \cdot & \cdot & 0 \\ 0 &
0 & 1 & 0 & \cdot & \cdot & 0
\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot &
\cdot & \cdot & \cdot &
\cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & 0 & \cdot &
\cdot & \cdot & 0 & 1 \\
k_1 & k_2 & \cdot & \cdot & \cdot & \cdot & k_n \end{array} \right).
\ee
Let $\mathbb{I}_n$ denote the $n \times n$ identity matrix. One checks easily that
det$(M - \mathbb{I}_n) = (-1)^{n+1}[(k_1 + k_2 + \cdots + k_n) - 1]$, hence, since
$M$ acts non-trivially and irreducibly, we have
\be
k_1 + k_2 + \cdots + k_n \not\equiv 1 \; ({\hbox{mod $p$}}).
\ee
Raising both sides of (2.7) to the $p$-th power for each $i$ yields
\be
\alpha_{1}^{p} = \alpha_{2}^{p} = \cdots = \alpha_{n}^{p}.
\ee
Doing the same for (2.8) and inserting (2.11) yields
\be
\alpha_{n}^{p} = z^{p} \alpha_{n}^{(k_1 + \cdots + k_n)p}.
\ee
By (2.10) there is a substitution $a_n := \sigma_n \alpha_n$, for some $\sigma_n \in
Z$, such that
$a_{n}^{p} = 1$. Working back through equations (2.7) one sees that one can
make a sequence of similar
substitutions $a_i := \sigma_i \alpha_{i}$, $i = n-1,n-2,...,1$, such that
\be
a_{1}^{p} = a_{2}^{p} = \cdots = a_{n}^{p} = 1.
\ee
With respect to the elements $a_1,...,a_n$, there exist $z_1,...,z_n \in Z$ such
that the action of
$b$ is given by
\begin{eqnarray}
b^{-1}a_i b = z_i a_{i+1}, \;\;\; i = 1,...,n-1, \\
b^{-1} a_n b = z_n a_{1}^{k_1} \cdots a_{n}^{k_n}.
\end{eqnarray}
Now let $\zeta_1,...,\zeta_n$ be any elements of $Z$. By (2.13), the mapping
\be
z \mapsto z \; (z \in Z), \;\;\; a_i \mapsto \zeta_i a_{i}^{-1},
\ee
extends to a well-defined automorphism $\phi$ of $A$ of order $2$. Note that
$\phi$ is trivial on $Z \supseteq A \cap B$. Thus, by Lemma 2.11,
this extends to an
automorphism of $G$ if $[\phi,I_b] = 1$. From (2.14)-(2.16) this reduces to a system
of $n$
equations in the $n$ unknowns $\zeta_1,...,\zeta_n$, namely
\begin{eqnarray}
\zeta_i \zeta_{i+1}^{-1} = z_{i}^{2}, \;\;\; i = 1,...,n-1, \\
\zeta_{1}^{-k_1} \cdots \zeta_{n-1}^{-k_{n-1}} \zeta_{n}^{1-k_n} = z_{n}^{2}.
\end{eqnarray}
From (2.10) one deduces that this system has a unique solution. Thus $G$ possesses
an automorphism of
order $2$, as desired.
\\
\\
{\sc Case 2} : Suppose that
$A^B = \psi_1 \times \psi_2$ where $\psi_2$ is one-dimensional
$(n_2 = 1)$ and trivial.
\\
\\
Let $\psi_1 = \; <I_b>$. From the previous case we can deduce that there is a basis
$Za_1,...,Za_{n-1},Za=Za_n$ for $A/(Z \cap A)$ such
that
\be
a_{1}^{p} = a_{2}^{p} = \cdots = a_{n-1}^{p} = 1
\ee
and that the action of $I_b$ is given by
\begin{eqnarray}
b^{-1}a_i b = z_i a_{i+1}, \;\;\; i = 1,...,n-2, \\
b^{-1} a_{n-1} b = z_{n-1} a_{1}^{k_1} \cdots a_{n-1}^{k_{n-1}}, \\
b^{-1}ab = a,
\end{eqnarray}
for some $z_1,...,z_{n-1},z \in Z$ and non-negative integers $k_1,...,k_{n-1}$
satisfying (2.10), with $n$ replaced by $n-1$.
The map $\phi : A \rightarrow
A$ should now satisfy
\be
z \mapsto z \; (z \in Z), \;\;\; a \mapsto a, \;\;\; a_i \mapsto \zeta_i a_{i}^{-1}
\;\; (i = 1,...,n-1).
\ee
We will have a well-defined automorphism of $A$ provided
\be
[a,a_i] = 1, \;\;\; {\hbox{for $i = 1,...,n-1$}}.
\ee
But (2.24) follows from (2.20)-(2.22) and (2.10). For inserting (2.22) into each of
the $n$ equations in (2.20) and (2.21) yields
\begin{eqnarray}
[a,a_i] = [a,a_{i+1}], \;\;\; i = 1,...,n-2, \\
\; [a,a_{n-1}] = \prod_{j=1}^{n-1} [a,a_j]^{k_j}.
\end{eqnarray}
Then (2.25) and (2.26), combined with (2.10), yields (2.24). The map $\phi$ extends
to $G$ for a unique choice of $\zeta_1,...,\zeta_{n-1}$, as in Case 1. This deals
with Case 2.
\\
\\
We now prove Theorem 2.13 in full generality. Suppose $A^B$ decomposes as in
(2.6),
where $\psi_i = \; <I_{b_i}>$ acts irreducibly on the $n_i$-dimensional subspace $W_i$,
and suppose that $\psi_{1},...,\psi_{s}$ are non-trivial.
Then for each $i = 1,...,s$ there is a basis $Za_{i1},...,Za_{in_i}$ for $W_i$, and
bases $Za_1,...,Za_{r-s}$ for $W_{s+1},...,W_r$ respectively such that
\be
a_{ij}^{p} = 1, \;\;\; i = 1,...,s, \; j = 1,...,n_i,
\ee
and that the action of $B$ is fully described by
\begin{eqnarray}
b_{i}^{-1}a_{ij}b_i = z_{ij} a_{i,j+1}, \;\;\; i = 1,...,s, \; j = 1,...,n_i - 1, \\
b_{i}^{-1}a_{in_i}b = z_{in_i} a_{i1}^{k_{i1}} \cdots a_{in_i}^{k_{in_i}}, \;\;\; i =
1,...,s, \\
b_{i}^{-1}a_{lj} b_i = a_{lj}, \;\;\; {\hbox{whenever $i \neq l$}}, \\
b^{-1} a_m b = a_m, \;\;\; \forall \; b \in B, \; m = s+1,...,r,
\end{eqnarray}
for some $z_{ij} \in Z$ and non-negative integers $k_{ij}$ satisfying
\be
\sum_{j=1}^{n_i} k_{ij} \not\equiv 1 \; ({\hbox{mod $p$}}), \;\;\; i = 1,...,s.
\ee
Then the mapping
\be
z \mapsto z \; (z \in Z), \;\;\; a_{ij} \mapsto \zeta_{ij}
a_{ij}^{-1}, \;\;\; a_m \mapsto a_m,
\ee
extends, for a unique choice of the elements $\zeta_{ij} \in Z$, to an
automorphism of $A$ of order $2$, which acts trivially on $Z$ and
commutes with $A^B$. Thus $G$ has an automorphism of order $2$ and the
theorem is proved.
\end{proof}

\setcounter{equation}{0}

\section{Case-by-case analysis}

Throughout this section, $G$ denotes a potential counterexample to Theorem 1.1,
whose non-existence we shall establish. By Lemma 2.8, we have $\Omega(|{\hbox{Aut}}(G)|) \geq 5$ and, together with
Corollary 2.10, one readily checks that this leaves the following 19 possibilities
for $|{\hbox{Aut}}(G)|$ :
\begin{eqnarray}
3^5 \cdot p, \;\;\; p \in \{5,7\}, \\
3^4 \cdot p, \;\;\; p \in \{5,7,11,13,17,19,23\}, \\
5^4 \cdot 3, \\
3^4 \cdot 5^2, \\
3^3 \cdot p^2, \;\;\; p \in \{5,7\}, \\
5^3 \cdot 3^2, \\
3^3 \cdot 5 \cdot p, \;\;\; p \in \{7,11,13\}, \\
3^3 \cdot 7 \cdot 11, \\
3^2 \cdot 5^2 \cdot 7.
\end{eqnarray}
The order of $G/Z$ must divide one
of those on the above list. Let $p \in \pi(G/Z)$ and suppose that $p^2$ does not
divide $|G/Z|$. Then any
Sylow $p$-subgroup of $G$ is abelian. So if $G/Z$, and hence $G$, has a normal Sylow
$p$-subgroup, then Lemma
2.5(ii) and Corollary 2.12 imply that $G$ possesses an automorphism of order 2. Using
Lemmas 2.5(i) and 2.7 this fact, together with Lemma 2.9, is
easily checked to already rule out all but the following 21 possibilities for
$|G/Z|$ :
\begin{eqnarray}
3^i \cdot 5, \;\;\; i \in \{4,5\}, \\
3^i \cdot 13, \;\;\; i \in \{3,4\}, \\
3 \cdot 5^i, \;\;\; i \in \{2,3,4\}, \\
3^i \cdot 5^2, \;\;\; i \in \{2,3,4\}, \\
3^i \cdot 7^2, \;\;\; i \in \{1,2,3\}, \\
3^2 \cdot 5^3, \\
3^i \cdot 5 \cdot 7, \;\;\; i \in \{2,3\}, \\
3^i \cdot 5^2 \cdot 7, \;\;\; i \in \{1,2\}, \\
3^i \cdot 5 \cdot 11, \;\;\; i \in \{2,3\}, \\
3^3 \cdot 5 \cdot 13.
\end{eqnarray}
For these remaining possibilities we may thus assume that $G/Z$ contains no normal
subgroup of prime order. In what follows, there is no loss of generality in
assuming that $\pi(G) = \pi(G/Z)$. For let $\pi$ be the set of primes dividing
$|G|$ but not $|G/Z|$. By Lemma 2.5, we have $G = A \times G_1$, where $A$ is an
abelian Hall $\pi$-subgroup of $G$. Then, by Corollary 2.12, $G$ has an
automorphism of order $2$ unless $A \cong C_1$ or $C_2$ and Aut$(G) =$
Aut$(G_1)$. \par
As our next step, we claim that in all but three of the above 21 cases, namely
\be
3^4 \cdot 5^2, \;\;\; 3^2 \cdot 5^2 \cdot 7, \;\;\; 3^4 \cdot 13,
\ee
we can identify a prime $p \in \pi(G/Z)$ such that the Sylow $p$-subgroup of $G/Z$
is normal of order $p^i$,
for some $i \in \{2,3,4,5\}$. More precisely we have the following table of values :
\\ \\

\begin{tabular}{|c|c|c|} \hline
$i$ & $p$ & $|G/Z|$ \\ \hline \hline
$2$ & $3$ & $3^2 \cdot 5 \cdot 7 \;\;$ $3^2 \cdot 5 \cdot 11$ \\ \hline
$2$ & $5$ & $3 \cdot 5^2 \;\;$ $3^2 \cdot 5^2 \;\;$ $3^3 \cdot 5^2 \;\;$ $3 \cdot
5^2 \cdot 7$ \\ \hline
$2$ & $7$ & $3 \cdot 7^2 \;\;$ $3^2 \cdot 7^2 \;\;$ $3^3 \cdot 7^2 \;\;$ \\ \hline
$3$ & $3$ & $3^3 \cdot 13 \;\;$ $3^3 \cdot 5 \cdot 7 \;\;$ $3^3 \cdot 5 \cdot 11\;\;$ $3^3 \cdot 5 \cdot 13$ \\
\hline
$3$ & $5$ & $3 \cdot 5^3 \;\;$ $3^2 \cdot 5^3$ \\ \hline
$4$ & $3$ & $3^4 \cdot 5$ \\ \hline
$4$ & $5$ & $3 \cdot 5^4$ \\ \hline
$5$ & $3$ & $3^5 \cdot 5$ \\ \hline
\end{tabular}

$\;$ \\
In 11 of the 18 cases listed in the table, the normality of the identified
$p$-subgroup is established by a direct
application of Lemma 2.5(i). In the case of $3^3 \cdot 13$ we also need to use our
assumption that $G_{13}$ is
not normal, being of prime order. The remaining 6 cases are those where $\omega(|G|)
= 3$. Here we appeal both to
the aforementioned assumption and to Lemma 2.7. One example will suffice to
illustrate the idea, say
$3^2 \cdot 5 \cdot 7$. By assumption neither $G_5$ nor $G_7$ is normal. Thus, by
Lemma 2.7, some $3$-subgroup
must be characteristic. If this is $G_3$, we are done, otherwise it is isomorphic to
$C_3$.
The quotient group then has order $3 \cdot 5 \cdot 7$. Applying Lemma 2.7 to the
quotient, we find that
$G/Z$ has a characteristic subgroup of order $9$, $15$ or $21$. In the first
instance we are done. In the
latter two, Lemma 2.5(i) and eq. (2.4) imply that $G_5$ resp. $G_7$ is characteristic
after all, contradicting our
assumptions.
\par Of the three numbers (3.20) omitted from the table, we can still apply Lemma 2.7
as above to conclude in the case of both $3^4 \cdot 5^2$ and $3^2 \cdot 5^2 \cdot 7$
that either the $G_3$ or $G_5$-subgroup must be normal,
though we don't know which one a priori. This leaves $3^4 \cdot 13$, where applying
the same kind of
analysis allows us only to conclude that either $G_3$ or a subgroup of
order $3^3$ must be normal. The case when $|G/Z| = 3^4 \cdot 13$ will be called the
{\em exceptional} case.
\\
\\
In the next part of our analysis, we ignore the exceptional case. This will be dealt
with at the finish.
In each of the remaining 20 cases, we are guaranteed to have a splitting
$G = P \rtimes B$, where
\\
\\
(i) $P$ is a Sylow $p$-subgroup for some prime $p \in \{3,5,7\}$,
\\
(ii) $|P/(Z \cap P)| = p^i$ for some $i \in \{2,3,4,5\}$,
\\
(iii) $B$ is a $p^{\prime}$-group, either abelian or of class 2,
\\
(iv) If $p = 3$ then $\pi(B) \subseteq \{5,7,11,13\}$,
\\
(v) If $p = 5$ then $i \leq 4$ and $\pi(B) \subseteq \{3,7\}$,
\\
(vi) If $p = 7$ then $i=2$ and $\pi(B) = \{3\}$.
\\
\\
We now divide the analysis into 2 steps, according as to whether $P/(Z \cap P)$ is
elementary abelian or not.
\\
\\
{\sc Step 1} : $P/(Z \cap P)$ is elementary abelian.
\\
\\
If $B$ acts trivially on $P/(Z \cap P)$ then $G = P \times B$. It is easy to check,
using the tools of Section 2,
that in all cases $B$ possesses an automorphism of order $2$, hence so does $G$.
If $B$ acts non-trivially then all the conditions of Theorem 2.13 are satisfied, so
we are done here too.
\\
\\
{\sc Step 2} : $P/(Z \cap P)$ is not elementary abelian.
\\
\\
Then $P/(Z \cap P)$ must have order $p^3$ at least. In all but 4 cases, namely
\be
5^3 \cdot 3, \;\;\; 5^3 \cdot 3^2, \;\;\; 5^4 \cdot 3, \;\;\; 3^5 \cdot 5
\ee
we can apply Lemmas 2.4 and 2.6 directly to find that $B$ acts trivially on $P$, and
then argue as in Step 1. We illustrate with
the example of $3^4 \cdot 5$. By Lemma 2.3(ii), there is a proper subgroup $P_2$ of
$P$, strictly containing
$Z \cap P$, and such that $P_{2}/(Z \cap P) \subseteq Z[P/(Z \cap P)]$. Thus $P_2$
is invariant under the action
of $B$, so $B$ acts on both $P_2/(Z \cap P)$ and
$P/P_2$, and both these groups are of order $3^i$ for some $i < 4$. Now Lemma 2.6
implies that
$|{\hbox{GL}}(i,3)|$ is not divisible by $5$ for any $i < 4$. It follows that $B$
acts trivially
on both $P_{2}/(Z \cap P)$ and $P/P_2$, and hence on $P/(Z \cap P)$ by Lemma 2.4.
\\
\\
This leaves us with the four cases in (3.21).
First suppose $|G/Z| \in \{5^3 \cdot 3, \; 5^3 \cdot 3^2\}$. In both cases, we
can choose a characteristic subgroup $P_2$ of $P$ as above such that $|P_2/(Z \cap
P)| = 5$. Then
$B$ centralises $P_2$, since $|{\hbox{GL}}(1,5)|$ is not divisible by $3$. But
$C_{G}(P_2)$
must then be a proper, characteristic subgroup of $G$ containing $P_2$ and $B$, and hence of
index
dividing $5^2$. Thus $G^{\prime} \subseteq C_{G}(P_2)$, which means that $B$ must
also act
trivially on $P/P_2$, thus on all of $P/(Z \cap P)$, by Lemma 2.4.
\par If $|G/Z| = 5^4 \cdot 3$ and $P/(Z \cap P)$ is non-abelian we can reach the
same conclusion,
for either the commutator subgroup or center of $P/(Z \cap P)$ must have order $5$
and yields
the characteristic subgroup playing the role of $P_2$ above. If $P/(Z \cap P)$ is
abelian, it must
be isomorphic to $C_{25} \times C_{25}$. In this case, we can still use the method of
Theorem 2.13. For one easily checks that $3 \parallel |{\hbox{Aut}}(C_{25}
\times
C_{25})|$, and that if $B/(Z \cap B) := \; <Zb>$ acts non-trivially on $P/(Z \cap
P)$, then
there is a generating set $Za_1,Za_2$ for the latter such that
\be
b^{-1}a_1 b = a_2, \;\; b^{-1} a_2 b = z a_{1}^{-1} a_{2}^{-1}, \;\;\;\;\; \exists
\; z \in Z.
\ee
This was the starting point of the method of proving Case 1 of Theorem 2.13, and
the same argument
carries through here.
\par Finally, suppose $|P/(Z \cap P)| = 3^5 \cdot 5$. Let $P_{2}/(Z \cap P)$ be
a non-trivial
characteristic subgroup of $P/(Z \cap P)$ of index at least $3^2$, whose existence
is guaranteed
by Lemma 2.3(ii). As noted previously, $|{\hbox{GL}}(i,3)|$ is not divisible by $5$,
for any
$i < 4$. Hence we can deduce from Lemma 2.3 that $B$ acts trivially on $P$, unless
$|P_{2}/(Z \cap P)| = 5$ and $B$ acts irreducibly on $P/P_2$. But the latter is
impossible, since
$P_2 B$ is contained in the characteristic subgroup $C_{G}(P_2)$ of $G$. This completes
Step 2.
\\
\\
All that now remains in order to establish Theorem 1.1 is to handle the exceptional
case where $|G/Z| = 3^4 \cdot 13$. We can assume that the Sylow $3$-subgroup of $G$
is not
normal, as otherwise
Theorem 2.13 can be applied directly. As noted earlier, some $3$-subgroup $P$ such
that
$|P/(Z \cap P)| = 3^3$ must then be normal. Since $13$ does not divide
$|{\hbox{GL}}(2,3)|$,
then $P/(Z \cap P)$ must be elementary abelian and acted upon irreducibly by any
Sylow $13$-subgroup, or else the latter would also be normal in $G$. Moreover, the
quotient $G/P$ must be non-abelian of order $39$, isomorphic to $N_{G}(Q)/(Z \cap
Q)$, for
any Sylow $13$-subgroup $Q$ of $G$. These facts combined imply that
\par (i) $G/Z$ is a centerless group,
\par (ii) $|G/G^{\prime}|$ is a multiple of $3$,
\par (iii) $G = PB$ for some subgroup $B$ with $P \cap B \subseteq Z$.
\\
By Lemma 2.1, (i) implies that any non-trivial central automorphism of $G$ is outer.
But if $G$ possesses
any outer automorphisms, then $|{\hbox{Aut}}(G)|$ will have to be at
least $3^5 \cdot 13$, which is greater than $3^7$, a contradiction. Thus $G$ cannot
have any
non-trivial central
automorphisms. But then (ii) and Lemma 2.2 force the conclusion that $Z$ contains no
elements of order $3$. This and (iii) in turn imply that $G = P \rtimes B$ and that
$P$ is elementary abelian of order $27$. But now we can apply Corollary 2.12 to
conclude that $G$ has an automorphism of order 2. The theorem is proved.

\setcounter{equation}{0}

\section{Discussion}

A careful reading of the foregoing proof will show that we have established
somewhat more than we have stated. In fact, if $G/Z$ has odd order less than
$3^7$, then either $G$ has an automorphism
of order $2$, or $G/Z$ is a group of order $3^4 \cdot 13$ with a very specific
structure :
namely, it is the split extension of an elementary abelian group of order $3^3$ by
a non-abelian group of order $39$. In the latter case we could only show that either $G$ has
an automorphism of order $2$, or a non-trivial central automorphism which was
necessarily outer. It would be interesting to establish if this case is a
genuine exception or whether a group with this structure cannot be N.I. either.
This might in turn shed light on the following questions :
\\
\\
1. What is
the smallest order of a non-nilpotent N.I. group and, conversely,
the smallest order of a non-nilpotent automorphism group of odd order ?
Theorem 1.1 and the result in \cite{MS} give the answers for nilpotent groups.
Martin's result \cite{M} that almost all $p$-groups have automorphism
group a $p$-group also motivates giving special attention to non-nilpotent
N.I. groups.
\\
\\
2. It should be particularly interesting to study complete groups, as these
have no non-trivial outer or central automorphisms. Several papers in
the literature deal with the construction of complete groups of odd order
(see \cite{D}, \cite{H}, \cite{HR}, \cite{Sch} and \cite{So} for example),
but the smallest possible order
of such a group is unknown. The smallest example in the
literature seems to be a complete group of order $3^{12} \cdot 5$
constructed in \cite{H}.
\\
\\
3. In \cite{SZ} the authors conjectured that if
$\Omega (|{\hbox{Aut}}(G)|) \leq
5$ then $|{\hbox{Aut}}(G)|$ is even (recall Lemma 2.8 above). This problem
remains open. Resolving it would also shed light on the previous two questions.
\\
\\
Finally, we remark that Corollary 2.12 and Theorem 2.13 are of some
interest in their own right. They give sufficient
conditions for an automorphism of order $2$ of a normal subgroup to be
extendible to the whole
group, when
the subgroup is nilpotent of class at most $2$. It would be interesting to
find similar sufficient conditions when the subgroup
has class 3.

\section*{Acknowledgement}

We thank John Curran for helpful comments on an earlier draft and the referee
for his suggestions.
The research of the first author is partly supported by a grant from the
Swedish Research Council (Vetenskapsr\aa det).

redpajama_set_name: RedPajamaArXiv


text:
Q: Google Charts Update Category Filter based on other Category Filter I have these two filters below, both are implemented into google charts. When I change filter1, the data changes in the visualization, but the values in filter2 do not change. I was wondering if that was possible with Google Charts.

*

*filter1 has: 1 and 2 for options

*filter2 has: A,B,C,D,E,F for options
Selecting 1 from filter1 needs to trigger the data-source of filter2 to filter to A,B,C.
Selecting 2 from filter1 needs to trigger the data-source of filter2 to filter to D,E,F.
Filters for google charts:
var filter_1 = new google.visualization.ControlWrapper({
                'controlType': 'CategoryFilter',
                'containerId': 'filter1',
                'options': {
                    'filterColumnIndex': 0,
                    'ui': {
                    'allowTyping': false,
                    'allowMultiple': false,
                    'selectedValuesLayout': 'belowStacked',
                    'label': 'Filter 1:'
                }
                }
            });

var filter_2= new google.visualization.ControlWrapper({
            'controlType': 'CategoryFilter',
            'containerId': 'filter2',
            'options': {
                'filterColumnIndex': 1,
                'ui': {
                'allowTyping': false,
                'allowMultiple': false,
                'selectedValuesLayout': 'belowStacked',
                'label': 'Filter 2:'
            }
            }
        });

A: bind(filter_1, filter_2)

This will make any filter_2 change its options based on filter_1. Change the order to have the reverse effect. Once that is done, to get both filters to your chart...
bind([filter_1, filter_2], chart)
draw(chart)

redpajama_set_name: RedPajamaStackExchange


text:
Biography & Autobiography /
Social Activists /
Disarmed
The Radical Life and Legacy of Michael Mj Sharp
By Marshall V. King
A powerful story of one man's radical commitment to peacemaking.
At 34 years old, Michael "MJ" Sharp was working for the United Nations Group of Experts in the Democratic Republic of the Congo--urging rebels to lay down their weapons--when he was murdered, likely assassinated alongside his colleague Zaida Catalán by those with government ties. This compelling account of MJ's life, death, and legacy from longtime journalist Marshall V. King explores what compelled Sharp to travel the world working for peace and the ongoing impact of his life and death in the ongoing story of Christian peacemaking in a war-torn world.
MJ was a modern Mennonite armed with wit and intellect, but not a gun. The son of a Mennonite pastor, he demonstrated a gift for listening and persuading early in life. His efforts to approach others with acknowledgement rather than judgement gave him the ability to connect on a level very few managed. He also honed a deep commitment to peace, and after college he joined the Mennonite Mission Network and moved to Germany, where he persuaded soldiers to choose peace and free them of their violent systems.
Manufacturer: Herald Pr
Author: Marshall V. King

redpajama_set_name: RedPajamaCommonCrawl


I am quite sceptical when it comes to celebrities and social media. I always assume that something is fake until I find a credible source that tells me otherwise. But when I stumbled upon Swedish football striker Zlatan Ibrahimovic on Twitter, I got a little curious. He hasn't been very active online and to my knowledge he still doesn't have an official webpage, so Zlatan being on Twitter would be somewhat ground breaking.
But the very same photo can be found here, and along with several other photos it comes from the press and photo agency Stella Pictures. If Zlatan would be tweeting photos from his house, why would be be posting photos that belong to a photo agency and not just snap a picture with his mobile? No, I think we can safely say that this story has been debunked. Zlatan is not on Twitter.
Tags: twitter, zlatan ibrahimovic, sociala medier, zlatan, zlatan ibrahimovic, twitter. Ping.
Previous Previous post: News sites or ad sites?


By Polly Kreisman | March 22, 2012 View as "Clean Read"
Congressional redistricting has claimed its first victim.
Mark Rosen, a Larchmont Republican who had announced his intention to seek the Congressional seat now held by Democrat Rep. Nita Lowey, dropped out Wednesday.
He lives in Larchmont, which like other areas of Westchester County, moves into Democratic Rep. Eliot Engel's district.
Lowey is getting all of Rockland County, which Engel had mostly represented, as well as the Bronx.
Rosen will endorse Joe Carvin, Rye Town Supervisor, who dropped out of a run for U.S. Senate against Democratic Sen. Kirsten Gillibrand last week.
"More Republicans in communities like Bronxville, Eastchester and Yonkers and Democrats in New Rochelle and White Plains were artfully carved out," he told theLoop.


text:
Levi's Stops Costumed Volleyball Antics to Honor Sen. Kennedy
Adrants reader Gareth sent over Levi's tribute to Senator Edward Kennedy. It appeared in the Sunday editions of the Times and the Globe and sports a rambling and whimsical quote from the Sen. ("The work goes on, the cause endures, the hope still lives, and the dream shall never die"), then a message from Levi's itself: "Let us continue his legacy of faith in the people and faith in the work that has yet to be done."
Wordy but loaded with gravity. We like that it remains sparing and casual; apart from the text, a hand-sketched version of the Levi's logo appears over its current tagline: "Go forth," part of a campaign primarily targeted to meatheads at uni.
The ad reminds us of Kenneth Cole's "Different Shoes" campaign, and it's a wonder to us that KC hasn't seized this opportunity to add Kennedy to its list of creative-friendly quotables.
It could just be that we didn't look hard enough though, so if we're wrong, send that bad-boy over.
by Angela Natividad Aug-31-09 Click to Comment
Topic: Brands, Campaigns, Magazine, Newspaper, Trends and Culture
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STONY BROOK, N.Y. -- Junior outfielder Sydney Harbaugh and freshman designated hitter Tia Smith each homered, leading Binghamton (21-18) to a 5-3 win over Albany (30-17) in an America East Softball Tournament first-round game on Thursday afternoon at University Field. With the victory, the Bearcats move on to face Stony Brook in the winner's bracket on Friday at 11 a.m.
Harbaugh led off the game by drilling a 3-2 pitch over the center field fence. Two innings later, she scored on a wild pitch, making the score 2-0. Later in the frame, freshman second baseman Stephanie Bielec scored on a fielding error.
Smith, meanwhile, smacked a two-run home run in the top of the fourth inning, stretching the lead to 5-0. Albany, however, answered in the bottom of the fourth inning on a three-run home run by Jess Tate.
Freshman pitcher Sarah Miller, who pitched the first four innings, got the win, improving her record to 8-6. Junior pitcher Cara Martin, meanwhile, tossed three scoreless innings to nail down her first save of the year. Devin Durnado was tagged with the loss, dropping her record to 14-6.
"I am so proud of our team today" head coach Michelle Burrell said. "It was another true team effort. We played nice and relaxed and came out in the bottom of the seventh inning and went after some outs. Sarah and Cara both pitched great for us today."
Binghamton advanced to the winner's bracket of the America East Tournament for the second time in three seasons.
Martin has now allowed just one run in her past 12 innings of work.


The City of Santa Clara is home to 116,000 residents and sits just to the west of San Jose. While it is 45 minutes south of San Francisco, Santa Clara will in fact soon be home to the NFL 49ers' new stadium. The city's roots go back to 1777 and the founding of the Mission Santa Clara de Asís shortly after explorer José Francisco Ortega first mapped out the area. In 1846 Santa Clara College, the state's oldest institution of higher learning, was established, and in 1862 the Town of Santa Clara was officially chartered.
The economic history of Santa closely matches that of its neighbor to the east, San Jose, as well as Silicon Valley as a whole. For most of its history the local economy was largely based on agriculture. World War II marked a transition to a more industrially-based economy, which then shifted rapidly towards technology.
Today, while San Jose, Cupertino, Mountain View, and Palo Alto are most often associated with "Silicon Valley," Santa Clara is actually home to many major high tech companies — including Agilent Technologies, Applied Materials, Intel, McAfee, National Semiconductor, OmniVision, Trident Microsystems, and Namco Bandai Games.
Architecturally, Santa Clara reflects a modern, suburban look. Santa Clara is largely a residential and commercial community. The Santa Clara City Library and CalTrain stations are recent, award-winning architectural developments. The NFL Levi's stadium, of course, will be the city's most significant architectural development. And, many corporate residents of the city occupy large, stunning office complexes, such as Mission City Center, McCandless Towers, and the WebEx Tower.
Demographically, Santa Clara is a young and diverse community. Twenty-one percent of its population is under the age of 18, eleven percent is between the ages of 18 and 24, thirty-six percent falls between the ages of 25 and 44, twenty two percent are aged 45 to 64. The city is home to excellent public and private schools as well as many vibrant community, arts and non-profit organizations.


text:
Scotch Irish Emigration To America
The following is abstracted from The Scotch-Irish, A Social History by James G. Leyburn, published by The University of North Carolina Press, Chapel Hill, copyright 1962, ISBN 0-8078-4259-1, LOC Cat.#62-16063. This publication covers the whole migration of lowland Scots from Scotland to Ireland beginning in 1610, then to America in the 1700's, and finally, across the mountains to the Pennsylvania frontiers and down the valleys into Virginia and the Carolinas. This abstract sketches the waves of migration from Ulster to America. There were five great waves of emigration, with a lesser flow in the intervening years. An analysis of the tides of 1717-18, 1725-29, 1740-41, 1754-55, and 1771-75 provides, in effect, a chart of the economic health of northern Ireland.
See also The Presbyterian Church below.
1717-18. This first movement, so significant as a path-opener, had as its immediate cause the years of drought; but it was the opinion of Archbishop King and Dean Swift that not even the dire effects of bad crops and high prices would have been enough to make the people move if they had not had the added goad of rack-renting*, still such a novel practice that it caused intense resentment. In a letter of 1718 to the Archbishop of Canterbury, King summed up the causes and tried to persuade his colleague to use his influence to arouse the English conscience to a realization of the effects of what was happening. He charged: "I find likewise that your Parliament is destroying the little Trade that is left us. These & other Discouragements are driving away the few Protestants that are amongst us. ...No Papists stir except young men that go abroad to be trained to arms, with intention to return with the Pretender. The Papists being already five or six to one, & a breeding People, you may imagine in what conditions we are like to be." . . .
In a sense, the emigrants of 1717 would be explorers whose report on their experiences could guide those who came after. The Ulstermen who went to Boston found unexpected difficulties and a welcome that lacked warmth. Those who followed them in the next two years were made to understand that they were not at all welcome. The people who entered America by the Delaware River, on the other hand, found a land of the heart's desire. Their enthusiastic praise of Pennsylvania persuaded others to follow them, and then still others, until by 1720 "to go to America" meant, for most emigrants from Ulster, to take ship for the Delaware River ports and then head west. For the entire fifty-eight years of the Great Migration, the large majority of Scotch-Irish made their entry to America through Philadelphia or Chester or New Castle.
*Rack-rent was simply raising the rent on the land after the period of the lease had expired, and renting to the highest bidder. Lease terms in Ulster were usually 31 years, much longer than they had been in Scotland, and were reasonable in the 17th century. As more and more immigrants came in and land became scarce landlords could get more for use of their land. However, the disposessed, who had been there for a generation or two, were outraged.
1725-29. The second wave was so large that not merely the friends of Ireland but even the English Parliament became concerned. Parliament appointed a commission to investigate the causes of the departures, for they had reached proportions that portended a loss of the entire Protestant element in Ulster.
Letters from immigrants themselves spoke of rack-rents as a determining cause of this second wave; but the Pennsylvania Gazette mentioned these as only one of the "unhappy Circumstances of the Common People of Ireland" that had resulted in so great an exodus. An article in that journal (November 20, 1729) reported "that Poverty, Wretchedness, Misery and Want are become almost universal among them; that . . . there is not Corn enough rais'd for their Subsistence one Year with another; and at the same Time the Trade and Manufactures of the Nation being cramp'd and discourag'd, the labouring People have little to do, and consequently are not able to purchase Bread at its present Rate; That the Taxes are nevertheless exceeding heavy, and Money very scarce; and add to all this, that their griping, avaricious Landlords exercise over them the most merciless Racking Tyranny and Oppression. Hence it is that such Swarms of them are driven over into America."
1740-41. Famine struck Ireland in 1740* and was certainly the principal occasion for the third large wave, which included numbers of substantial Ulstermen. An estimated 400,000 persons died in Ireland during 1740-41; for the next decade there was a tremendous exodus to America. This third wave marked, on the American side, the first movement of Scotch-Irish in any numbers beyond the confines of generous Pennsylvania to the southwest. Following the path through the Great Valley, many Ulstermen now went into the rich Shenandoah Valley of Virginia, whose southern extremity opens out toward North and South Carolina. Arthur Young, writing in 1779, estimated that between 1728 and 1750 Ulster lost a quarter of her trading cash and probably a quarter of her population that had been engaged in manufacture. His comment, if accurate, suggests the caliber of men now leaving the country.
*Not to be confused with the potato crop failure that was the cause of the great Catholic Irish migration in 1845-47.
1754-55. The fourth exodus had two major causes; effective propaganda from America and calamitous drought in Ulster. A succession of governors of North Carolina had made a special effort to attract to that province colonists from Ulster and from Scotland. That two of these officials were themselves Ulstermen lent persuasiveness to their invitation and appeal. As drought ravaged the countryside, testimony of Scotch-Irish success in American struck a particularly responsive chord in hearts back home. ...
At this moment, however, the Scotch-Irish pioneers had their first taste of real trouble with the Indians. The French and Indian wars broke out in the colonies and were to last for more than seven years. For the time being, these violent disturbances effectively dried up the source of new immigration. More than this, Ulster was just now undergoing a true economic recovery. Her prosperity was so pronounced that the vacuum left by emigrants began to be filled by arrivals of people from the south of Ireland and from Scotland. Her population began to increase apace; indeed, it was the pressure of numbers, combined with a new economic depression, that caused the final large wave of migration.
1771-75. Young, writing in 1779, when the outbreak of the American Revolutionary War had eliminated the possibility of further emigration, said that the people of Ulster had by 1770 become very poor, living chiefly "on potatoes and milk and oat bread," and that their little farms had been divided and subdivided until "the portions were so small they cannot live on them." More than this, the shipowners at the ports of Belfast and Derry were in distress because their "passage trade, as it was called," which had long been a regular branch of commerce, was now cut off.
There was, however, a special reason for the departure of this final wave. In 1771, when the leases on the large estate of the Marquis of Donegal in county Antrim expired, the rents were so greatly advanced that scores of tenants could not comply with the demands and so were evicted from farms their families had long occupied. This aroused a spirit of resentment so intense that an immediate and extensive emigration was the consequence. During the next three years nearly a hundred vessels sailed from the ports in the North of Ireland, "carrying as many as 25,000 passengers, all Presbyterian." Froude gives an even larger figure: "In the two years which followed the Antrim evictions, thirty thousand Protestants left Ulster. ...
Throughout the fifty-eight years of the Great Migration, religious liberty had been a motive only at the beginning. It is nevertheless significant, both for Ireland and America, that those who left Ulster were almost all Presbyterians. Members of the Established Church rarely went, nor did Roman Catholic Irishmen. ...
All of the thirteen original American colonies received Scotch-Irish settlers. By comparison with the main stream that flowed through Pennsylvania, the Valley of Virginia, and the Carolina Piedmont, however, Scotch-Irish settlement in other colonies was insignificant in numbers. The strength of Presbyterianism in many of the colonies (New Jersey, for example) was not, as might be supposed, evidence of Scotch-Irish settlement, on the contrary, most of these churches had been founded by English and Welsh Presbyterians and many by immigrants directly from Scotland.
A clear distinction should be made at this point between colonists from Scotland and those from Ulster, for the two have often, to the complete distortion of events, been thought identical. It has already been noted that by 1717 Scots and Ulstermen were two different nationalities. Extensive emigration from Scotland to America occurred during the eighteenth century, possibly a fourth or a fifth as large as that from Ulster; but the reasons for Scottish emigration were distinct. Before the union of the two Crowns in 1707, many Scots were exiled as criminals and many more came as indentured servants or as merchants to America. After the Union, since Scots had equal rights with Englishmen, including the right of moving to the colonies, thousands came over to escape the grinding poverty at home. Defeat of the Highlanders in 1746, after the collapse of the Stuart cause, with the determination of the government to "civilize" these people, caused a large exodus; and the enclosure of lands, the dispossession of tenants, and the consequent dissolution of ties of personal loyalty binding man to chief, sent thousands of others to America. The pull from the colonies was, as usual, the opportunity for a better life. At times during the nineteenth century there came to be a positive "rage for emigration" throughout both Lowlands and Highlands.
Scots in America from the first showed traits clearly different from those of the Scotch-Irish. Scots were seldom explorers, Indian fighters, or frontier traders; they played only a minor role as pioneers, preferring to settle in the east and to carry on business enterprises. Their greatest difference from their Ulster cousins, however, was seen at the time of the American Revolution: whereas the Scotch-Irish were usually ardent patriots and notable fighters in the cause of the colonies, the Scots were, with notable exceptions, Loyalists faithful to the Crown. Only in their Presbyterianism and a few of their traits of personality did they resemble the Scotch-Irish. In North Carolina the Highland Scots for a long while retained their Gaelic language and even their Highland dress.
Children and grandchildren of the original Scotch-Irish settlers in America were always among the leaders in the move to the new West; but they were no longer Scotch-Irish in their social characteristics and outlook. Just as they were likely to become Methodists and Baptists instead of remaining Presbyterians, so they were likely to marry persons whose background may have been English or German. The memory of Ulster and its respectabilities and distinctions meant little or nothing to these constant pioneers. They were Americans.
[The Scotch-Irish] moved immediately upon arrival to a region where there was neither a settlement nor an established culture. He held land, knew independence, had manifold responsibilities from the very outset. He spoke the language of his neighbors to the East through whose communities he had passed on his way to the frontier. Their institutions and standards differed at only minor points from his own. The Scotch-Irish were not, in short, a "minority group" and needed no Immigrant Aid society to tide them over a period of maladjustment so that they might become assimilated in the American melting pot. Like all people, whether immigrants or stay-at-homes, they must have known individual discouragement and disappointment; some may even have had a heightened feeling of inner lonliness, a quality of mind Weber attributes to most Calvinists who reflect upon the implications of the doctrine of predestinatiion. But to the extent that their neighbors shared similar experiences and attitudes, without pressure from other Americans to be different, the Scotch-Irish were not ... marginal men. They were, on the contrary, full Americans almost from the moment they took up their farms in the back-country.
The Presbyterian Church
The following is abstracted from The Scotch-Irish, A Social History by James G. Leyburn, published by The University of North Carolina Press, Chapel Hill, copyright 1962, ISBN 0-8078-4259-1, LOC Cat.#62-16063.
[See also History of Early American Presbyterian Churches]
The course of Presbyterianism in American between 1717 and 1789 neatly reflects the transformations of the mind and the social life of the Scotch-Irish as they became Americans. The eighteenth century in the colonies was a period whose currents of thought had inevitable effects upon church as well as state. Presbyterianism changed much during the century, and in three aspects its changes were significant for the Scotch-Irish: the church became Americanized; it enlarged its conception of service to the common man; and it made tentatives toward democracy. Yet during the very century that saw its increase in vision and effectiveness, the Presbyterian Church lost its hold upon thousands of Scotch-Irish for whom it had been a birthright. ...
In spite of all the expansion of education and the remarkable missionary accomplishment, the church could not begin to meet the religious needs of the two hundred thousand Scotch-Irish, who by 1776 were filling the back-country and steadily increasing their large families. Certainly the church was vividly aware by now of the spiritual needs of the people. Presbyteries ordered pastors to leave their congregations to make missionary journeys among the settlements--preaching, performing marriages, administering the sacraments, consoling the ill and bereaved. Young men who wished to enter the ministry were not ordained until they had visited the frontier. So persistent were the calls that a good part of the time of each Presbytery's meetings was taken up by consideration of appeals from Scotch-Irish settlements. Yet the Church's best was not enough: thousands of the Scotch-Irish people were without the care of a church or minister, and had been for years.
What Presbyterians could not do, Baptists accomplished. All the ardor and adaptability displayed by the former following the Great Awakening [George Whitefield's evangelism in colonial America beginning in 1738] could not overcome the major obstacle of insufficient numbers of ministers. One fundamental Presbyterian commitment stood in the way: the clergy must be well educated. Baptists had no such requirements. To them the gospel was simple, uncomplicated, within the reach of all. Neither Christ nor his disciples had been university men, and his final command had directed ordinary persons to preach the gospel to all men. More than this, it required no complex organization to form a Baptist church; the approval of no Presbytery or other ecclesiastical court was involved. A group of like-minded Christians could form a congregation and select as their minister a dedicated Baptist who felt the "call." He was forthwith a minister, endowed, as he felt, by God's grace to perform all the functions of his office. While Presbyterians were spending six years or more at great expense getting ready to preach, Baptists were already at work--and more of them every year.
At times the zealous young Baptist ministers and missionaries and exhorters could not even read or could read only haltingly, but they knew many passages of the Bible from memory and could speak directly to the hearts of their ready listeners about the great issues of life and death, sin and hell, faith and heaven. ...
Late in the eighteenth century the Methodist Church, reflecting the zeal of the Wesleys and the far-sighted direction of its first American bishop, Francis Asbury, began to share the Baptist success. After independence, when the Appalachians began to be traversed and the Ohio Valley to be filled, the progress of these two denominations was accompanied by methods truly sensational. Whitefield's meetings may be said to have been forerunners of the "revival meeting," which both Baptists and Methodists eagerly adopted; but by 1802, in Kentucky, the revival had lead to the still more fervid and dramatic "camp meeting.." The two sects were evangelical and assiduous in a way that no Protestants had ever been before.
The Methodists devised one of the truly effective adaptations to frontier conditions of life, the circuit rider. A minister, instead of being tied to a single church, rode hundreds of miles each month to visit pioneers on their remote farms. If there were neighbors, he would preach; in any case, he could perform all the services of a pastor to a scattered flock, comforting, counseling, marrying young couples, burying the dead. The devotion and indefatigability of these circuit riders became proverbial: Kentuckians remarked of a day of foul weather that no one would be abroad in it "but crows and Methodist ministers."
Scottish Migration to Ulster
McClelland Home Page


Benjamin de Bivort does not work for, consult, own shares in or receive funding from any company or organization that would benefit from this article, and has disclosed no relevant affiliations beyond their academic appointment.
After many years of research, we still do not completely understand the brains of even the simplest organisms. The human brain with its 80 billion neurons is largely a mystery. But with better tools we are learning to unravel this complexity.
Current methods used to study human brains can only highlight large areas of the brain responding to activity. With each neuron connected to hundreds of others, the sheer complexity makes it difficult to pinpoint the role of single neurons. Only simple organisms such as a simple round worm (Caenorhabditis elegans), which has only 302 neurons, can be understood at the level of each individual neuron. The next animal on the ladder of increasing complexity that will be understood at this level is the fruit fly (Drosophila melanogaster), which has about 100,000 neurons.
My lab at Harvard University investigates why some animals behave one way, while closely related animals behave another. To understand this, we study the humble fruit fly. The hope is that studying how flies move and scrutinising their behaviour will inform our understanding of the fly's brain at the level of single neurons.
Within neuroscience, one field that has advanced faster than others is behavioural neuroscience, in which the behaviour of an organism is correlated with its brain activity. Previously, researchers would have to be present in dark rooms watching flies and clicking tally counters when a fly made a certain move. Instead today, cheaper cameras and faster computing power allow scientists to automate many of these activities. A modern lab features imaging rigs feeding data into computers for real-time processing.
In our earlier investigations of the behaviour of flies, such as their response to light, we have tried to answer a series of questions: What path does a fly follow as it moves toward light? What elementary runs and turns make up that path? What are the fly's individual legs doing as it executes those runs and turns? Now, in a paper published this week in the journal Nature Communications, we report a new method to study this behaviour at the finest scale possible.
Flies are tiny - 2.5mm long - and their legs tinier. To record their position and motion, we borrowed a technology applied to humans: motion capture, which is commonly used in computer-generated imagery for movies. Instead of dressing flies in black suits adorned with ping pong balls, we glue a very small piece of dye to each of their legs.
The fluorescent dye shines only under infrared light, which is completely invisible to the fly. This means we can track its behaviour without it even noticing. This requires dextrous coordination of tiny pins and forceps under the microscope.
Once the fly is marked with the dye, we glue its back to a wire and mount it on top of a clear plastic ball. The ball itself is floating on a cushion of air, flowing from below, which allows it to turn with very little friction – so little that the fly can easily turn the ball with its tiny legs. In effect, the fly is now walking on a boundless two-dimensional treadmill, and we can record its motion using laser sensors extracted from technology found in your computer mouse.
We are not the first to experiment by putting an insect on a ball. Scientists have been doing it since the 1950s. Our insight was that a clear ball would allow us to image leg positions, as indicated by the spots of infrared dye, by looking up through the ball, which now acts as both a lens and a treadmill simultaneously.
With leg-tracking methods in place, we were able to conduct a number of experiments. First, we examined whether a computer could automatically classify what the fly was doing from moment to moment - running, turning, or grooming (which, like cats, flies do a lot). Our results using simple machine learning algorithms showed that computers could evaluate what the fly was doing essentially as well as human observers.
Second, we asked what differed in the leg-motion patterns between individual flies. Surprisingly, we found that individual flies have persistent, idiosyncratic patterns in how they move from one behaviour to another. For instance, one fly routinely started grooming its abdomen after resting motionless, while another fly never performed these behaviours in sequence. It is these sorts of differences, now that they have been observed, which our research is focused on understanding.
Finally, we found that despite these behavioural differences, when flies were presented with coherent visual feedback – such as when they turned to the left, the environment correspondingly moved to the right – their behaviour became more similar. This suggests that their brains are not simply sending rote commands to their legs. Instead, the visual information was continuously being used to make the motor patterns robust to the peculiar traits of each animal's nervous system.
One night we accidentally left a fly mounted to the rig. The next day, there it was, happily running on the ball after 16 hours of no food or water. They are creatures of surprising resilience. We hope this resilience means we can push the limits of our imaging system by simultaneously recording their leg motion, ball motion and neural activity in the the brain. This would involve opening the fly's "skull" and imaging its brain with a specially-built microscope. Hopefully then we will be able to identify which single neurons become active as flies implement each of the detailed behaviours the leg-tracker can record.


text:
Stop-at-nothing zealots exploit the ADA
http://www.jeffjacoby.com/4099/stop-at-nothing-zealots-exploit-the-ada
TO THE BEST of Wayne Crosby's knowledge, nobody in a wheelchair has ever tried to enter the Christy's Market convenience store in downtown Brockton, Mass., one of several that he manages.
"But if anybody ever had trouble getting into one of my stores," he tells me, "I'd be the first one to help him in."
When one of his customers, "a very, very large lady," found the aisles too narrow to navigate, Crosby rearranged the store to accommodate her.
"I'm always looking to see what I can do for my customers. I make eye contact, try to anticipate what a customer needs. Say an old lady needs help getting to her car -- I do it. I have one lady, has a piece of glass in her brain. It causes her to have seizures sometimes. So whenever she comes in, I have somebody walk her home, just to be sure she's safe."
Crosby is telling me these things after I called to ask him about a federal lawsuit brought under the Americans with Disabilities Act that accuses his store of being inaccessible to disabled people. The suit has been filed by the Disability Law Center, a group of lawyers specializing in disability-rights litigation. It names as defendant not Crosby, but Massachusetts Treasurer Joe Malone, chairman of the state Lottery Commission. The lawyers want the lottery to order all 7,000 of its ticket agents, as a condition of keeping their licenses, to spend whatever it takes to become fully handicapped-accessible.
For reasons only the DLC lawyers can explain, they decided to single out Wayne Crosby's store. The complaint charges that his Christy's Market "is inaccessible to persons with disabilities because this business has a step at the entrance which prevents individuals who use wheelchairs from entering."
I read this accusation to Crosby. He's shocked. His store, as it happens, is accessible -- there's a ramp a few feet to the left of the door, marked in fading yellow paint with the familiar "handicapped" symbol. Evidently the DLC's lawyers don't bother to look before they litigate. But that isn't what stuns Crosby. He can't believe any customer with a problem would rather go to court to resolve it than come to him.
"To sue someone?" he asks disbelievingly. "Without even calling first to ask if we can make a change? Even someone in a wheelchair can pick up the phone to make a request. Why start a lawsuit?"
Any plaintiffs' lawyers reading this are by now laughing uproariously. What a sap, they're thinking. Doesn't Crosby know that outfits like the DLC exist to sue -- that they were formed for the express purpose of using lawsuits as blunt weapons with which to reshape society to fit their views? Doesn't he know the Americans with Disabilities Act was designed to generate massive amounts of litigation? Why does he think it was nicknamed the "Lawyers Full Employment Act?"
The ADA is a terrible law enacted with the best of intentions. Most of us are instinctively sympathetic to the hardships faced by the disabled and would like to see their lives made easier. Reasonable people also recognize that there are some activities in which the disabled, by definition, will never be able fully to take part. But the ADA proceeds from the unreasonable premise that equal access for the disabled -- which is defines so broadly as to include one-sixth of the population -- is a matter of "civil rights," and that to secure those rights, no cost or burden is too heavy to bear. Or rather, to impose.
The suit against the lottery is a case in point. If it succeeds, any tavern or convenience store wanting to sell lottery tickets will have to redesign its place of business so that no person with a disability would need help to get in or get around (as if there is something shameful about helping a human being unfortunate enough to be handicapped). And if some of these businesses can't afford the cost of a retrofit? Tough, say the DLC lawyers. And if some will go bankrupt without their income from lottery sales? Tough, say the DLC lawyers.
Odds are this lawsuit won't succeed. The ADA exempts companies with fewer than 25 employees, and while it does cover activities of a "public entity," a private business doesn't become a public entity merely by getting a license to sell lottery tickets. (Memo to the Lottery's legal eagles: Check out Tyler vs. Manhattan, 849 F.Supp. 1429 (1994).)
But this is just one case in a thundering avalanche of litigation. As often as not, the goal in these cases isn't to win in court, but to bludgeon defendants with the threat of such crushing legal costs that they agree to settle. And to the outlandishness of the demands that are being made in the name of civil rights for the disabled, there is no limit in sight.
Just ask David Freeman's ex-wife.
Freeman was the Duxbury, Mass., firefighter who beat his sleeping wife with a baseball bat, fracturing her skull in six places and nearly severing her ear. Then he beat the rap by pleading insanity. The fire department, not wanting an "insane" wife-beater on its force, dismissed Freeman. Whereupon he sued for handicap discrimination -- the disability being his insanity -- and won. The town was ordered to reinstate Freeman, complete with back pay and damages. And who were his lawyers? The Disability Law Center.
Eventually the ruling was reversed. But the case suggests just how far the disability-rights bar intends to take the ADA. A dangerous law is being exploited by stop-at-nothing zealots. It isn't going to end with the Massachusetts Lottery.
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When terrorists killed his daughter, his long crusade for justice began
Lower the voting age? Let's raise it instead


text:
package com.planet_ink.coffee_mud.Abilities.Spells;
import com.planet_ink.coffee_mud.core.interfaces.*;
import com.planet_ink.coffee_mud.core.*;
import com.planet_ink.coffee_mud.core.collections.*;
import com.planet_ink.coffee_mud.Abilities.interfaces.*;
import com.planet_ink.coffee_mud.Areas.interfaces.*;
import com.planet_ink.coffee_mud.Behaviors.interfaces.*;
import com.planet_ink.coffee_mud.CharClasses.interfaces.*;
import com.planet_ink.coffee_mud.Commands.interfaces.*;
import com.planet_ink.coffee_mud.Common.interfaces.*;
import com.planet_ink.coffee_mud.Exits.interfaces.*;
import com.planet_ink.coffee_mud.Items.interfaces.*;
import com.planet_ink.coffee_mud.Locales.interfaces.*;
import com.planet_ink.coffee_mud.MOBS.interfaces.*;
import com.planet_ink.coffee_mud.Races.interfaces.*;

import java.util.*;

@SuppressWarnings("rawtypes")
public class Spell_Frailty extends Spell
{
	@Override public String ID() { return "Spell_Frailty"; }
	private final static String localizedName = CMLib.lang()._("Frailty");
	@Override public String name() { return localizedName; }
	private final static String localizedStaticDisplay = CMLib.lang()._("(Frailty)");
	@Override public String displayText() { return localizedStaticDisplay; }
	@Override public int abstractQuality(){return Ability.QUALITY_MALICIOUS;}
	@Override protected int canAffectCode(){return CAN_MOBS;}
	@Override public int classificationCode(){ return Ability.ACODE_SPELL|Ability.DOMAIN_ENCHANTMENT;}

	@Override
	public boolean okMessage(final Environmental myHost, final CMMsg msg)
	{
		if(!super.okMessage(myHost,msg))
			return false;

		if(!(affected instanceof MOB))
			return true;

		final MOB mob=(MOB)affected;
		if((msg.amITarget(mob))
		&&(msg.targetMinor()==CMMsg.TYP_DAMAGE)
		&&((msg.tool()==null)
				||(!(msg.tool() instanceof Ability))
				||((((Ability)msg.tool()).classificationCode()&Ability.ALL_ACODES)==Ability.ACODE_SKILL)
				||((((Ability)msg.tool()).classificationCode()&Ability.ALL_ACODES)==Ability.ACODE_THIEF_SKILL)
				))
		{
			final int recovery=(int)Math.round(CMath.div((msg.value()),3.0));
			msg.setValue(msg.value()+recovery);
		}
		return true;
	}

	@Override
	public void unInvoke()
	{
		// undo the affects of this spell
		if(!(affected instanceof MOB))
			return;
		final MOB mob=(MOB)affected;

		super.unInvoke();

		if(canBeUninvoked())
		{
			if(mob.location()!=null)
				mob.location().show(mob,null,CMMsg.MSG_OK_VISUAL,_("<S-NAME> seem(s) less frail."));
		}
	}

	@Override
	public boolean invoke(MOB mob, Vector commands, Physical givenTarget, boolean auto, int asLevel)
	{
		final MOB target=this.getTarget(mob,commands,givenTarget);
		if(target==null) return false;

		// the invoke method for spells receives as
		// parameters the invoker, and the REMAINING
		// command line parameters, divided into words,
		// and added as String objects to a vector.
		if(!super.invoke(mob,commands,givenTarget,auto,asLevel))
			return false;
		final boolean success=proficiencyCheck(mob,(mob.phyStats().level()+(2*getXLEVELLevel(mob)))-target.phyStats().level(),auto);

		if(success)
		{
			// it worked, so build a copy of this ability,
			// and add it to the affects list of the
			// affected MOB.  Then tell everyone else
			// what happened.
			invoker=mob;
			final CMMsg msg=CMClass.getMsg(mob,target,this,verbalCastCode(mob,target,auto),auto?"":_("^S<S-NAME> incant(s) to <T-NAMESELF>.^?"));
			if(mob.location().okMessage(mob,msg))
			{
				mob.location().send(mob,msg);
				if(msg.value()<=0)
				{
					mob.location().show(target,null,CMMsg.MSG_OK_VISUAL,_("<S-NAME> seem(s) frail!"));
					maliciousAffect(mob,target,asLevel,10,-1);
				}
			}
		}
		else
			return maliciousFizzle(mob,target,_("<S-NAME> incant(s) to <T-NAMESELF>, but the spell fades."));
		// return whether it worked
		return success;
	}
}

redpajama_set_name: RedPajamaGithub


The British royal vaults contain lots of hidden treasures, and today, we're looking at a piece that hasn't been worn in public since before World War II: the Teck Crescent Tiara.
The tiara's first wearer was Princess Mary Adelaide of Cambridge, the daughter of Prince Adolphus, Duke of Cambridge and Augusta of Hesse-Kassel. She was the granddaughter of one king (George III) and the grandmother of two more (Edward VIII and George VI). Mary Adelaide married Francis, the Duke of Teck, and they became the parents of Queen Mary, linking them back once more to the main line of the royal family.
But Mary Adelaide kept close ties to her British family long before her daughter married a Windsor. Her godmother was her aunt, Princess Mary, Duchess of Gloucester. When Mary Gloucester died, she left Mary Adelaide a collection of jewels, which she combined into this rose and crescent tiara. Hugh Roberts identifies one item from Mary Gloucester's will ("the diamond bandeau for the head and three diamond roses") as the likely forerunner of the crescent tiara.
Mary Adelaide often wore the tiara with two extra rows of diamonds at its base. The tiara exists today as a row of twenty diamond crescents hugging three diamond-set roses, with a slim band of diamonds at its base. The crescents on the tiara can actually be worn facing either backwards or forwards, and there are photos of its wearers sporting it both ways.
When Mary Adelaide died in 1897, the tiara was inherited by her eldest son, Prince Adolphus. He had succeeded his father as Duke of Teck, but he renounced that German title during World War I and was made Marquess of Cambridge instead. The tiara was worn by Adolphus's wife, Margaret (who was a daughter of the 1st Duke of Westminster). In 1901, a new frame was made for the piece. At some point -- this part is a little unclear -- the tiara ended up in the hands of Margaret's sister-in-law, Queen Mary. We don't know precisely how she got it, but by the 1930s, it was in her collection.
To my knowledge, Queen Mary was never photographed wearing the tiara. Instead, she passed it along to her daughter-in-law, Queen Elizabeth (later the Queen Mum). It was never one of the Queen Mother's favorites; in fact, she made only one known public appearance in the piece, at a banquet in Canada in 1939. The tiara didn't surface again until it was included in the major exhibition of tiaras at the Victoria and Albert Museum in 2001. The Queen Mum kept the tiara in her possession until her death in 2002, when it became the property of the current Queen.
Today, the tiara is currently in the hands of another member of the family. It's one of the tiaras that was given by the Queen to the Duchess of Cornwall as a long-term loan after her wedding to Prince Charles. Camilla has worn both of the other tiaras lent to her by the Queen (that would be the Greville and the Delhi Durbar), but she has yet to wear this tiara at a public event. We got to see wonderful photographs of the piece in Hugh Roberts's book, but alas, it's sitting in the vaults right now. Here's hoping we get to see this heirloom get an airing someday soon!