# Thread:

1. ## Group Theory - Sylow Theory and simple groups

I'm unsure on the following points:

(i) Prove that if a group G has order $p^nq$, where p and q are prime numbers and $p > q$, then G contains a unique Sylow p-Subgroup. Hence conclude that there are no simple groups of orders 55, 57 or 58.

(ii) Prove that every group of order 59 is simple.

(iii) Prove that there are no simple groups of order 56.

(iv) Give an example of a simple group of order 60, as well as an example of a non-simple group of order 60.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For (i) I know that the number of Sylow p-Subgroups of G divides the order of G and that it has the form $pk+1$( $k \geq 0$). I figured $pk+1$ does not divide $p^nq$ for $k \geq 1$ (is this right?If so, why?) so there must be only one(unique) Sylow p-Subgroup. $55=11.5$, $57=19.3$, $58=29.2$. If a group is not simple then it has a proper normal subgroup. So groups of these orders have a unique Sylow p-Subgroup and for some reason this means that the Sylow p-Subgroup in question is normal(proper).

For (ii) 59 is a prime so for some reason there is no unique sylow Subgroup.(no subgroup by Lagrange) My question is this: If there is no Sylow Subgroup then is there no subgroup in general?
So therefore there is no proper subgroup of a group of order 59, so no proper normal subgroup, so it's simple?

For (iii) $56=2^3.7$ , so there exists a subgroup in this group, say G, of order 2,4,8, and 7. I guess I need to prove that G contains a unique Sylow p-Subgroup but I don't know how.

For (iv) A simple group of order 60 is $A_5$. Is a non-simple group of order 60 the dihedral group $D_{30}$, if so why is this non-simple?

Thanks for any help!

2. 2) A prime ordered group is always cyclic. Lagrange's theorem suffices to show it is simple because the order of a subgroup must divide the order of the group, so the only subgroups are the trivial one and the group itself. \

Sylow's theorem guarantees the existence of a sylow p subgroup for each prime dividing the order of the group. It is just in this case the only prime dividing the group's order is 59. So it is the only subgroup.

4) $d_30$ has a cyclic subgroup of order 30 (the group of rotations) any subgroup of index 2 is normal. So it has a normal subgroup

1 and 3 are easily done if you actually understand what Sylow's theorem is.

3. Originally Posted by Jason Bourne
I'm unsure on the following points:

(i) Prove that if a group G has order $p^nq$, where p and q are prime numbers and $p > q$, then G contains a unique Sylow p-Subgroup. Hence conclude that there are no simple groups of orders 55, 57 or 58.

(ii) Prove that every group of order 59 is simple.

(iii) Prove that there are no simple groups of order 56.

(iv) Give an example of a simple group of order 60, as well as an example of a non-simple group of order 60.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For (i) I know that the number of Sylow p-Subgroups of G divides the order of G and that it has the form $pk+1$( $k \geq 0$). I figured $pk+1$ does not divide $p^nq$ for $k \geq 1$ (is this right?If so, why?) so there must be only one(unique) Sylow p-Subgroup. $55=11.5$, $57=19.3$, $58=29.2$. If a group is not simple then it has a proper normal subgroup. So groups of these orders have a unique Sylow p-Subgroup and for some reason this means that the Sylow p-Subgroup in question is normal(proper).

$kp + 1 \mid p^n q$ and $\gcd(kp + 1, p)=1.$ thus we must have $kp+ 1 \mid q,$ which is possible only if $kp+1=1,$ because $p > q.$

For (ii) 59 is a prime so for some reason there is no unique sylow Subgroup.(no subgroup by Lagrange) My question is this: If there is no Sylow Subgroup then is there no subgroup in general?
So therefore there is no proper subgroup of a group of order 59, so no proper normal subgroup, so it's simple?

by Lagrange, a group of prime order has no proper non-trivial subgroup and hence no proper non-trivial normal subgroup. so it's simple.

For (iii) $56=2^3.7$ , so there exists a subgroup in this group, say G, of order 2,4,8, and 7. I guess I need to prove that G contains a unique Sylow p-Subgroup but I don't know how.

suppose Sylow 7-subgroup is not normal in G. then by Sylow theorem, we must have 8 Sylow 7-subgroups. since every two Sylow 7-subgroups intersect at the identity element only, the union of

these 8 Sylow 7-subgroups contains exactly 49 elements. so we'll have 56 - 49 = 7 elements left and hence we can have only one Sylow 2-subgroup, i.e. Sylow 2-subgroup is normal in G.

For (iv) A simple group of order 60 is $A_5$. Is a non-simple group of order 60 the dihedral group $D_{30}$, if so why is this non-simple?

Thanks for any help!
$D_n$ has an element, say $a,$ of order $n$ (see the definition of a dihedral group again!). thus $<a>$ is a subgroup of index 2 in $D_n,$ and so it's normal. hence $D_n$ is not simple for any $n > 1.$

4. Originally Posted by NonCommAlg
suppose Sylow 7-subgroup is not normal in G. then by Sylow theorem, we must have 8 Sylow 7-subgroups. since every two Sylow 7-subgroups intersect at the identity element only, the union of

these 8 Sylow 7-subgroups contains exactly 49 elements. so we'll have 56 - 49 = 7 elements left and hence we can have only one Sylow 2-subgroup, i.e. Sylow 2-subgroup is normal in G.
Thanks, this helps very much.

The argument relies on supposing that either of the sylow p-subgroups is not normal in G, can we know for certain which is actually the case?

5. Originally Posted by Jason Bourne
Thanks, this helps very much.

The argument relies on supposing that either of the sylow p-subgroups is not normal in G, can we know for certain which is actually the case?
either of Sylow 7 or 2 subgroup can be normal. it depends on the group. what i proved was that at least one of them is normal and this means the group cannot be simple.

6. Let $G$ be the general linear group $GL(2,7)$ consisting of all 2 $\times$ 2 matrices with entries from $\bold{Z}_{7}$ which have a non-zero determinant.

I know that the order of $G$ is 2016.

Define the sets: $H = \{ \left(\begin{array}{cc}1&a\\0&1\end{array}\right) : a \in \bold{Z}_{7} \}$

$K = \{ \left(\begin{array}{cc}1&0\\a&1\end{array}\right) : a \in \bold{Z}_{7} \}$

(i) Prove that both H and K are Sylow 7-subgroups of G.

(ii) Find a matrix $A \in G$ such that $A^{-1}HA=K$.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For (i) should I just show that H and K are subgroups of G and that they have order 7 which divides 2016?

For (ii) I know that such a matrix A exists by one of Sylow's Theorems but I don't know how to get there.

7. ## Partial solution

1) $2016=2*7*144= 2*7*12^2=2^5*3^2*7$
So any subgroup with order 7^1 is a sylow 7 subgroup. Those look like they have order 7.
I trust that you can show why they are in fact a subgroup

2)I think you are just gonna have to play with this, I am not sure of a good way to find this. I know if you do the JCF of K you would get
[11]
[01]
which gets you close, but I am afraid I do not know an eloquent way of doing it other than some brute force.

8. Originally Posted by Gamma
1) $2016=2*7*144= 2*7*12^2=2^5*3^2*7$
So any subgroup with order 7^1 is a sylow 7 subgroup. Those look like they have order 7.
I trust that you can show why they are in fact a subgroup

2)I think you are just gonna have to play with this, I am not sure of a good way to find this. I know if you do the JCF of K you would get
[11]
[01]
which gets you close, but I am afraid I do not know an eloquent way of doing it other than some brute force.
Yeah I guess so. I'm not familiar with "JCF".

9. Got it.
multiply by
[01]
[10]

It is its own inverse.

10. Hi Jason Bourne.

Originally Posted by Jason Bourne
For (ii) 59 is a prime so for some reason there is no unique sylow Subgroup.(no subgroup by Lagrange) My question is this: If there is no Sylow Subgroup then is there no subgroup in general?
So therefore there is no proper subgroup of a group of order 59, so no proper normal subgroup, so it's simple?
Of course there is a unique Sylow 59-subgroup, namely the whole group itself. It is just that this information is useless in showing that the group is simple, so for this question you need to use Lagrange’s theorem instead. As a general rule, every finite group of prime order is simple.

Originally Posted by Jason Bourne
For (iv) A simple group of order 60 is $A_5$. Is a non-simple group of order 60 the dihedral group $D_{30}$, if so why is this non-simple?
You could also have chosen the cyclic group of order 60 as an example of a non-simple group of order 60. That would have been a more simple non-simple group.

11. Originally Posted by Jason Bourne
(ii) Find a matrix $A \in G$ such that $A^{-1}HA=K$.
Hi Jason Bourne.

The elements in $H$ and $K$ have determinant 1; $\therefore\ \det(A)=\pm1.$ Let’s take $+1$ first.

Let $A=\begin{pmatrix}g&h\\j&k\end{pmatrix}$ where $gk-hj=1$ and $\begin{pmatrix}1&x\\0&1\end{pmatrix}\in H.$ Then

___ $\begin{pmatrix}g&h\\j&k\end{pmatrix}\begin{pmatrix }1&x\\0&1\end{pmatrix}\begin{pmatrix}k&-h\\-j&g\end{pmatrix}$

$=\ \begin{pmatrix}g&h\\j&k\end{pmatrix}\begin{pmatrix }k-jx&gx-h\\-j&g\end{pmatrix}$

$=\ \begin{pmatrix}1-gjx&g^2x\\-j^2x&1+gjx\end{pmatrix}$

If this is to be in $K$ then $g=0.$ $\therefore\ gk-hj=1\ \implies\ j=-h^{-1}.$

Hence $A=\begin{pmatrix}0&h\\-h^{-1}&k\end{pmatrix}.$

If $\det(A)=-1,$ then we would have $A=\begin{pmatrix}0&h\\h^{-1}&k\end{pmatrix}.$

So any $A=\begin{pmatrix}0&h\\\pm h^{-1}&k\end{pmatrix}$ where $h,k\in\mathbb Z_7$ and $h\ne0$ will conjugate $H$ into $K.$

12. I have another question:

Let p be a prime. Prove that every group of order $p^2$ is abelian.
~~~~~~~~~~~~~~~~~~~~

Is this done by saying something like that the sylow p-subgroup is uniqe, then it is normal in the group, if this group has a normal subgroup then it is abelian?

13. ## Class Equation

Use the class equation to see that the center is non trivial.

Then consider the various possibilities for the factor groups. $G/Z(G)$. By LaGrange $|Z(G)|=1,p,p^2$
1) Can't be 1 by the class equation

p^2) This means the center is the whole group so it is abelian

p) This is a really elementary proof and it is in pretty much every book, if you can't get this let us know and I am sure someone will write it up explicitly. If $G/Z(G)$ is cyclic, then G is abelian.

$|G/Z(G)|=\frac{p^2}{p}=p$ so it is cyclic.

14. Originally Posted by Gamma
Use the class equation to see that the center is non trivial.

Then consider the various possibilities for the factor groups. $G/Z(G)$. By LaGrange $|Z(G)|=1,p,p^2$
1) Can't be 1 by the class equation

p^2) This means the center is the whole group so it is abelian

p) This is a really elementary proof and it is in pretty much every book, if you can't get this let us know and I am sure someone will write it up explicitly. If $G/Z(G)$ is cyclic, then G is abelian.

$|G/Z(G)|=\frac{p^2}{p}=p$ so it is cyclic.
so the class equation is some thing like $|G| = |Z(G)| + \sum_{i=1}^{r} [G:N(x_i)]$

I don't think I understand the Class Equation properly. So I don't have this proof anywhere, $G/Z(G)$ is cyclic so this implies G is abelian, how?

15. ## Class Equation

Sorry I don't have much time to write it out myself for you, I have to get to a final exam.

basically p divides |G| and each term of the sum, but if |Z(G)|=1 this would be impossible (move the sum over to the other side and p divides the left side, but not the right. Z(G) is a subgroup, so by lagrange its order has to divide $p^2$ and we see it is not 1.

Here is a proof of the G/Z(G) cyclic thing.
http://crypto.stanford.edu/pbc/notes...quotient.xhtml

Sorry again for the brevity.
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