## Groups of order 4,312 are not simple: An application of Sylow Theory

Today we will solve a problem from Wisconsin’s 2012 Algebra Qualifying Exam. The main tools used in the problem are Sylow Theory and group actions; we will take these tools for granted, but for great introductions to each topic I would recommend Keith Conrad’s notes here and here. The solution as written is not the cleanest; instead I have tried to add in some of my thoughts when solving the problem to highlight why the solution is (somewhat) natural to stumble upon. The problem is in three parts:

• #### Conclude that $G$ is not simple.

Let’s do this. A natural place to start the first part is with Sylow Theory. Since $77=7\cdot 11$, we might try to show that a Sylow-11 subgroup is unique and hence normal, and then form a product subgroup with an element of order $7$ to get the desired subgroup.

Thus let $n_{11}$ be the number of Sylow-11 subgroups. By the third Sylow Theorem, $n_{11}\equiv 1 \mod 11$ and $n_{11}$ divides $2^3\cdot 7^2$. By simply checking all possibilities, we conclude $n_{11}\in \{1,56\}$. If $n_{11}=1$, then there is a unique Sylow-11 subgroup which is normal. We may then take a subgroup of size $7$ (which exists by Cauchy’s theorem or by the extended first Sylow Theorem), form a product subgroup, and create a subgroup of size $77$. Otherwise, we have $n_{11}=56$. In this case we note that (miraculously) $|G|=56\cdot 77$. Furthermore, the second Sylow Theorem tells us that all Sylow-11 subgroups are conjugate to one another; hence we consider the action of $G$ on a Sylow-11 subgroup via conjugation. The orbit of this action has size $56$, and thus the Orbit-Stabilizer Theorem tells us that the stabilizer subgroup has size $77$ as desired. Thus in either case we have found a subgroup of size 77, and we move to part 2. (NB: The fact that there are possibly $56$ Sylow-11 subgroups and that $4312=56\cdot 77$ is really just a happy accident that makes for a good test problem; it would be hard to plan for this from the outset, and hence this is somewhat of a “black magic” solution; however, it does follow quite organically from the consideration of Sylow Theory (and highlights the power of the theory) which is partly why I like this problem so much).

For part two, we need a subgroup of size $7$ with a large normalizer. Recall the definition of the normalizer of a subgroup $K$ as

$N_K=\{g\in G:gKg^{-1}=K\}.$

Given that we just found a subgroup $H$ of size $77$, the obvious choice is to explore this subgroup a little bit more. Now $H$ has size $77=7\cdot 11$; a simple application of the third Sylow Theorem shows that $H$ has a unique Sylow-7 subgroup $K$, and hence $K$ is normal in $H$ (in fact, one may show that $H$ must be cyclic of order $77$ by pursuing the Sylow Theory a little bit more, but that is irrelevant to the current problem). So we have a chain of subgroups

$K\subset H\subset N_K\subset G$.

Thus we already know $77$ divides $|N_K|$. Now $K$ is a $p$-group of size $7$. Again by Sylow Theory we know that $K$ must be contained in a Sylow-7 subgroup, say $Syl_7$. Now $|Syl_7|=7^2$; a simple fact of $p$-groups is that groups of order $p^2$ are abelian, and hence $Syl_7$ must also normalize $K$ (in fact it centralizes it, but again, not important). So we also have $Syl_7\subset N_K$. So $|N_K|$ is divisible by both $77$ and $49$, and hence $539$ divides $|N_K|$. Looking at indices, we get $|G:N_K|$ divides $8$, as desired.

For the last part we are supposed to conclude the $G$ is not simple. To do this, we must find a normal subgroup contained in $G$. The easiest normal subgroups to find in groups are kernels of homomorphisms; given that this is the last part of a problem, we should try to use the previous part to construct some sort of homomorphism with a kernel and we will be done.

Let us take the normalizer subgroup $N_K$ which we found above. There are two natural homomorphisms we can consider when given a subgroup such as $N_K$, both arising from natural group actions. One is given by conjugation on $N_K$, and the other is the multiplication action on cosets of $N_K$. Since we have no idea about the number of subgroups which are conjugate to $N_K$, but we do know that there are relatively few cosets of $N_K$, let’s try the action on the cosets. Say that $|G:N_K|=8$, so that there are $8$ cosets of $N_K$ (the proof with fewer cosets is identical). Then the action via left multiplication of $G$ on the set of cosets gives a map $G\to S_8$, the symmetric group on $8$ symbols.  If this map was injective then we could identify $G$ with a subgroup of $S_8$. But $11$ divides $|G|$, and $11$ does not divide $|S_8|=8!$; hence we would arrive at a contradiction. So this map cannot be injective, and hence must have a nontrivial kernel. Thus we are done, right? Well, not quite. It could be that the entire homomorphism was trivial, and hence the kernel is $G$ itself. But this means that $N_K$ must be equal to $G$ (why?). In particular, the normalizer of our original subgroup $K$ is all of $G$, and hence $K$ itself is normal. So in any case $G$ has a proper normal subgroup, and hence $G$ cannot be simple.

This concludes this problem, which I hope exhibits the strength of the Sylow Theorems and also the importance of keeping natural group actions in the back of your head when tackling problems of this type. As always, thanks for reading!