How to show that Group of order $2376$ is not simple,

Now I know that $2376=2^3.3^3.11$

So, $n_{11}=1,12$(Are there any other possibilities? According to my calculation; these are the all)

Now if I have $12$ Sylow-11 subgroups then counting the elements would not help me.

Even if I consider $2$ Sylow-11 subgroups say $H,K$ then their intersection will contain $1$ element only. So their normalizer is the whole group so this way will also not work.

Now I was thinking that here $N_G(H)/C_G(H) \cong Aut(H) \cong \Bbb Z_{10} $ could that help as if I can show that $N_G(H)=G$ then I am done. Here $N_G(H)$ is the normalizer of the group $H$ in $G$, $C_G(H)$ is the centralizer of the group $H$ in $G$.

I would like to know if there would be any other way as well.

  • $\begingroup$ Did you mean to show that a group of this order is not simple? $\endgroup$ – ahulpke Sep 27 '18 at 21:30
  • $\begingroup$ yes, you are right $\endgroup$ – user561073 Sep 27 '18 at 21:32

I must admit I have never seen such a hard exercise of proving a group of a specific order is not simple. Here is a solution. Assume $G$ is a simple group of order $2376$. Then $n_{11}=12$. Let $S$ be an 11-Sylow subgroup. Then by Sylow theorems the index of $N_G(S)$ is $12$ and hence $|N_G(S)|=198$. Next we use the normalizer-centralizer theorem and get that $|N_G(S)/C_G(S)|$ divides $10$. As $|C_G(S)|$ must be an integer we can see that it is either $99$ or $198$, and anyway it is divisible by $9$.

Now let's look at $C=C_G(S)$. It is a group of size $99$ or $198$, and anyway its $3$-Sylow subgroups have size $9$. So let $P$ be a $3$-Sylow subgroup of $C$, and let $H=N_G(P)$. We know that $P\leq C=C_G(S)$. That means every element of $P$ commutes with any element of $S$. So we also have $S\leq C_G(P)\leq N_G(P)=H$, so $|H|$ is divisible by $|S|=11$. Also, note that $P$ is a $3$-group in $G$ and hence is contained in a $3$-Sylow subgroup of $G$. So let $Q$ be a $3$-Sylow subgroup of $G$ that contains $P$. It is known that if $p$ is prime and we have a group of order $p^k$ then any subgroup of order $p^{k-1}$ is normal in it. Hence $P$ is normal in $Q$, and that implies $Q\leq N_G(P)=H$. So $|H|$ is also divisible by $|Q|=27$.

So we have that $H$ is a subgroup of $G$ which is divisible by $11$ and by $27$, so it is divisible by their lcm which is $297$. That means the index of $H$ in $G$ is at most $8$. Now, if $H=G$ then it means that $N_G(P)=G$ and hence $P$ is normal in $G$ which contradicts our assumption that $G$ is simple. So it means $H$ must be a proper subgroup of $G$ and its index is at most $8$. So now we can define an action of $G$ on $G/H$ (the left cosets) by $g.xH=gxH$. The action gives us a homomorphism $\varphi:G\to S_{G/H}$ by $\varphi(g)(xH)=gxH$. As $G$ is simple the homomorphism must be either trivial or injective. It is easy to see that it isn't trivial, so it must be injective. But then $G$ is isomorphic to a subgroup of $S_{G/H}$ and hence $|G|$ divides $|S_{G/H}|$. From here we get that $11$ must divide $|S_{G/H}|$. But that is a contradiction because $|S_{G/H}|=k!$ when $k$ is a number which is at most $8$, so it can't be divisible by $11$. So here is the contradiction, $G$ cannot be simple.

  • 1
    $\begingroup$ I think you can shorten this. The conjugation action of $G$ on the $12$ conjugates of $S$ induces a homomorphism $G \to S_{12}$ which must be injective because $G$ is simple. But $3$ divides $|C_G(S)|$, so $G$ has an element of order $33$ and $S_{12}$ does not. $\endgroup$ – Derek Holt Sep 28 '18 at 5:15
  • $\begingroup$ Oh, you say we take $g\in S$ of order $11$ and $x\in C_G(S)$ of order $3$, and as they commute we get that $gx$ has order $33$? You're right. Very good idea! $\endgroup$ – Mark Sep 28 '18 at 8:12

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