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

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.

• Did you mean to show that a group of this order is not simple? – ahulpke Sep 27 '18 at 21:30
• yes, you are right – 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.
• 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. – Derek Holt Sep 28 '18 at 5:15
• 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! – Mark Sep 28 '18 at 8:12