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A group is $p$-closed if it has only one Sylow $p$-subgroup.

Theorem: Let $G$ be a group of order 24 that is not 3-closed. Then either $G\cong S_4$ or $G/Z(G)\cong A_4$.

Proof. $G$ acts on $$ \Omega:=\operatorname{Syl}_3 G$$ by conjugation. Since $G$ is not 3-closed Sylow's Theorem gives $|\Omega|=4$. Thus, there exists a homomorphism $\varphi:G\to S_4$ such that $$\operatorname{Ker}\varphi =\bigcap_{S\in\Omega} N_G(S)=: N.$$ $G/N$ is a subgroup of $S_4$ and $|N|$ a divisor of $\frac{24}{4}=6$. If $|N|\in\{3,6\}$, then $N$ and thus also $G$ is 3-closed, a contradiction. The case $N=1$ yields $G\cong S_4$, and the case $|N|=2$ implies $N=Z(G)$ and $G/N\cong A_4$. $\square$

I don't understand the case $|N|=2$. Obviously $Z(G)\leq N_G(S)$ for every $S\in Syl_3 G$ and so $Z(G)\leq N$. Is it that $Z(G)=2$? And why is $G/N$ isomorphic to $A_4$? This seems even more difficult to prove!

Alejandro Tolcachier
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stf91
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  • This is from "The Theory of Finite Groups: An Introduction," by Kurzweil and Stellmacher. – Shaun Aug 30 '20 at 19:44
  • Also, please refrain from using pictures like yours here. They're difficult to search and are less user-friendly than typed-up $\LaTeX$. – Shaun Aug 30 '20 at 19:46

1 Answers1

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If $|N|=2$, you proved that $Z(G)<N$. But any normal subgroup of order 2 must be contained in the center (see for example, Let $K$ be a normal subgroup of order 2 in group $G$, show that $K$ lies in the centre of $G$), so $N<Z(G)$ and thus $N=Z(G)$.

Now $G/N$ is a subgroup of $S_4$ that has order 12. Therefore $[S_4:G/N]=2$, and it is a very common exercise (see $A_n$ is the only subgroup of $S_n$ of index $2$. ) that the unique subgroup of $S_n$ of index 2 is $A_n$.

We conclude $G/N\cong A_4$.

Alejandro Tolcachier
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