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My original question was:

Suppose $G$ is a group of order $24$ such that there does not exist any element of order $6.$ Then $G\cong S_4.$

I tried solving the problem as follows:

We note that the number of Sylow 3 subgroups of $G$ is $n_3=1,4.$

If $n_3=4,$ then let $S=\text{Syl}_3(G).$ Suppose $G$ acts on $S$ by left conjugation then this induces a homomorphism $f$ from $G$ into $A(S)$ such that $f(g)=\tau_g,\forall g\in G$ where, $\tau_g(H_3)=gH_3g^{-1},\forall H_3\in S.$ Hence, $\text{Ker}(f)=K=\cap_{H_3\in S}N_G(H_3).$

Now, $|N_G(H_3)|=\frac{|G|}{n_3}=6.$ So, $|K|=1,2,3,6.$ If $|K|=1$ then $G\cong A(S)\cong S_4,$ and we are done. Also, $N_G(H_3)\cong S_3\text{ or } Z_6.$ But if $N_G(H_3)\cong Z_6,$ then $G$ has an element of order $6,$ a contradiction. So, $N_G(H_3)\cong S_3.$ But then, $|K|\neq 2.$ This is because, $|K|=2\implies$ that there $K\unlhd N_G(H_3)\cong S_3$ and $S_3$ has a normal subgroup of order $2.$ This is a contradiction. So, $|K|\neq 2\implies |K|=1,3,6.$ The case when $|K|=1$ is already presented. So, let $|K|=6.$ This means, if $H\in S$ we have, $H_3'\unlhd N_G(H),\forall H_3'\in S.$ But this means, $|N_G(H)|=6\geq 8,$ a contaduction. If, $|K|=3,$ then we know that $N_G(K)/C_G(K)\cong \text{Aut}(Z_3)\cong Z_2\implies 2||C_G(K)|$ as $|C_G(K)|=12.$ This means, $C_G(K)$ has an element of order $2.$ Also, this implies $\exists a\in C_G(K)$ such that $o(a)=2.$ Now, $|K|=3\implies K=<b>$ such that $o(b)=3$ and then $o(ab)=6,$ a contradiction, since $ab\in G.$

But I can't proceed with the case when $n_3=1,$ hence the question in the title. Can someone help me with this remaining case? I think with $n_3=1$ we will get some contradiction, but I don't know how to go about it. Any help regarding this will be gladly appreciated.

Thomas Finley
  • 3,625

1 Answers1

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If there is only one Sylow 3-subgroup $H$ (which is isomorphic to $\mathbb{Z}/3$), then it is a normal subgroup. On the other hand, any Sylow 2-subgroup $H'$ acts on the Sylow 3-subgroup by conjugation. So we have the group homomorphism

$$\begin{align*}\varphi: H'&\to \text{Aut}(H)\cong\mathbb{Z}/2\\ h'&\mapsto (g\mapsto h'gh'^{-1})\end{align*}$$

Since $|H'|=4$, the kernel of $\varphi$ must be nontrivial, meaning that there exists $h'\in H'$ which is of order 2 and commutes with $H$. Then $hh'$ is a group element of order 6, where $h$ is a generator of $H.$

Thomas Finley
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Alex Fok
  • 5,256
  • I get that $H\times \langle h'\rangle\cong \Bbb Z_6$ but how can you say that $\exists $ a subgroup $S$ of $G$ which is isomorphic to $H\times \langle h'\rangle$? – Thomas Finley May 07 '25 at 08:55
  • @ThomasFinley $H$ and $\langle h' \rangle$ are both subsets of $G$ and the multiplication is inherited from the whole group? – zyy May 07 '25 at 12:43
  • @zyy What theorem are you using? – Thomas Finley May 07 '25 at 16:17
  • @ThomasFinley $H \times \langle h' \rangle$ here is itself a subgroup of $G$. – zyy May 07 '25 at 17:57
  • @ThomasFinley Also, notice that the Sylow subgroup $H'$ is now of order 8 after your correction of typo. However, the proof still works. – zyy May 07 '25 at 17:59
  • @zyy You say that, $H\times \langle h'\rangle $ is itself a subgroup of $G.$ But how do you know that $G$ is written as a direct product of two groups ? – Thomas Finley May 07 '25 at 18:14
  • @ThomasFinley It is not necessary that $G$ is a direct product. Let us put it in another way. Denote by $h$ a generator of $H$ which is of order $3$. Since $h'$ is of order $2$ and commutes with $h$, the element $hh'$ is of order $6$. – zyy May 07 '25 at 18:56
  • @ThomasFinley $H \times \langle h' \rangle$ coincides with $\langle hh' \rangle$, so my last comment is just a paraphrase of Mr. Fok's answer. – zyy May 07 '25 at 18:59
  • @AlexFok I edited your answer a little to make it more precise. Is it okay with you? +1 from me for your answer! – Thomas Finley May 08 '25 at 03:04
  • $24=2^3.3$, so why $|H'|=4$? – Kan't May 08 '25 at 05:31
  • @Kan't The question originally is about $A_4$ at that time Mr. Fok wrote down this answer, and later changed to be about $S_4$. However, I think this proof still works for the case $\mid H' \mid = 8$. – zyy May 08 '25 at 17:06
  • @ThomasFinley For Fok's original argument, my instinct tells me there could be some proposition claiming that if $H$ and $K$ have coprime orders, and commute with each other, than $HK = H\times K$ is a subgroup. However, I am not very sure since I have been long away from finite group theory. – zyy May 08 '25 at 17:20