I'm trying to solve the following problem:

Let $G$ be a group of order 12. Assume the 3-Sylow subgroups of $G$ are not normal. Prove that $G\cong A_4$.

Here's my attempt: let $\mathscr S$ be the set of 3-Sylow subgroups of $G$. Since the elements of $\mathscr S$ are not normal, by Sylow's theorem, $\# \mathscr S > 1$. Again by Sylow's theorem, $\#\mathscr S = 4$ and the elements of $\mathscr S$ are conjugate to each other. Hence, one can define a group action of $G$ on $\mathscr S$ by conjugation, and this defines a homomorphism $\phi : G\rightarrow \mathrm{Sym}(\mathscr S)\cong S_4$. Thus, it suffices to show that $\phi$ is injective and its image is $A_4$.

But I'm stuck at this last step. I tried to find the kernel of $\phi$ and found that $a\in\mathrm{Ker}\phi\Leftrightarrow \forall H\in\mathscr S\ aHa^{-1} = H$, but I do not understand what this leads to.

I would be most grateful if you could provide a clue (not necessarily a complete solution).

  • $\begingroup$ You could always investigate the Sylow 2-subgroups. $\endgroup$ – Mark Bennet Sep 8 '13 at 13:54
  • $\begingroup$ @MarkBennet Could you please elaborate on that? $\endgroup$ – Pteromys Sep 8 '13 at 14:49
  • $\begingroup$ Well how many elements are there left once you have filled all your Sylow 3-subgroups? $\endgroup$ – Mark Bennet Sep 8 '13 at 14:51

Let $S_1$ and $S_2$ be two of the Sylow $3$-subgroups. If $a \in \ker \phi$, then in particular

$$aS_1a^{-1} = S_1 \Rightarrow a \in N_G(S_1).$$

The same holds for $S_2$, so

$$\ker \phi \subset N_G(S_1) \cap N_G(S_2).$$

Now, since the Sylow $3$-subgroups aren't normal, what is $N_G(S_i)$?

  • $\begingroup$ By orbit-stabilizer theorem, $\#N_G(S_i)$ are all 3, so $N_G(S_i) = S_i$. They are all of prime order, so $\mathrm{ker}\ \phi$ is contained in the trivial group. Am I correct? $\endgroup$ – Pteromys Sep 8 '13 at 15:00
  • $\begingroup$ Yes. That's exactly the idea. $\endgroup$ – Daniel Fischer Sep 8 '13 at 15:02

If you are still asking why the image of $G$ is (contained in) $A_4$, consider the homomorphism $G \to S_4 \to \mathbb{Z}/2$, where the latter map is the parity morphism. If it were surjective, there would be a normal subgroup of order 6 in $G$. This leads to a contradiction.


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