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I'm trying to prove that if $G$ is a finite group of order $2^{k}\cdot3$, with $k ≥ 1$, then $G$ is not simple.

The idea is to use the permutation representation associated to the conjugation action of $G$ on the Sylow $2$-subgroups, presumably to illustrate that some such subgroup is normal. However, I'm not really sure where to start. What's the general procedure to follow?

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Here's the sketch without details. The number of Sylow 2-subgroups is either 1 or 3. If it is 1 then that group is normal. So assume there are 3. G acts transitively on these 3 subgroups and that induces a homomorphism from G to $S_3$. If it is injective then the order of G is 6. Then consider the number of Sylow-3 subgroups.

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  • $\begingroup$ How exactly do the 3-subgroups fit into the picture? $\endgroup$ – cloudchamber Oct 27 '16 at 1:56
  • $\begingroup$ If k=1 then the number of 3-subgroups divides 2 and is congruent to 1 mod 3. So how many are there? $\endgroup$ – Grad student Oct 27 '16 at 1:58
  • $\begingroup$ Ahh, OK: One, which implies normality trivially. If I understand the proof correctly -- the scheme is to say that the homomorphism is injective iff |G|=6, effectively reducing it to that case, and then just show that that specific case doesn't work (with the 3-subgroups)? $\endgroup$ – cloudchamber Oct 27 '16 at 2:01
  • $\begingroup$ Basically. But these cases all are possible. We're not giving a proof by contradiction. The idea is to break it up into several cases. If the number of 2-subgroups is 1 we are done. If not then it is 3. Then we get a homomorphism from G to S3. If the kernel is nontrivial we are done. If the kernel is trivial then G injects to S3. But by the condition on the order k must be 1 so G has order 6. Then the number of sylow 3's is 1 and we are done. These are all possibilities. $\endgroup$ – Grad student Oct 27 '16 at 2:07
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Let $G$ act on Sylow $2$-subgroups by conjugation. This action permutes the three Sylow $2$-subgroups. Therefore, there exists a homomorphism $\phi$ from $G$ to $S_3$, where $S_3$ is symmetric group over three letters. If $k>1$ then $\ker \phi$ is non trivial and so $G$ is not simple. For $k=1$, the answer is obvious.

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  • $\begingroup$ Maybe I'm missing something here -- but how exactly are you concluding that there are three distinct Sylow subgroups? Is that just from the congruence conditions? $\endgroup$ – cloudchamber Oct 27 '16 at 1:53
  • $\begingroup$ Number of Sylow 2-subgroups must divide 3 by sylow theorem. $\endgroup$ – Ashar Tafhim Oct 27 '16 at 1:54
  • $\begingroup$ It is either 1 or 3 from congruence conditions. If it is 1 then it is normal. $\endgroup$ – Grad student Oct 27 '16 at 1:55
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Here's another proof that doesn't make use of homomorphisms and group actions.

Obviously $G$ has a subgroup of order $2^k$, it's Sylow $2$-subgroup. By Sylow's Theorems $n_G(2) \equiv 1 \pmod 2$ and $n_G(2) \mid 3$. If $n_G(2) = 1$, we're done, as $G$ has a normal Sylow $2$-subgroup.

Now assume that $n_G(2) = 3$. Then $G$ has two (in fact three) distinct subgroups of order $2^k$, denote them by $H_1$ and $H_2$. Then:

$$2^k \cdot 3 = |G| \ge |H_1H_2| = \frac{|H_1||H_2|}{|H_1 \cap H_2|} \implies 3 \ge \frac{2^k}{|H_1 \cap H_2|}$$

So from this we can conclude that $|H_1 \cap H_2|=2^k$ or $|H_1 \cap H_2|=2^{k-1}$ The first case is impossible, as $H_1$ and $H_2$ are distinct. So therefore $|H_1 \cap H_2|=2^{k-1}$. Now we have that $[H_1:H_1 \cap H_2] = [H_2:H_1 \cap H_2] = 2$, so $H_1 \cap H_2$ is normal in both of them and so $H_1 \cap H_2 \unlhd \langle H_1, H_2 \rangle $. But now:

$$|\langle H_1, H_2 \rangle| \ge |H_1H_2| = 2^{k+1}$$

Also $|\langle H_1, H_2 \rangle|$ divides $2^k \cdot 3$ by Lagrange's Theorem and the only divisor of it greater than or equal to $ 2^{k+1}$ is itself. Hence $\langle H_1, H_2 \rangle = G$ and so $H_1 \cap H_2 \unlhd G$ and $G$ isn't simple.

Unlike the other proof this one tells us something more. That is that a group $G$ of order $2^k \cdot 3$ has a normal subgroup of order either $2^k$ or $2^{k-1}$

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