# Proving that a group of order $2^{k}\cdot3$ isn't simple?

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?

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.

• How exactly do the 3-subgroups fit into the picture? – cloudchamber Oct 27 '16 at 1:56
• If k=1 then the number of 3-subgroups divides 2 and is congruent to 1 mod 3. So how many are there? – Not a grad student Oct 27 '16 at 1:58
• 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)? – cloudchamber Oct 27 '16 at 2:01
• 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. – Not a grad student Oct 27 '16 at 2:07

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.

• 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? – cloudchamber Oct 27 '16 at 1:53
• Number of Sylow 2-subgroups must divide 3 by sylow theorem. – Ashar Tafhim Oct 27 '16 at 1:54
• It is either 1 or 3 from congruence conditions. If it is 1 then it is normal. – Not a grad student Oct 27 '16 at 1:55

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 $2^{k-1}$