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? 
 A: 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.
A: 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.
A: 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 $2^{k-1}$
