Group of order 96. Show there exists a normal subgroup of order 16 or 32. Let $G$ be a group. Show that there exists a normal subgroup of $G$ such that its order is either 16 or 32.
Attempt at a solution:
If $n_{2}=1$ (the number  of Sylow $2$-subgroup of G) then we have a unique Sylow 2-subgroup of order 32 which is normal in $G$. So suppose $n_{2}\neq 1$. So we have $n_{2}=3$.
Let $H$ and $K$ be distinct Sylow 2-subgroups of G.
$|H|=|K|=32$ and $|HK|=\frac{|H||K|}{|H\cap K|}=\frac{32^{2}}{|H\cap K|}\implies |H\cap K|$ is a power of 2.
We know that $|HK|$ is between 32 and 96. Testing the possible values of $|H\cap K|$ we get $|H\cap K|=16$ and $|HK|=64.$
Since $H$ and $K$ are Sylow 2-subgroups, they are conjugates of each other. Thus we can say that $HK\subseteq N_{G}(H)$ and $HK\subseteq N_{G}(K)$ and thus $HK\subseteq N_{G}(H\cap K)$.
We know that $H\cap K \leq G$ since $H, K\leq G$. So $N_{G}(H\cap K)$ is also a subgroup of $G$. 
So $|N_{G}(H\cap K)|$ is greater than or equal to $|HK|$ and must divide $|G|$. The only possible value is $96$. Thus $|N_{G}(H\cap K)|=|G|\implies N_{G}(H\cap K)=G\implies H\cap K$ is normal in $G$.
We now have a normal subgroup of $G$ of order 16.
Am I right in assuming that since $H$ and $K$ are Sylow 2-subgroups, they are conjugates of each other. Thus we can say that $HK\subseteq N_{G}(H)$ and $HK\subseteq N_{G}(K)$?
Any other ways how to solve this?
Thank you! :)
 A: Hint: We have $n_2=3$. $G$ acts by conjugation on the set of Sylow 2-subgroups. Hence there is a homomorphism $f:G\to S_3$ and $G$ has a normal subgroup (the kernel of $f$) of order $48, 32$ or $16$.
A: As Boris says, G acts by conjugation on the 3 Sylow 32-groups if $n_2=3$.
However, we also know some of the permutations in the image. The 3 32-groups
act by conjugation on each other, and no group is in the normalizer of any of the others. This means that the permutations (12), (13), and (23) must be in the image.
Hence the image is all of $S_3$.
The kernel of $f : G \rightarrow S_3$ is then a normal subgroup of order 16.
Your argument that $HK \subset N_G(H)$ and $HK \subset N_G(K)$ is not correct.
The 32-groups do not normalize each other.
A: For the second part there is a solution in your method:
After finding that $ |HK|=64$ and $|H \cap K|=16$ we do the following.
$(H \cap K)$ is a subgroup of H and K.
Here $ N_G(H \cap K)$ contains $ H $ and $K \implies N_G(H \cap K)$ contains $|HK|$.
Therefore, $N_G(H\cap K)\implies 64 $and $N_G(H \cap K)$ divides $|G| $ together gives us than $N_G(H \cap K)=96 $i.e.,$ N_G(H \cap K)=G \geq (H \cap K) $ is a normal subgroup of order $16 $ in $G$.
