TITLE: Proving that a group of order $112$ is not simple
QUESTION [7 upvotes]: So I'm proving that a group $G$ with order $112=2^4 \cdot 7$ is not simple. And I'm trying to do this in extreme detail :) 
So, assume simple and reach contradiction. I've reached the point where I can conclude that $n_7=8$ and $n_2=7$. 
I let $P, Q\in \mathrm{Syl}_2(G)$ and now dealing with cases that $|P\cap Q|=1, 2^2, 2^3$ or $2^4$. 
I easily find contradiction when $|P\cap Q|=2^4$ and $2$. 
Um, got stuck REAL bad on the case $|P\cap Q|=2^3$ and $2^2$. 
If $|P \cap Q |=2^3= 8$ and $|P|=|Q|=16$, is there any relationship between $P,Q$ and their intersection that can help me?

REPLY [5 votes]: I wanted to write out Mikko Korhonen's first idea for proof in detail as a separate answer, since it is not trivial at all, and provoked some questions in the comments.
As mentioned in the original question, we assumed $n_2=7$. From Sylow's second theorem, we know that all the 2-Sylow subgroups are conjugate, and we can look at $G$'s action on them by conjugation. This action implies a homomorphism:
$$f:G\rightarrow S_7$$
$\ker(f)$ cannot be $G$ since the action is a transitive action. Then, if $\ker(f)$ is non-trivial (meaning $ker(f)\neq\{e\}$) then it is a non-trivial normal subgroup of $G$, and therefore $G$ isn't simple.
Otherwise, we get $ker(f)=\{e\}$ and therefore $f$ is injective, meaning that $G$ is isomorphic to a subgroup of $S_7$. For convenience purposes, we'll write $G\leq S_7$. $G$ cannot be contained in $A_7$ because 112 doesn't divide $|A_7|$. In that case $GA_7=S_7$ and using the second isomorphism theorem we get:
$$G/(G\cap A_7)\cong GA_7/A_7\cong S_7/A_7\cong\mathbb{Z}_2$$
and therefore $[G:(G\cap A_7)]=2$ meaning that $G\cap A_7$ is a normal subgroup of $G$.