Proving that a group of order $112$ is not simple 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? 
 A: 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 $G$.
A: If $G$ is a simple group, it must have exactly $7$ Sylow $2$-subgroups. Thus $G$ embeds into $S_7$, and in particular into $A_7$ since $G$ does not have a subgroup of index $2$. But the order $A_7$ is not divisible by $112$.
If you want to go along the lines of your original idea, you can rule out the case $|P \cap Q| = 2^3$ by noticing that then $P \cap Q$ is normal in $P$ and $Q$ (as a subgroup of index $2$), so $N_G(P \cap Q)$ contains $P$ and $Q$, which implies that $N_G(P \cap Q) = G$. 
ADDED: I'm not sure if there is an easy way to deal with rest of the cases. However, there is a nice argument which also works for proving that every group of order $p^n q$ ($p$, $q$ distinct primes) is nonsimple. I believe the idea of the proof goes back to G. A. Miller (around 1900-1910). Here's an illustration of it in this case.
Suppose that $G$ is a simple group of order $112$. Then $G$ has exactly $7$ Sylow $2$-subgroups. Let $P, Q \in Syl_2(G)$ be such that $P \neq Q$ and that $D = P \cap Q$ has largest possible order. Steps for the proof:


*

*Using the fact that $D < N_P(D)$ and $D < N_Q(D)$ (proper inclusion), prove that $N_G(D)$ cannot be a $2$-group.

*Thus $D$ is normalized by an element $g \in G$ of order $7$. Prove that $P, gPg^{-1}, \ldots, g^6Pg^{-6}$ are distinct. Conclude that $D$ is contained in every Sylow $2$-subgroup.

*Since the intersection of all Sylow $2$-subgroups is normal, $D$ is trivial.

*By counting elements in Sylow $2$-subgroups, prove that $G$ contains exactly one Sylow $7$-subgroup. 
This same argument works for proving the statement for groups of order $p^n q$.
A: Sylow's theorems require that $n_2=1$ or 7, $n_7=1$ or 8, so $G$ has a chance to be simple only if $n_2=7$ and $n_7=8$. Note that the Sylow 7-subgroups can only intersect at the identity, any two Sylow 2-subgroups can share a subgroup of order at most 8, and a Sylow 7-subgroup and a Sylow 2-subgroup can share only the identity.
Hence the union of Sylow 7-subgroups has $1+8\cdot(7-1)=49$ elements, and the union of the Sylow 2-subgroups has at least $8+7\cdot(16-8)=64$ elements, which happens precisely when they all share a subgroup $H$ of order 8. In this case, the union of all Sylow 7-subgroups and 2-subgroups has $64+49-1=112$ elements, so we learn that no other scenario (one with a greater union of the Sylow 2-subgroups) is allowed.
Now notice that $H$ is a normal subgroup of $G$ since conjugation by $g\in G$ permutes the Sylow 2-subgroups and so preserves their intersection. Hence $G$ is not simple.
Edit: This answer is wrong because the union of the Sylow 2-sbgps can be smaller than 112, see the comments below.
