I posted this because Alex Clark asked in chat and I'm not sure how to proceed. Let $G$ be a group such that it has a fixed subgroup isomorphic to $A_7$, which we denote simply by $A_7$. Let $H$ be a subgroup of $G$ that has order $112$. Prove that $H\cap A_7 \lhd H$.
Since $|H|=2^47$ and $|A_7|=2^3\cdot 7 \cdot d$ with $d$ an odd number, we can see the order of $H\cap A_7$ is a divisior of $2^3\cdot 7$.
Some trivial cases are when the order of $H\cap A_7$ is $1$ or $2^37$. That is the only stuff I can see.
I think the following is a counterexample.
Let $G=S_{10}$. As expected let $A_7$ consist of the even permutations of $G$ that have $8,9,10$ as fixed points.
Let $\alpha=(1234567)$. The elementary 2-abelian group $$ E=\langle(18)(24)(37)(56),(28)(14)(35)(67),(38)(46)(25)(17)\rangle $$ is stable under conjugation by $\alpha$. Unless I fumbled it, this is just yet another instance of the multiplicative group of the field $\Bbb{F}_8$ acting on its additive group. So with $C=\langle \alpha\rangle$ the subgroup $$ K=\langle E,\alpha\rangle=E\rtimes C $$ is a subgroup of order $56$ acting transitively on $\{1,2,3,4,5,6,7,8\}$.
Let $H=\langle K,(9;10)\rangle\le G$. Then $H\cong K\times C_2$ so $|H|=112$. We see that $K$ has eight Sylow $7$-subgroups (point stabilizers of one of $1,2,\ldots,8$ inside $K$), and these account for all its elements outside of $E$. Therefore $H\cap A_7=K\cap A_7=C$.
But $C$ is not a normal subgroup of $H$ because it has those eight conjugates in $K$.