Let $G$ be a finite group of order $2^aq^b$ for odd prime $q$ and $a,b\geq 2$. Suppose $G$ has at least $2$ subgroups $A$ and $B$ of order $2^{a}$. I suspect that $A\cap B$ is a normal subgroup of size $2^{a-1}$. Any hint to the proof or give a counter example if you don't believe the proposition is true.
This question is answered with a counter example below.
Counterexample:
Let $S_3$ be the symmetry group on $[1,3]$.
Let $G=S_3\times S_3$. $|G|=2^23^2$.
Let $\tau_1 = (1,2) \in S_3$ and $\tau_2 = (1,3) \in S_3$
Let $A$ be the subgroup generated by $(\tau_1,e)$ and $(e,\tau_1)$
Similarly let $B=\langle (\tau_2,e) , (e,\tau_2) \rangle$.
Then $A \cap B=\{e\}$.
This example can be extended to make $|A\cap B|=2^k$ for any $0 \leq k < a$.