Let $G$ be a group, $N\triangleleft G$ an index-two (normal) subgroup, and $H_1,H_2<G$ two subgroups. Is it true that $$H_1\cap N = H_2\cap N \Rightarrow H_1 = H_2\ ?$$ If no, is it true with the extra hypothesis that $H_2=gH_1g^{-1}$ for some $g\in G$?
Proofs or couterexamples would be appreciated!
It's not true, even with the additional hypothesis.
For a counterexample, take $G = S_3$.
Let $N$ be the subgroup of order $3$, which is normal because its index is $2$.
Let $H_1$ and $H_2$ be two of the three subgroups of order $2$. Then $H_1$ and $H_2$ are conjugate because they are Sylow (or use the fact that any elements with the same cycle structure are conjugate).
We have $H_1 \cap N = H_2 \cap N = 1$ by Lagrange, but $H_1 \neq H_2$.
No, take $G=V_4=\{1,a,b,c\}$, the Klein 4-group. Let $H_1=\{1,a\}$, $H_2=\{1,b\}$ and $N=\{1,c\}$.
