Suppose we have a group $G$, and two subgroups $H$ and $J$ of $G$. Further, suppose $J$ has index $2$ in $G$. Consider the homomorphism $f:G \to G/J$ satisfying $f(x)=xJ$. Now, consider the restriction of $f$ to the set $H$, say $f(H)$. Show that if $f(H)$ contains two elements (i.e. two distinct left cosets), then $H \cap J$ has index $2$ in $H$.
How does one solve this problem? Perhaps we can find a bijection between $H \cap J$ and $H \smallsetminus (H \cap J)$ defined as $\phi(a)=xa$, where $x\in H \smallsetminus (H \cap J)$ (we know such an $x$ exists because if not, then there would only be one left coset in $f(H)$, namely, $J$). I can show that this is injective, but cannot show that this is surjective. Or how would one solve this problem anyway?
Use the isomorphism theorems (two or three):
$$H/\left(H\cap J\right)\cong HJ/J$$
but $\;(i)\;$ since there there are two different elements of $\;H\;$ modulo $\;J\;$ and $\;(ii)\;$ since $\;HJ/J\le G/J\;$, we get what we want:
$$2\stackrel{(i)}\le\left|H/\left(H\cap J\right)\right|= \left|HJ/J\right|\stackrel{(ii)}\le \left|G/J\right|= 2$$