Why is $\frac{G}{H \cap K}$ not cyclic if G is finite and H and K are 2 different subgroups of index 2? I was making some exercises on group theory when I came across this problem:
Suppose G is a finite group and K and H are 2 different subgroups of index 2. 
Now first prove $(H \cap K) \triangleleft G$ 
and then prove $\frac{G}{H \cap K}$ is not cyclic.
The first part went well (I used the fact that $H \triangleleft G$ and $K \triangleleft G$ because the indices of these subgroups are 2, the rest is quite easy). The second part however, didn't.
Thank you for your help!

As mentioned in the comments, we need $H \neq K$. Let $\alpha \in G$. Then it is easy to see that $\alpha^2 \in H$ and $\alpha^2 \in K$ as $[G : H] = [G : K] = 2$. Thus, $\bar{\alpha}^2 = \bar{1}$. If $\bar{\alpha}$ generates $G/(H \cap K)$, then it must be that $[G : H \cap K] = 2$ which is not possible.
