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!

This is Jack's comment (hint) fleshed out: suppose $\,G/(H\cap K)\;$ is cyclic and $\,H\ne K\;$ :
$$H/(H\cap K)\;,\;K/(H\cap K)\le G/(H\cap K)\;,\;\;\text{and we also have that}$$
$$\left|H/(H\cap K)\right|\cong \left|HK/K\right|\stackrel{\text{why?}}=2\stackrel{\text{why?}}=\left|HK/H\right|\cong\left|K/(H\cap K)\right|$$
Thus we have a finite cyclic group with two subgroups of the same order, and this is impossible.
