I have been trying without success to do the following:
"Suppose $H$ and $K$ are distinct subgroups of $G$ of index $2$. Prove that $H \cap K$ is a normal subgroup of $G$ of index $4$ and that $G/(H \cap K)$ is not cyclic"
I have the hint to use Second Isomorphism Theorem: If $K$ is a subgroup of $G$ and $N$ is a normal subgroup of $G$, prove that $K/(K \cap N)$ is isomorphic to $KN/N$.
Since $|G:H|=|G:K|=2$, $H$ and $K$ are normal in G. Since the intersection of two normal subgroups is again normal, $H \cap K$ is normal. For the index $4$ part, I tried using the fact that the orders of $K/(K \cap N)$ and $KN/N$ are equal, plugging in $|H|=\dfrac{|G|}{2}$ and $|K|=\dfrac{|G|}{2}$, but couldn't derive anything useful.
I abandoned the S.I.T. and tried playing with the indices, since $|G:(H \cap K)|=|G:K||K:(H \cap K)|$. Unfortunately, this didn't get me anywhere either.
I'd appreciate a hint on how to show this. Thanks.
Best Answer
For a proof not using the Second Isomorphism Theorem, let $\pi\colon G\to C$, with kernel $H$ and $C$ cyclic of order two. Similarly, let $\pi'\colon G\to C'$ with kernel $K$ and $C'$ cyclic of order two. Now define $\Pi\colon G\to C\times C'$ by $g\mapsto(\pi(g),\pi'(g))$. Clearly a homomorphism, clearly onto since there are elements of $H$ not in $K$ and vice versa. Now the only thing is to check that $\ker(\Pi)=H\cap K$, and this is almost immediate.