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.


2 Answers 2


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.

  • $\begingroup$ I accepted the other answer because it was more along the lines of what I was trying to do, but I really like your answer. I like the idea of combining quotient maps like that. Thanks. $\endgroup$ Commented Sep 4, 2013 at 14:36

Since $H$ and $K$ are distinct subgroups of index 2, there exists $x \in K - H$. Since $H$ has index $2$ we must have $G=H \cup x H$ and this implies that $G=HK=KH$. Hence $2=(G:K)=(HK:K)=(H:H\cap K)$. This proves that $(G:H \cap K)=4$.

  • $\begingroup$ I don't understand G = HK part. $\endgroup$
    – Bluey
    Commented Jul 4, 2018 at 17:28
  • $\begingroup$ @blueboy, $H,xH\subseteq KH\Rightarrow G=H\cup xH\subseteq KH,$ while $KH\subseteq G$ trivially. Therefore $G=KH$ and $KH$ is a subgroup of $G$ if and only if $KH=HK,$ so we have $G=KH=HK.$ $\endgroup$
    – PinkyWay
    Commented Nov 13, 2023 at 18:54

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