Suppose $G$ is finite and $G=H\cup K\cup L$ for proper subgroups $H,K,L$. Show that $|G:H|=|G:K|=|G:L|=2$.

What I did: so if some of $H,K,L$ is contained in another, then we have $G$ being a union of two proper subgroups, which is impossible due to another result. So none of $H,K,L$ is contained is each other. If some element $h$ belongs to only $H$ and $k$ belongs to only $K$, then its product $hk$ cannot be in $H$ or $K$, so must belong to only $L$.


1 Answer 1


At least one of $H, K$ and $L$ must have index $2$ in $G$ by counting. Let's say $|G:H| = 2$. Then, $H$ is normal in $G$. Then, $|K:K\cap H| = 2 = |L:L\cap H|$.

Now, for $G = H \cup (K - K\cap H) \cup (L - L\cap H)$, this union must be disjoint by counting and we must have $|G:K| = 2 = |G:L|$.

  • $\begingroup$ I see. So $|K:K\cap H|=2$ follows from the inequality $|K:K\cap H|\le |G:H|$ and the fact that $K\ne K\cap H$. Can we derive it directly from the fact that $H$ is normal? $\endgroup$
    – Paul S.
    May 25, 2013 at 21:19
  • 2
    $\begingroup$ Maybe I am wrong, but I think your proof is not accurate, cause you said the union must be disjoint but isn't it possible that there would be an element in $(K-H)\cap L$?And how do you get the index of $H$ in $G$ two by counting. Please do concern $\endgroup$
    – Aref
    Aug 5, 2014 at 13:01
  • $\begingroup$ If $(K-H)\cap L\ne\varnothing$, then $K\cap L\ne H\cap L$, which contradicts claim 1 in this answer by Arturo. The proof of the claim should be in this answer. (Note: this answer is from a deleted user.) $\endgroup$
    – Frenzy Li
    Aug 8, 2018 at 10:57

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