Suppose $G$ is a finite group, $H\leq G$, and $[G:H]=2$. Prove that for every $g\in G$, $g^2\in H$.

Question

I am having trouble with the question above.

Since $[G:H]=2$, I know there are 2 cosets of $H$ in $G$, which partition $H$. I know I'm supposed to use the index of $H$ and potentially Lagrange's Theorem, but I am not seeing how I can get $g^2\in H$ for any $g\in G$.

Thanks for your help

Alternatively, a case or two: $g \in H$ (easy) or $g \notin H$ (not too bad). — Randall, Oct 22 '18 at 19:49
and https://math.stackexchange.com/q/1938717/29335 too. Duplicates found by searching for "$[G:H]=2$ and $x^2\in H$." You should try this too, in the future. — rschwieb, Oct 22 '18 at 20:02

score 3 · Answer 1 · answered Oct 22 '18 at 19:52

Split into cases. If $g\in H$ then obviously $g^2\in H$ because $H$ is itself a group. Now suppose $g\notin H$. Then $gH\ne H$. Now look at $g^2H$. As there are only $2$ left cosets we have either $g^2H=H$ or $g^2H=gH$. Suppose $g^2H=gH$. That means $g=g^{-1}g^2\in H$ which is a contradiction to our assumption. Hence $g^2H=H$ which implies $g^2\in H$.

Suppose $G$ is a finite group, $H\leq G$, and $[G:H]=2$. Prove that for every $g\in G$, $g^2\in H$.

1 Answers1