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Claim : If $H \le G$ and $H$ has index 2 in $G$ then $a^2 \in H, \forall a\in G$

My attempt : There will be exactly two cosets lets call $H$ and $aH$ respectively. If $a\in H$ then $a^2\in H$ because of closure property of $H$. Second case when $a\in aH$ then $aHaH = a^2H$ so now how to prove that $a^2 \in H$

Please Help.

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Hint: If $a^2H = aH$, then $aH = H$.

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  • $\begingroup$ NB: To show this, multiply on the left by $a^{-1}$. $\endgroup$ – Shaun Apr 13 '18 at 15:48