# Is this proof of $H\le G, [G:H] =2 \implies a^2\in H \forall a\in G$ correct?

If $$H$$ is a subgroup of a group $$G$$ and the index (number of right cosets) of $$H$$ is $$2$$, then $$a^2 \in H$$ for all $$a\in G$$.

My attempt: if $$a\in H$$ then $$a^2\in H$$ directly. If $$a\notin H$$ and $$a^2\notin H$$ then $$a(a^{-1})^{-1} = a^2 \notin H$$ then $$G = Ha \cup Ha^{-1}$$ (the union being disjoint). But $$e\notin Ha$$ because $$e=a^{-1}a$$ and $$a^{-1}\notin H$$. Also $$e\notin Ha^{-1}$$ because $$e = aa^{-1}$$ and $$a\notin H$$. Then we have a contradiction. So $$a^2$$ must be in $$H$$ in both cases.

I think it's ok but I dont feel the arguments with the identity $$e$$ are right and don't see how to justify them more rigorously.

Your proof looks fine, just a bit complicated. Here is another way. If $$a\in H$$ then we are done. If $$a\notin H$$ then $$Ha\ne H$$. Since there are only $$2$$ right cosets we conclude that $$H$$ and $$Ha$$ are all the right cosets, and $$a^2$$ must be in one of them. But if we assume that $$a^2\in Ha$$ then there is an element $$h\in H$$ such that $$a^2=ha$$. We multiply by $$a^{-1}$$ from the right side and get $$a=h\in H$$ which is a contradiction. So $$a^2\in H$$.