If $H \leq G$ such that $[G:H]=2$ then $a^2 \in H $ for every $a\in G $

Question

If $H \leq G$ such that $[G:H]=2$ then $a^2 \in H $ for every $a\in G $

If $a \in H $ then $a^2 \in H $ so I took $a\in G-H $ and wrote $G=Hx \cup Hy $ for some $x,y \in G $ but wasn't able to arrive anywhere. Any hint?

Answer 1

Better write $G=H \cup aH$, where $ a \notin H$. Try to argue that $a^2 \notin aH$.

Answer 2

Such a subgroup $H$ must be normal, since the action $h.g = h^{-1}gh$ of $H$ on the two-element coset space $G/H$ clearly preserves $H$ itself. Thus any $g\in G/H = \mathbb{Z}_2$ has $g^2 = 1$; that is, $g^2\in H$.

Answer 3

If $a \in H$, you're done. Suppose $a \not \in H$. Then $a^2 H = a (aH)$. Suppose to the contrary $a^2 \not \in H$. Then $a^2 H = a H \Rightarrow aH = H \Rightarrow a \in H \Rightarrow a^2 \in H$, a contradiction.