If there exists a subgroup of index 2, show that any element squared is in the subgroup. I am trying to understand the following:
Let $H$ be a subgroup of index 2 in a group $G$. Show that $g^2 \in H$ for all $g \in G$.
I don't really know how to approach this. I can see that $H$ is normal since there are only 2 cosets, but then I'm stuck. Could somebody give me a hint?
 A: $G/H$ has order two, so anything squared is the identity.
A: Hint: Of the two cosets, which one is $g^2H$?
A: Hint:
Obviously if $g\in H$ there's nothing to prove.
But what if $g\notin H$? Then the two cosets are $\{H, gH\}$.  Can it be the case that $g^2H=gH$?
Once you conclude that's not possible, $g^2H=H$ says exactly that $g^2\in H$.

Using your idea that $H$ is normal:
$G/H$ is a group of order $2$, so $(gH)^2=H$ for every $g\in G$. But that leads to exactly what you are looking for...
A: Since $i_G (H) = 2$, $H \unlhd G$ as you correctly pointed out. The result is a lemma to the following proof. 
Consider $h \in H$. Then $h^2 \in H$ by closure in $H$. Now consider some $g \in G$ s.t. $g \notin H$. Since $H \unlhd G$, then $G$ is partitioned by the cosets $He$ and $Hg$ i.e. $G = He \cup Hg$ and $He \cap Hg = \emptyset$. 
Now $Hg Hg = Hgg = Hg^2$, once again since $H \unlhd G$, and thus $g^2 \in Hg^2$. 
It follows that $Hg^2 \cap Hg = \emptyset$ (since $g \neq e$, otherwise $g \in H$), and there are only two cosets of $H$ in $G$. Thus it follows that $Hg^2 = He$ and hence $g^2 \in H$. 
A: You understand the notion of normal subgroup, so I'm assuming that you know what a quotient is.
$G/H$ is a group of order $2$, therefore any element has order $1$ or $2$ there. What does it mean in terms of cosets that $$(gH)^2=H\text{ (the identity in the quotient)?}$$
