I find this problem on a local contest, but i didn't solved it. Maybe someone can give me some points how to continue?
We consider $G$ a finite group for such that exists a proper subgroup $H$ and an element $x \in G$ with property $G = H \cup xH$, where $xH = \{ xh \mid h \in H \}$ and subset $K = \{ y^2 \mid y \in G \}$. Show that:
a) $K \subseteq H$
b) If $|G| = 2n$ with $n$ odd, then $K=H$.
What I've got is $H \neq xH$ hence if that was true we get by hypothesis $G = H$ so $H$ is not a trivial group, contradiction. Is well knowed $|H| = |xH|$ but more I don't know what to do.
$\text{IMPORTANT: I don't want a full solution, just some ideas!}$
I think it should be an assumption that $H$ is a proper subgroup of $G$. Otherwise $H=G$ would qualify, and (b) would not hold.
Hint for (a): $y\in G$ is either of the form $h\in H$ or $xh\in xH$. So look at each of these squared.
Hint for (b): Note $|H|=n$ is odd. Show that $h\mapsto h^2$ is injective (use that $h^{n+1} = h$). Conclude that every $h\in H$ is a square.