Let $H$ be a subgroup of a group $g$. Show that if $g_1H = Hg_2$ then $g_2H=Hg_1$.
My attempt is the following:
$\forall h_1 \in H, \exists h_2 \in H$ such that $g_1h_1=h_2g_2$.
Since each element of a group has unique inverse, I obtain $ h_2^{-1}g_1=g_2h_1^{-1}$.
If I call $h_1^{-1}$ as $h_2^*$, $h_2^{-1}$ as $h_1^*$, then $\forall h_2^* \in H, \exists h_1^* \in H$ such that $h_1^*g_1=g_2h_2^*$.
Repeat this argument for every $h_2 \in H$, I obtain that $g_2H=Hg_1 \ \square$
Is my proof complete and rigorous?
Is there a better and more elegant proof you can provide?