Suppose that $H$ is a subgroup of $G$ and that $g_1H = Hg_2$ for some $g_1,g_2 ∈ G$. Show that $g_1H = Hg_1 =g_2H =Hg_2$.

Question

Suppose that $H$ is a subgroup of $G$ and that $g_1H = Hg_2$ for some $g_1,g_2 ∈ G$. Show that $g_1H = Hg_1 =g_2H =Hg_2$.
I have already shown that $g_1H = g_2H$. from the assumption, we have that $g_1H = Hg_2$.
To finish my proof I just need to show that $Hg_1$ is equal to one of these sets however I am stuck and I have no idea how to proceed.

Answer 1

If $g_1H=Hg_2$, then $g_1\in Hg_2$. Since we also have $g_1\in Hg_1$, then the cosets $Hg_1$ and $Hg_2$ are not disjoint, so they are equal.

Symmetrically, $g_2\in g_1H\cap g_2H$, so $g_1H=g_2H$.

So $g_2H=g_1H=Hg_2=Hg_1$.