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?

  • $\begingroup$ To show that $g_2H=Hg_1$, I would start with a $g_2h\in g_2H$, and show that it’s writable as $h'g_1$ for some other $h'\in H$. This will show that $g_2H\subset Hg_1$, and the reverse inclusion will be similar. I think your proof can easily be modified to do this, but to me, the logic is less clear in your current version than it might be. For my argument, see below. $\endgroup$
    – Lubin
    Nov 1, 2015 at 3:16

1 Answer 1


My argument: The set $g_1H=Hg_2$ is both a left coset and a right coset of $H$. It’s a right coset containing $g_1$, so it’s equal to $Hg_1$; and it’s a left coset containing $g_2$ so it’s equal to $g_2H$.


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