# Showing that a group $G$ as many right cosets as there are left cosets.

Let $G$ be a group. I want to show that there's a bijection $f$ from the set of $G$'s right cosets to the set of $G$'s left cosets, such that $f(Ha) = a^{-1}H$. I thought about it for a while, but I'm not sure where to begin. Any hints?

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Can you show that $aH=bH$ $\Leftrightarrow$ $b^{-1}a\in H$? Can you show that $Hc=Hd$ $\Leftrightarrow$ $cd^{-1}\in H$? I think this could help. – Martin Sleziak Dec 9 '11 at 15:50

You need this: $g_1H=g_2H\Leftrightarrow Hg_1^{-1}=Hg_2^{-1}$ Then your $f$ is a bijection.
That shows that your function is well-defined and 1-1. It is onto because $g\to g^{-1}$ is a bijection of $G$. – Thomas Andrews Dec 9 '11 at 16:55