0
$\begingroup$

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?

  • 3
    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.2011-12-09

1 Answers 1

10

You need this: $g_1H=g_2H\Leftrightarrow Hg_1^{-1}=Hg_2^{-1}$ Then your $f$ is a bijection.

  • 2
    That shows that your function is well-defined and 1-1. It is onto because $g\to g^{-1}$ is a bijection of $G$.2011-12-09