If $G$ is a group and $H \subset G$ is a subgroup, how would I show a map $\phi : G/H \longrightarrow$ $H\setminus G$ defined by $gH \mapsto (gH)^{-1}$ is well-defined?
I know we need to show that, for some $g_1,g_2 \in G$ such that $g_1H=g_2H$, we have $Hg_1^{-1}=Hg_2^{-1}$. I found that
$g_1H=g_2H \Longrightarrow H =g_1^{-1}g_2H \Longrightarrow Hg_1^{-1} = g_1^{-1}g_2Hg_1^{-1}$.
However, I am not sure how to show that $g_1^{-1}g_2Hg_1^{-1} = Hg_2^{-1}$.
Best Answer
Suppose
$$gH=xH\Longleftrightarrow x^{-1}g\in H\Longleftrightarrow Hx^{-1}=Hg^{-1}$$
and voila: going from left to right you get the map is well defined, and going from right to left you get the map is $\,1-1\,$