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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 $g_1^{-1}g_2Hg_1^{-1} = Hg_2^{-1}$.

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    What are $\,G/H\,\,,\,H/G\,$? I'd say these two are one and the same set: the set of left cosets of H in G, but then why would you use two different notations?2012-09-28
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    @DonAntonio $G/H$ is for the left cosets of $H$ in $G$, while $H\G$ is for the right cosets of $H$ in $G$ ...that's also clear from my work following the definition of the map2012-09-28
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    Then what you actually meant is to define $\,gH\to Hg^{-1}\,$ , right?2012-09-28
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    @DonAntonio actually $(gH)^{-1} = Hg^{-1}$, so not really2012-09-28
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    @DonAntonio to be precise, $(gH)^{-1} = H^{-1}g^{-1} = Hg^{-1}$ where the last equality came from the fact that subgroups are closed under inversion2012-09-28

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