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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? – DonAntonio Sep 28 '12 at 3:31
@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 map – afedder Sep 28 '12 at 3:40
Then what you actually meant is to define $\,gH\to Hg^{-1}\,$ , right? – DonAntonio Sep 28 '12 at 3:42
@DonAntonio actually $(gH)^{-1} = Hg^{-1}$, so not really – afedder Sep 28 '12 at 3:43
@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 inversion – afedder Sep 28 '12 at 3:45


$$gH=xH\Longleftrightarrow x^{-1}g\in H\Longleftrightarrow Hx^{-1}=Hg^{-1}$$

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 $\,1-1\,$

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Assume $$(g_{1}H)^{-1}=(g_{2}H)^{-1}$$ then $$(g_{1}H)(g_{1}H)^{-1}=(g_{1}H)(g_{2}H)^{-1}\iff e=(g_{1}H)(g_{2}H)^{-1}$$ thus $$e(g_{2}H)=((g_{1}H)(g_{2}H)^{-1})(g_{2}H)\iff g_{2}H=g_{1}H$$

So to conclude: $$(g_{1}H)^{-1}=(g_{2}H)^{-1} \iff g_{2}H=g_{1}H$$

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that would be to show injectivity, not well-definedness – afedder Sep 28 '12 at 3:37
I guess, would that show well-definedness AND injectivity? – afedder Sep 28 '12 at 3:42
yes, it shows both – Belgi Sep 28 '12 at 3:53
what about surjectivity? would that just come from the fact that given any right coset $Hg$, $g^{-1}H \mapsto Hg$? – afedder Sep 28 '12 at 4:02
Indeed @afedder , surjectivity follows almost immediately. – DonAntonio Sep 28 '12 at 4:04

In general, for this sort of argument it helps to use the following fact: $$ g_1H= g_2 H\ \text{ iff }\ g_1^{-1}g_2\in H.$$

This is because both are equivalent to $H = g_1^{-1}g_2 H$. Similarly, we have $$Hg_1^{-1}= Hg_2^{-1}\ \text{ iff }\ g_1^{-1}g_2\in H,$$ because both are equivalent to $Hg_1^{-1}g_2 = H$. Put those together and you're done.

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