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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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    @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

3 Answers 3

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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 $\,1-1\,$

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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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    Indeed @afedder , surjectivity follows almost immediately.2012-09-28
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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.