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Let $H$ be a subgroup of a group G. Show that if $aH=bH$ then $Ha^{-1}=Hb^{-1}$. Prove that the map $\alpha$ defined by $\alpha(gH)=Hg^{-1}$ is a bijection.

For the first part of the proof:

Since $H$ is a group itself, $e \in H$.

Thus $\exists h_e \in H$ such that $a=bh_e$, hence $b^{-1}=h_ea^{-1}$.

Therefore $\forall h_1 \in H, h_1b^{-1}=h_1h_ea^{-1}=h_2a^{-1}$ and hence $Hb^{-1}=Ha^{-1}$

I think to prove $\alpha$ is a bijection, I need to prove that it is both surjective and injective. But since I am self-studying maths, I feel uncertain of how to do it rigorously, could someone provide a complete solution for this?

Rescy_
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  • Do you know the general scheme for proving that a function $f:X\to Y$ is injective? and surjective? – Lee Mosher Nov 01 '15 at 02:40
  • @LeeMosher How about the proof in the answer? – Rescy_ Nov 01 '15 at 02:54
  • Re: I think to prove $\alpha$ is a bijection, I need to prove that it is both surjective and injective. You can prove instead that it has an inverse. (Assuming that had a proof of the equivalence "$f$ is bijective $\Leftrightarrow$ $f^{-1}$ exists". See here and in other related questions. – Martin Sleziak Nov 01 '15 at 05:57
  • BTW I do not like the way the statement is phrased. The statement: "$f$ is a bijection" is unclear without saying what are the sets $X$ and $Y$ such that $f\colon X\to Y$. That's why I would prefer the formulation: "Show that the map $\alpha$ defined by $\alpha(gH)=Hg^{-1}$ is a bijection between the sets of the left cosets and the sets of the right cosets. Or some similar formulation which clearly states the domain and the codomain of $\alpha$. – Martin Sleziak Nov 01 '15 at 06:02

1 Answers1

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For $\alpha: GH \mapsto HG$

  1. Prove injectiveness

If $Ha^{-1} = Hb^{-1}$ then from the first part (actually we need to show if and only if) we see that $aH=bH$.

  1. Prove surjectiveness

$\forall g^{-1} \in G, \alpha (gH)=Hg^{-1} \ \ \ \square$

Rescy_
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