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Note that $H$ is a subgroup of a group $G$ and $L_{H}$ is the left cosets of $H$ in $G$ and $R_{H}$ is the right cosets of $H$ in $G$.

How do I show that this map is well-defined or not?

What I tried:

If $g_{1}H=g_{2}H$, then $\phi(g_{1}H)=Hg_{1}=Hg_{2}=\phi (g_{2}H)$, but I don't think this is true in general.

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    $\begingroup$ Try a non-normal subgroup. $\endgroup$ Feb 26 '18 at 15:53
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    $\begingroup$ You have to prove the step $g_1H=g_2H\Rightarrow \phi(g_1H)=\phi(g_2H)$ $\endgroup$ Feb 26 '18 at 15:55
  • $\begingroup$ @Fakemistake Sorry, I made a typo in my question. I meant if $g_{1}H=g_{2}H$... $\endgroup$
    – user482939
    Feb 26 '18 at 15:57
  • $\begingroup$ You have to justify the $Hg_1=Hg_2$ step. $\endgroup$
    – Shaun
    Feb 26 '18 at 16:03
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    $\begingroup$ I don't think you can. Like @LordSharktheUnknown said, consider when $H$ is not normal. $\endgroup$
    – Shaun
    Feb 26 '18 at 16:08
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Your question is equivalent to the statement that "$gH=hH\Rightarrow Hg=Hh$". This does not hold in general. To see this, recall that $gH=hH$ if and only if $g^{-1}h\in H$, while $Hg=Hh$ if and only if $gh^{-1}\in H$. Then it is not necessarily the case that "$gh^{-1}\in H\Leftrightarrow g^{-1}h\in H$".

For example, take $H=\langle (12)\rangle\leq S_3$. Then $(123)H=(23)H$ while $H(123)\neq H(23)$. (Instead, $H(23)=H(132)$.)

The map you actually want is $gH\mapsto Hg^{-1}$. Then this map is well defined.

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  • $\begingroup$ Thank you. I was trying to prove that this relation is not well-defined. I guess a counterexample such as yours is sufficient to show it is not a map? $\endgroup$
    – user482939
    Feb 26 '18 at 16:29
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    $\begingroup$ If you want to show that something does not hold then you need a counter-example. My counter-example shows that your map is not well-defined, as the coset $(123)H$ is mapped to two different places when viewed as $(123)H$ and as $(12)H$. $\endgroup$
    – user1729
    Feb 26 '18 at 16:36

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