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Prove that if $H$ is a subgroup of a group $G$ and $g_1,g_2 \in G$ then if $Hg_1^{-1}=Hg_2^{-1}$ then $g_1 H \subseteq g_2 H$. My attempt:

Obviously $g_2^{-1} \in Hg_2^{-1}=Hg_1^{-1}$ so $g_2^{-1} = hg_1^{-1}$ for some $h\in H$. Then $g_2^{-1}g_1=h\Rightarrow g_1 = g_2 h \Rightarrow g_1 h' = g_2 h h' $ for some $h' \in H$. As $hh' \in H$ this means that $g_1 H \subseteq g_2 H$. Is there any flaw in this?

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    $\begingroup$ Looks ok to me. $\endgroup$ – marwalix Feb 19 '18 at 3:57
  • $\begingroup$ More simply. $(g_i H)^{-1} = H g_i^{-1}$. $\endgroup$ – anomaly Feb 19 '18 at 5:21
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you need multiply g1 from right: $$ Hg_{1}^{-1}*g_{1} = Hg_{2}^{-1}* g_{1} $$ $$H=Hg_{2}^-{1}*g_{1}$$ so the group is normal and the property hold

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  • $\begingroup$ I haven't learned what a normal group is yet. $\endgroup$ – darmendarizp Feb 19 '18 at 4:02
  • $\begingroup$ $H$ need not be normal. This does not help at all. $\endgroup$ – ancientmathematician Feb 19 '18 at 9:17

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