Prove that if $Hg_1^{-1}=Hg_2^{-1}$ then $g_1 H \subseteq g_2 H$ 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?
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
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