Let H be a subgroup of G. Let $g_1,g_2\in G$. Prove that the following conditions are equivalent:
i.$g_1H=g_2H$
ii.$Hg_1^{-1}=Hg_2^{-1}$
iii.$g_2\in g_1H$
iv.$g_1^{-1}g_2\in H$
This was a question given in a book. However, I am not been able to get how can we say that i. implies ii.. If $g_1H=g_2H$, then we can say for, $h_1,h_2\in H$, we have $g_1h_1=g_2h_2$ and hence, $g_2=g_1h_1h_2^{-1}$. Now, $h=h_1h_2^{-1}\in H$ and thus, $g_2=g_1h$ . So, we can say $g_2\in g_1H$. Now, if $g_2=g_1h$, then $g_1^{-1}g_2=h\in H$. So, we can say that, $g_1^{-1}g_2=\in H$. Thus , we see that i.implies iii. and iii. implies iv.. But how to prove i. implies ii. ? Also then if we prove i. implies ii. , then can we say ii. imples iii. since i. imples ii. ? How to show that ii. implies iii. ? If we can show that i. implies ii. and ii. implies iii. , I think we can say that all the following conditions are equivalent, right? I think we dont need to show that ii. implies i. (which may not be the case also) as the question asks us to show that i,ii,iii,iv are equivalent .But how to prove it ? I am not quite getting it.
Best Answer
$$g_{1}H=g_{2}H \implies g_{1}=g_{1}e=g_{2}h \implies g_{1}^{-1}=h^{-1}g_{2}^{-1} $$ So $$\forall k\in H, kg_{1}^{-1}=kh^{-1}g_{2}^{-1}\implies Hg_{1}^{-1} \subset Hg_{2}^{-1}$$ In the same way u can prove the reverse inclusion.