Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent:

(a) $g_1H = g_2H$,

(b) $Hg_1^{-1}=Hg_2^{-1}$,

(c) $g_1H \subset g_2H$,

(d) $g_2 \in g_1H$,

(e) $g_1^{-1}g_2 \in H$.

I'm beyond confused with this problem, all I know is that to prove the conditions are equivalent, I need to show that (a) implies (b), (b) implies (c), (c) implies (d), (d) implies (e), and (e) implies (a).

I have now gotten answers for (a) implying (e) and (e) implying (d). I'm overthinking all of this and am still confused about (d) implying (c) and (c) implying (b). When it comes to (b) implying (a), I thought I was getting somewhere but it doesn't seem to be working.

  • $\begingroup$ Have you started with anything? What have you tried to get from (a) to (b)? $\endgroup$ – amsmath Mar 28 '19 at 1:01

I'll get you started, maybe you're having trouble with the order of the clauses, but you should take the effort to do the rest yourself, as this is a basic question:

Assume $g_1H = g_2H$. Then $g_1^{-1}g_2H = g_1^{-1}g_1H = H$ so $g_1^{-1}g_2 \in H$. So (a) implies (e).

Assume (e). Then there is $h \in H$ such that $g_1^{-1}g_2 = h$. Thus $g_2 = g_1h\in g_1H$. So (e) implies (d).

  • $\begingroup$ it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order $\endgroup$ – Claire Mar 28 '19 at 1:10
  • $\begingroup$ @Claire I suspected so, as going from a to b is weird without e :) $\endgroup$ – Mariah Mar 28 '19 at 1:12
  • $\begingroup$ ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah $\endgroup$ – Claire Mar 28 '19 at 2:00
  • $\begingroup$ @Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^{-1}g_2 \in H$ then also $g_2^{-1}g_1 \in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance. $\endgroup$ – Mariah Mar 28 '19 at 2:06
  • $\begingroup$ okay yes i do see how we can assume that $\endgroup$ – Claire Mar 28 '19 at 2:11

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