# $RS$ is a subgroup if $RS=SR$

I'm stuck trying to prove that the product of two subgroups $R$ and $S$ is a subgroup too when $RS=SR$. I see I need to show that $RS$ is closed under its product and inverses. The first one I've proved by assuming that $RS = SR$: $$r_1s_1\cdot r_2s_2 = r_1(s_1 r_2) s_2 = (r_1r_3) \cdot (s_3 s_2)$$, $s_1 r_2 = r_3 s_3$ since $RS=SR$.

The problem is to prove $(r_1 s_1)^{-1}\in RS$.

UPD:

The answer $(r_1 s_1)^{-1} = s_1^{-1} r_1^{-1}$ looks nice but it would be great if somebody would explain me why it is so.

UPD2:

The proof of the statement $(r_1 s_1)^{-1} = s_1^{-1} r_1^{-1}$ is here.

• That's because $(r_1s_1)^{-1} = s_1^{-1}r_1^{-1} \in SR=RS$. Apr 17, 2016 at 15:15
• What do you mean by "why it is so"? That's just how it works out algebraically. Apr 17, 2016 at 15:34
• @DustanLevenstein I mean to give me a reference to an axiom or a property. It doesn't look so obvious for me as for the mathematicians who answered the question. Apr 17, 2016 at 15:41
• math.stackexchange.com/questions/1119626/… Apr 17, 2016 at 15:59
• @flipback For future reference, "why it is so" is an extremely vague question to ask. You should try to identify which part is confusing you. You could very easily have had a confusion related to misunderstanding the meaning of $RS=SR$, which would have required a completely different answer. Apr 17, 2016 at 16:12

You have $(rs)^{-1}=s^{-1}r^{-1}\in SR=RS$.