$aH=Ha$ implies $a$ commutes with every element in H

Question

I was trying to prove this question:

Let $G$ be a group and $H \le G$. Let $a,b \in G$. Now, if $aH=Ha$ and $bH=Hb$, then show that $(ab)H=H(ab)$.

It appears to me that if I can show that $a,b$ commutes with every element of $H$, then it is quite easy to show $(ab)H=H(ba)$.

However, how do I prove that $aH=Ha$ implies $a$ commutes with every element in $H$? Suppose $x\in H$ be arbitrary. Then $ax \in aH$ and $xa \in Ha$. Also by their equality, it follows that $xa \in aH$ and $ax \in Ha$. But does it necessarily mean $ax=xa$ in any of the cosets?

Answer 1

No, it doesn't mean that. Given any $h_1\in H $, there exists $h_2\in H $ such that $bh_1=h_2b $. There exists $h_3\in H $ such such that $ah_2=h_3a $. Then $$abh_1=ah_2b=h_3ab. $$ As you can also do this backwards, you get $abH=Hab $.

Answer 2

We don't have $a$ commutes with every elements of $H$. But, we have the following statement:

If $A,B,C$ are subsets of a group $G$, then $(AB)C=A(BC)$. Here, $AB = \{ ab \in G| a \in A, b \in B \}$.(This fact is easy to prove by proving they have same elements form $abc$)

So,we have $(ab)H = a(bH)$(as above) $= a(Hb) = (aH)b = (Ha)b = H(ab)$ (use above statement twice.)