Let G be a group and H a subgroup of G. Let a, b in G. Prove the following:
If aH = Ha and bH = Hb, then (ab)H = H(ab).
Can someone show this please? Thanks a lot in advance!
So we want to show that $\displaystyle (ab)H \subset H(ab)$
let $\displaystyle k \in (ab)H$
Then $\displaystyle k=(ab)h$ for some $\displaystyle h \in H$
By the relation $\displaystyle bH=Hb$ we know that there exists an $\displaystyle h_1 \in H$ such that $\displaystyle bh=h_1b$
and we know that by $\displaystyle aH=Ha$ there exists an $\displaystyle h_2 \in H$ such that $\displaystyle ah_1=h_2a$
Putting all of these together we get
$\displaystyle k=(ab)h=a(bh)=a(h_1b)=(ah_1)b=h_2(ab) \in H(ab)$
Now you just need to show or justify the other direction (it is exactly the same.)