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!

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- Oct 31st 2012, 05:38 PMjzelltLeft and Right Cosets
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! - Oct 31st 2012, 06:08 PMTheEmptySetRe: Left and Right Cosets
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.)