Cosets

• September 22nd 2009, 01:23 PM
Godisgood
Cosets
If every right coset of H in G is a left coset of H in G prove that
aHa^-1 = H for all a elements of G?
• September 22nd 2009, 02:23 PM
HallsofIvy
Quote:

Originally Posted by Godisgood
If every right coset of H in G is a left coset of H in G prove that
aHa^-1 = H for all a elements of G?

For every a in G, and every h in H, ah is in aH. If "every right coset of H in G is a left coset of H in G" then aH= Ha so there exist k in H such that ah= ka. Multiply on the left by $a^{-1}$.
• September 22nd 2009, 02:40 PM
Taluivren
Hi,

"every right coset of H in G is a left coset of H in G" means that there exists some b in G such that aH=Hb. so aHb^-1 = H and because the identity element e belongs to H we have ab^-1 in H. Since H is a group, we also have (ab^-1)^-1 = ba^-1 in H. In other words, Hba^-1 = H.
Now we have aHa^-1 = Hba^-1 = H.