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?
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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 byQuote:
Originally Posted by Godisgood
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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.