# Centralizers and elements of odd order

I was going over a proof in which the following is given:

if abb=bba and b is of odd order, then ab=ba (i.e. if b^2 centralizes a then so does b)

I'm not sure why this is so. Any clarification would be appreciated.

Cheers,
W. =)

Dick
Homework Helper
b^2 commutes with a. If b has odd order then b^(2n+1)=e for some n. So b^(2n+1)a=a(b^(2n+1)). Do you see it now?

I believe so...
b^(2n)ba = ab^(2n)b
b^2n(ba) = b^2n(ab) since abb = bba
ba = ab

If that's it, then cheers! =)

Dick
Homework Helper
I believe so...
b^(2n)ba = ab^(2n)b
b^2n(ba) = b^2n(ab) since abb = bba
ba = ab

If that's it, then cheers! =)

Cheers!