Group Theory: Prove o(b) even if ab = b^-1a

[Dromepalin]

Group Theory: Prove o(b)|2 if ab = b^-1a

Homework Statement

Suppose G is a group and a, b $$\in$$ G
a) If o(a) is odd and a*b = b^−1*a, prove that o(b)|2.
b) If o(a) is even and a*b = b^−1*a, does it follow that o(b)|2? Prove your answer.

Homework Equations

n/a

The Attempt at a Solution

a) since ab = b^-1a, bab = a
(bab)^o(a) = a^o(a)
so, b^o(a)a^o(a)b^o(a) = e
[[I now realize I can't do this since G is not necessarily abelian, so not commutative..but ploughing on...]]
b^2o(a) = e, thus o(b) is even

I don't know how to approach this one. Been stuck on it for a frustrating amount of time now.

 

[Citan Uzuki]

What is aba^-1? a²ba^-2? a³ba^-3? Can you find a general pattern for aⁿbaⁿ?

Incidentally, you aren't trying to prove that o(b) is even (indeed, this is impossible, as b=e satisfies the identity ab=b^-1a). You are trying to prove that o(b) divides 2, i.e. that b² = e

 

[Hurkyl]

When I see relations like

a*b = b^−1*a

I like to use them to normalize products. Given any product of a's and b's (and their inverses), you can use this relation to move all the a's to the left and the b's to the right.

So, I am inclined to try and write down products of a's and b's that might be interesting, and see what they normalize to.

 

[Dromepalin]

Thank you very much Citan Uzuki and Hurkyl :D
Here's how I did it:
ab = b^-1a
a = bab And a = b^-1ab^-1. Let me be the odd integer such that o(a) = m
Then: a^m = (bab)(b^-1ab^-1)(bab)...(bab)(b^-1ab^-1)(bab) = ba^mb
Thus b^2 = 1 and o(b)|2