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
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...