Recent content by Dromepalin

matt grime's method is elegant! I have never seen it done that way...

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...