1. ## Group Theory

Suppose $\displaystyle a$ and $\displaystyle b$ belong to a group, $\displaystyle a$ has odd order, and $\displaystyle aba^{-1} = b^{-1}$. Show that $\displaystyle b^2 = e$.

I have a feeling I'm missing something obvious, but I can't for the life of me figure out what $\displaystyle a$ having odd order has to do with this.

2. Originally Posted by spoon737
Suppose $\displaystyle a$ and $\displaystyle b$ belong to a group, $\displaystyle a$ has odd order, and $\displaystyle aba^{-1} = b^{-1}$. Show that $\displaystyle b^2 = e$.

I have a feeling I'm missing something obvious, but I can't for the life of me figure out what $\displaystyle a$ having odd order has to do with this.
1)$\displaystyle aba^{-1} = b^{-1}$.

2)$\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^{-1} a^{-1} = (aba^{-1})^{-1} = (b^{-1})^{-1}=b$.

3)$\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = aba^{-1} = b^{-1}$.

4)$\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^{-1}a^{-1} = (aba^{-1})^{-1} = b$.

The pattern is obvious: $\displaystyle a^{2n}ba^{-2n} = b$ and $\displaystyle a^{2n+1}ba^{-(2n+1)} = b^{-1}$.

Let $\displaystyle k$ be order of $\displaystyle a$ then $\displaystyle b = a^k b a^{-k} = b^{-1} \implies b^2 = e$.