# Thread: Element with odd order

1. ## Element with odd order

Suppose a and b are in a group, a has odd order, and $aba^{-1} = b^{-1}$. $Show that b^{2} = e$.

2. In the relation $aba^{-1} = b^{-1}$, take the inverse of both sides, to get $ab^{-1}a^{-1} = b$. It follows that $a^2ba^{-2} = ab^{-1}a^{-1} = b$. By the same calculation, $a^3ba^{-3} = b^{-1}$, and so on. In fact, $a^kba^{-k}$ is equal to $b^{-1}$ if k is odd, and $b$ if k is even.

Now take k to be the order of a, which is odd. Then $a^k=e$ (the identity element of the group), and $a^kba^{-k} = ebe = b$. But since k is odd, this is also equal to $b^{-1}$. So $b=b^{-1}$ and therefore $b^2=e$.