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

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- Sep 8th 2007, 10:04 AMtttcomraderElement with odd order
Suppose a and b are in a group, a has odd order, and $\displaystyle aba^{-1} = b^{-1} $. $\displaystyle Show that b^{2} = e$.

- Sep 9th 2007, 10:20 PMOpalg
In the relation $\displaystyle aba^{-1} = b^{-1} $, take the inverse of both sides, to get $\displaystyle ab^{-1}a^{-1} = b $. It follows that $\displaystyle a^2ba^{-2} = ab^{-1}a^{-1} = b $. By the same calculation, $\displaystyle a^3ba^{-3} = b^{-1}$, and so on. In fact, $\displaystyle a^kba^{-k}$ is equal to $\displaystyle b^{-1}$ if

*k*is odd, and $\displaystyle b$ if*k*is even.

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