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**Swlabr** Assume $\displaystyle b$ has even order, then find a contradiction (this is relatively easy, just re-arrange the formula you are given).

Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

You should get that the order of $\displaystyle b$ can be one of two things.

I should point out that I didn't use the fact that $\displaystyle a^5=1$, but I can't see a problem in my proof...