Let G be a finite group that has the properties that it contains 2 elements a and b such that:
1. $\displaystyle a^5=e$
2. $\displaystyle aba^- ^1 =b^2$
3. a ≠ e
4. b ≠ e
Find O(b).
Ok, I've tried applying the process for others such as $\displaystyle b^1^0$, or really any n where $\displaystyle b^n$ can be broken down into factors of $\displaystyle b^2$ and I think I'm just not seeing it. When attempting to find something such as $\displaystyle b^5$, if I broke that down into $\displaystyle b$ and $\displaystyle b^4$, then the order would matter since the group is not necessarily abelian, right? I may just be missing something or it's not catching my eye but I don't see right now how I'm going to get $\displaystyle b^n=e$, for some n. Any additional "hints" or points in the right direction are appreciated
What I meant was that you do this: $\displaystyle a^2ba^{-2}=b^4 \Rightarrow a^2b^2a^{-2}=b^8$ $\displaystyle \Rightarrow a^2(aba^{-1})a^{-2} = a^3ba^{-3} = b^8$.
Can you continue in this way until you reach $\displaystyle a^5ba^{-5}$ ?
You can also prove the more general theorem: if $\displaystyle aba^{-1} = b^i$ then $\displaystyle a^mba^{-m} = b^{(i^m)}$
I think I get it in that if $\displaystyle a^{-5}=e$, then we have $\displaystyle b=b^{32}$, which gives us $\displaystyle e=b^{31}$. Thus, the order of b must be a divisor of 31, meaning it is either 1 or 31. However it can't be 1 since b is a non-identity element by the given, thus $\displaystyle o(b)=31$.
However, can you please explain why $\displaystyle a^5=e$ gives $\displaystyle a^{-5}=e$? Thanks.