How can I show that the elements A * B and B * A have the same order? where A, B belong to a finite group G
How can I prove that 2 elements have the same order? I was thinking of showing that if (A * B)^n=id; and (B * A)^m=id then n=m; Any ideas or tips how I can do that? Thank you!
Hint:
$(ab)^n = ababab\cdots ab = abab\cdots abaa^{-1} = a(baba\cdots ba)a^{-1} = a((ba)^n)a^{-1}$
Hint: Suppose $(AB)^n=e$ where $e$ is the identity. Then $A^{-1}(AB)^n=A^{-1}$ by multiplying by $A^{-1}$ on the left. Then $A^{-1}(AB)^nA=e$ by multiplying by $A$ on the right. However, $A^{-1}(AB)^nA=(BA)^n$, so the order of $BA$ is less than or equal to the order of $AB$.
Hint : show that in a group $G$ if $g,h\in G$ then $g$ and $hgh^{-1}$ have same order (the order is an invariant by conjugation). Then find $h\in G$ such that $hABh^{-1}=BA$.
Suppose $(ab)^n=1$. Then $$a(ba)^n=ab\cdots aba = (ab)^na=1\cdot a=a$$ and this can happen if and only if $(ba)^n=1$.
