Need help with this question:
Prove: Let $x = a_1, a_2...a_n$, and let $y$ be a product of the same factors, permuted cyclically. (That is, $y = a_k, a_{k+1}...a_na_1...a_{k-1}$. Then ${\rm ord}(x)={\rm ord}(y)$.
I guess I don't really understand the "permuted cyclically" thing, and I don't understand how I'm supposed to make this relate to e (since the best definition for "order" than I have is the "least positive integer" $a^n=e$.
Cyclic permutations arise by conjugation, e.g. $\,a_3 a_4\cdots a_n\,a_1a_2 = (a_1 a_2)^{-1}(a_1 a_2 \cdots a_n) (a_1 a_2)$
But generally conjugation $\ g\mapsto a^{-1}ga,\, $ is a group isomorphism, with inverse $\ g\mapsto aga^{-1}.$ Isomorphisms preserve all "group-theoretic" properties, which includes the order of an element $\,g,\,$ since this equals the order (cardinality) of the cyclic group generated by $\,g.\,$ But an isomorphic image of a group has the same order (cardinality).