TITLE: If two elements belong to the same conjugacy class then they have the same order
QUESTION [8 upvotes]: How do we show that if two elements belong to the same conjugacy class then they have the same order? i.e. if G is a group and for $$ g,h\in G$$ $$\exists x\in G$$ s.t. $$h=xgx^{-1}$$
Thanks for any help

REPLY [12 votes]: The other answers are just fine! For fun here's another way to look at it.
Prove that if $f:G \to G'$ is an isomorphism, then $g$ and $f(g)$ have the same order. Now let $f:G\to G$ be defined by $f_x(g)=xgx^{-1}$. Check that $f_x$ is an isomorphism (with inverse $f_{x^{-1}}$). Hence $g$ and $xgx^{-1}$ have the same order.

REPLY [6 votes]: By Cayley's theorem, all group elements can be thought of as permutations.  The order of a permutation is a function of the lengths of its cycles (specifically, their least common multiple). Since conjugate elements have the same cycle structure, they must have the same order.