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
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
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.
Notice $(xgx^{-1})^n=xg^nx^{-1}$, and so $(xgx^{-1})^n=e$ if and only if $g^n=e$. To see this, $(xgx^{-1})^n=e$ implies $xg^nx^{-1}=e$, or $xg^n=x$, so $g^n=e$ by left-multiplying by $x^{-1}$. The other direction is clear.
It follows from this that the order of $g$ divides the order of $xgx^{-1}$ and vice versa, so the orders of $g$ and $xgx^{-1}$ must be equal.
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.
Show that
and