# If two elements belong to the same conjugacy class then they have the same order

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

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Hint: $h^k$ is a conjugate of $g^k$ –  Thomas Andrews Mar 17 '12 at 15:32
@ThomasAndrews $h=xgx^{-1}$ and assume that $o(h)=n$ (and assume wlog $o(h)\leq o(g)$) then $h^n=xg^nx^{-1}$ and so $xg^nx^{-1}=e\rightarrow xg^n=x$ and so $o(g)=n$ –  hmmmm Mar 17 '12 at 15:42
Yes, that's the idea. –  Thomas Andrews Mar 17 '12 at 15:49

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.

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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.

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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.

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Show that

• There is some $x\in G$ such that $h^k = x g^k x^{-1}$, for all $k\geq 1$.

and

• Let $e$ be the identity element in $G$. Then, $h^n=e$ if and only if $g^n=e$.
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