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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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2  
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
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Yes, that's the idea. –  Thomas Andrews Mar 17 '12 at 15:49

4 Answers 4

up vote 3 down vote accepted

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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