conjugacy question abstract algebra

**henderson7878** · March 12th 2013, 03:03 PM

If x~_c y in a group G, show that x and y have infinite order or o(x)=o(y)

(Note: the relation x~_c y is called conjugacy, and as for any equivalence relation, the set G is partitioned by its equivalence classes)

**Nehushtan** · March 12th 2013, 03:31 PM

Hint: Show that $\left(gxg^{-1}\right)^n=gx^ng^{-1}$ for all $g\in G$ and $n\in\mathbb Z^+$ . It will follow that $\left(gxg^{-1}\right)^n=e\Leftrightarrow x^n=e$ .

**jakncoke** · March 12th 2013, 03:33 PM

This stuff becomes easy if you realize $(gxg^{-1})^{n} = gx^ng^{-1}$ for $(gxg^{-1})^{n} = gxg^{-1}gxg^{-1}gxg^{-1}....= gx^ng^{-1}$

If |x| = $\infty$ .
if |y| = k ( a finite number)
then $y^k = e$
so since x ~ y, there exists a $g \in G$ s.t $gxg^{-1} = y$ then $(gxg^{-1})^{k} = y^k = e$
so $gx^{k}g^{-1} = e$
or $gx^k = g$
or $x^k = e$ which cannot be since order of x is $\infty$ .

Thus is x has order $\infty$ so does y.

If |x| = n (finite)
then $(gxg^{-1})^{n} = y^n$ so $e = y^n$ , so either |y| = n, so |y| is less than n. if |y| = k < n $y^k = (gxg^{-1})^{k}$
so $e = gx^kg^{-1}$ or $x^k = e$ contradiction so n = k.

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