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)

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- Mar 12th 2013, 03:03 PMhenderson7878conjugacy question abstract algebra
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) - Mar 12th 2013, 03:31 PMNehushtanRe: Conjugacy question in abstract algebra
Hint: Show that $\displaystyle \left(gxg^{-1}\right)^n=gx^ng^{-1}$ for all $\displaystyle g\in G$ and $\displaystyle n\in\mathbb Z^+$. It will follow that $\displaystyle \left(gxg^{-1}\right)^n=e\Leftrightarrow x^n=e$.

- Mar 12th 2013, 03:33 PMjakncokeRe: conjugacy question abstract algebra
This stuff becomes easy if you realize $\displaystyle (gxg^{-1})^{n} = gx^ng^{-1} $ for $\displaystyle (gxg^{-1})^{n} = gxg^{-1}gxg^{-1}gxg^{-1}....= gx^ng^{-1}$

If |x| = $\displaystyle \infty $.

if |y| = k ( a finite number)

then $\displaystyle y^k = e $

so since x ~ y, there exists a $\displaystyle g \in G $ s.t $\displaystyle gxg^{-1} = y $ then $\displaystyle (gxg^{-1})^{k} = y^k = e $

so $\displaystyle gx^{k}g^{-1} = e $

or $\displaystyle gx^k = g $

or $\displaystyle x^k = e $ which cannot be since order of x is $\displaystyle \infty $.

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

If |x| = n (finite)

then $\displaystyle (gxg^{-1})^{n} = y^n $ so $\displaystyle e = y^n $, so either |y| = n, so |y| is less than n. if |y| = k < n $\displaystyle y^k = (gxg^{-1})^{k} $

so $\displaystyle e = gx^kg^{-1}$ or $\displaystyle x^k = e $ contradiction so n = k.