# conjugacy question abstract algebra

• Mar 12th 2013, 03:03 PM
henderson7878
conjugacy 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 PM
Nehushtan
Re: 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 PM
jakncoke
Re: 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.