# Two conjugate elements of a group have the same order PROOF

Gold Member

## Homework Statement

Let x and y be conjugate elements of a Group G. Prove that x^n = e if and only if y^n = e, hence x and y have the same order.

## Homework Equations

Conjugate elements : http://mathworld.wolfram.com/ConjugateElement.html

## The Attempt at a Solution

Since y is a conjugate of x, there exists z ∈G such that y = (zxz^-1).

If x^n = e, then y^n = (zxz^-1)^n = (zx^nz^-1) = (zez^-1) = zz^-1 = e.

Similiarly, if y^n = e, then e=y^n = (zxz^-1)^n = (zx^nz^-1). Multiplying on the left by z^-1 and on the right by z we see
z^-1ez = z^-1zx^nz^-1z and so e = ex^ne = x^n

Done.

My concern is in the definition of conjugate elements. If x and y are conjugate elements of a group G, does that necessarily mean y is a conjugate of x?

I've supplied the definition of conjugate elements in the 'relevant equations' part of this thread.

andrewkirk
Homework Helper
Gold Member
I don't like that definition because it never clearly states the necessary and sufficient conditions for being conjugate. It also blurs the distinction between the - at first - very different statements 'a and b are conjugate' and 'a is conjugate to b'. Use this definition instead:

If a and b are elements of group G then we say that a is conjugate to b iff there exists ##g\in G## such that ##gag^{-1}=b##.

It is then simple to prove that a is conjugate to b iff b is conjugate to a.

Hence, we can refer to two elements as being conjugate, without specifying an order, ie the statement 'a and b are conjugate in G' is well-defined, and means 'a is conjugate to b', which is the same as 'b is conjugate to a'.

RJLiberator
Gold Member
So, when the question states "conjugate elements" we can think of the definition as

If a and b are elements of group G then we say that a is conjugate to b iff there exists ##g\in G## such that ##gag^{-1}=b##.

The way the question is worded, and by this definition, then I can safely assume that there exists z ∈G such that y = (zxz^-1) and my proof follows.