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Two conjugate elements of a group have the same order PROOF

  • #1
RJLiberator
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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.
 

Answers and Replies

  • #2
andrewkirk
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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'.
 
  • #3
RJLiberator
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So, when the question states "conjugate elements" we can think of the definition as

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##.
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.
 

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