0

I want to prove that in a group $G$, the order of an element $g$ of $G$ equals the order of its conjugate $xgx^{-1}$, for every element $x$of $G$. In what follows, $e$ will denote the unit of $G$.

So suppose that $|g|=n$, for some positive integer $n$. Then \begin{align*} (xgx^{-1})^n&=xg^nx^{-1}\nonumber\\ &=xex^{-1}\nonumber\\ &=xx^{-1}\nonumber\\ &=e\nonumber \end{align*} This proves that $|xgx^{-1}|\leq |g|$.

Now suppose that $g^n=y$, for some $y\neq e$. Then $$(xgx^{-1})^n=xg^nx^{-1}=xyx^{-1}$$ So if $xy=e$ then $$xg^nx^{-1}=x^{-1}$$ which is different form $e$, otherwise $xgx^{-1}$ would be just $g$. Similarly, if $yx^{-1}=e$ then $$(xgx^{-1})^n=x$$ which is not $e$.

So it seems I have proved that $|xgx^{-1}|\geq |g|$ in the two cases I have considered, namely $x=y$ and $x=y^{-1}$, but I am not convinced they are all possible cases to consider. Can you please help to complete the proof?

Note : I don't want to use the fact that conjugation is an automorphism of $G$ and that automorphisms obviously preserve orders, just trivial facts coming from the axioms of a group, thanks!

Maryam
  • 1,889
  • 5
    Just observe that if $g'=xgx^{-1}$, then $g=x^{-1}g'x$, so $g$ is a conjugate of $g'$, and you can apply your first result to conclude that $|g|\le|g'|$ and hence that $|g'|=|g|$. – Brian M. Scott Jun 01 '20 at 18:12
  • 1
    Suppose that the order of $xgx^{-1}$ were smaller than $n$ and do the same computation to obtain a contradiction. – Dietrich Burde Jun 01 '20 at 18:13
  • 1
    Just a piece of advice: $x^n=e$ if and only if $o(x)|n$. This is much more useful than $x^n=e$ implies $o(x)\leqslant n$. It makes no odds in this question, but I think you'll find it's the right way to think about orders. – ancient mathematician Jun 01 '20 at 18:43

1 Answers1

4

You have already shown that for any $g, x\in G$, we have $\lvert xgx^{-1} \rvert \leq \lvert g\rvert$.

To complete the argument, we have only to observe that $g = x^{-1}(xgx^{-1})x$. Applying the fact in the first line of this answer, we have $$ \lvert x^{-1}(xgx^{-1})x\rvert \leq \lvert xgx^{-1}\rvert $$ which means that $\lvert g \rvert \leq \lvert xgx^{-1}\rvert$. So we are done.

Sebastian Monnet
  • 3,871