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Let $(G, ·)$ be a group, and let $a, x$ ∈ $G$. Show that the orders of $x$ and $axa^{−1}$ are equal.

My attempt:

In order to show that the orders of $x$ and $axa^{−1}$ are equal, I have to show that $axa^{−1}$ = $x$.

Since $G$ is a group, we have that $axa^{−1}$ = $a.(x.a^{-1}) = (a.x).a^{-1}$ (associativity) = $a^{-1}.(a.x)$ (commutativity) = $(a^{-1}.a).x$ (associativity) = $e.x = x.e = x$ (property of the identity element).

Hence, I've shown that $axa^{−1}$ = $x$, and so their orders are equal.

Is my proof correct? Did I make a mistake somewhere? Do you have other suggestions?

JOJO
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    The hypotheses do not say $G$ is commutative. – Bernard Sep 12 '19 at 09:12
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    I don' think you can assume commutativity. In general $axa^{-1} \neq x$ but the orders are the same. – Kavi Rama Murthy Sep 12 '19 at 09:12
  • $axa^{−1}$ = $a.(x.a^{-1}) = (a.x).a^{-1}$ (associativity) = $(a.a^{-1}).x$ (associativity) = $e.x = x.e = x$ (property of the identity element). Can I do it this way? without bringing in the commutativity part? – JOJO Sep 12 '19 at 09:17
  • See e.g. here: https://math.stackexchange.com/a/3829008/810157 –  Oct 04 '20 at 20:27

4 Answers4

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The proof is not correct, since conjgate elements need not be equal. However, conjugate elements always have the same order, i.e., $\operatorname{ord}(x)=\operatorname{ord}(axa^{-1})$. This is because of $$ (axa^{-1})^n=ax^na^{-1}. $$ This has been proved in several duplicates here, e.g.

Proving order of certain elements in a group is the same

Dietrich Burde
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Well, note that "$axa^{-1} = x$ iff $ax=xa$" does not always hold.

Let $n$ be the order of $axa^{-1}$. Then $e = (axa^{-1})^n = ax^na^{-1}$ and so $x^n = a^{-1}a=e$. Thus the order of $x$ is a divisor of $n$.

Conversely, if $n$ is the order of $x$, then $(axa^{-1})^n =a x^na^{-1} = aea^{-1}=e$ and so the order of $ax^{-1}$ is a divisor of $n$. Equality follows.

amWhy
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Wuestenfux
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  • In the last sentence, you mean: "and so the order of $axa^{-1}$ is a divisor of $n$?" – JOJO Sep 12 '19 at 09:49
  • Indeed, if for a group element $g$, we have $g^n=e$, then the order of $g$ divides $n$ or $n$ is a multiple of the order of $g$. – Wuestenfux Sep 12 '19 at 09:51
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Hint: use induction to show that $(axa^{-1})^{n} =ax^{n}a^{-1}$ and then use the definition of order to finish the proof.

Kavi Rama Murthy
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In order to show that the orders of $x$ and $axa^{−1}$ are equal, I have to show that $axa^{−1}$ = $x$.

This is false. You do not have to show that two elements are equal. All you have to show is that the two elements have equal order.

Since $G$ is a group, we have that $axa^{−1}$ = $a.(x.a^{-1}) = (a.x).a^{-1}$ (associativity) = $a^{-1}.(a.x)$ (commutativity) = $(a^{-1}.a).x$ (associativity) = $e.x = x.e = x$ (property of the identity element).

False again. You cannot assume that a group is commutative. There are plenty of non-commutative groups out there, and for those groups, while $axa^{-1}=x$ does not always hold, it is still always true that $x$ and $axa^{-1}$ have the same order.

I suggest you retry your proof and first prove the following statement:

If $x^n=e$, then $(axa^{-1})^n = e$.

5xum
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