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So, I have a question that says for any group elements $a$ and $x$, prove that $|xax^{-1}| =|a|$, where $|\cdot|$ is the order of an element.

I know that $xx^{-1} = e$ and $ae=a$, but I'm not sure how to get the $x$'s on the same side.

Currently, for my proof, which is not much, I have:

Let $n = |xax^{-1}|$, then $(xax^{-1})^n = e$.

Honestly, I have no idea where to go from here, or if what I've got is even right. I would really appreciate just a point in the right direction.

Alexia Paskevicius
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3 Answers3

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Hint: Generally, when you want to prove that $o(a)=o(b)$ (assuming they're finite) in a group $G$, you show that

$$b^{o(a)} = a^{o(b)} = e$$

If you prove this, then it implies that $o(a) \mid o(b)$ and $o(b) \mid o(a)$. Since the order of an element is positive, that implies $o(a)=o(b)$.

Edit: As you figured, $(xax^{-1})(xax^{-1})=xa^2x^{-1}$. What happens exactly is this

$$(xax^{-1})(xax^{-1})=xa(x^{-1}x)ax^{-1}=xa(e)ax^{-1}=xa^2x^{-1}$$

If we calculate $(xax^{-1})(xax^{-1})(xax^{-1})$, we see the same pattern. Each time we multiply $xax^{-1}$ by itself, the term $x\cdot x^{-1}$ appears which is canceled. So, you can conclude that $(xax^{-1})^n=xa^nx^{-1}$.

Now, if you plug in $o(a)$ instead of $n$, you get $(xax^{-1})^{o(a)}=xa^{o(a)}x^{-1}=x(e)x^{-1}=e$. Hence, $o(xax^{-1}) \mid o(a)$. Conversely, $$xa^{o(xax^{-1})}x^{-1}=(xax^{-1})^{o(xax^{-1})}=e$$ $$xa^{o(xax^{-1})}x^{-1}=e \implies a^{o(xax^{-1})}=x^{-1}x=e$$

This shows that $o(a) \mid o(xax^{-1})$. Since they're both positive, $o(a) = o(xax^{-1})$

Can you handle the infinite case now?

stressed out
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  • This assumes Lagrange, which the OP may or may not have, but it can also be proven directly without. – Randall Sep 10 '18 at 02:04
  • @Randall Hmm. Why does it assume Lagrange's theorem? I think $a^n=e \implies o(a) \mid n$ can be proven using division algorithm which is really elementary. Divide $n$ by $o(a)$. Show that if $r$ is the remainder, then since $r < o(a)$, it must be $0$ or we'll have a contradiction. – stressed out Sep 10 '18 at 02:06
  • OK, true. Memory lapse. I'm 43. – Randall Sep 10 '18 at 02:06
  • Sorry, I'm a little slow. I've calculated (xax^-1)(xax^-1) and I got a^2. I'm still not sure how I would go about manipulating (xax^-1)^o(a) into a form where I can get the identity. – Alexia Paskevicius Sep 10 '18 at 02:14
  • @AlexiaPaskevicius see the edit. Is it clear now? – stressed out Sep 10 '18 at 02:25
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    Yes! Thank you. You explained it very nicely! Hopefully I'll start to see things this way as I learn, since this was really hard for me to get. The only thing I don't understand is after you say Conversely. Why are we plugging o(xax^-1) into the same equation that we plugged o(a) into? – Alexia Paskevicius Sep 10 '18 at 02:38
  • @AlexiaPaskevicius You're welcome. It takes time to get used to the concepts you learn in abstract algebra but you'll definitely learn them by practicing enough. Why did I consider $xa^{o(xax^{-1})}x^{-1}$? Well, because I had to prove that $a^{o(xax^{-1})}=e$. It turned out that everything worked nicely as we expected. One could say that It was just a calculation trick, which is more or less obvious if you play with the problem for a while. The question is, do you understand the converse part of the proof and why is it necessary? – stressed out Sep 10 '18 at 02:44
  • Yes, I've finally got it. I thank you so much for your patience! We haven't talked much about infinite order in class, but would I start the infinite part of the proof by just assuming that |a| = infinity? – Alexia Paskevicius Sep 10 '18 at 03:02
  • @AlexiaPaskevicius You're welcome. Usually when they ask you to prove that the orders of two elements are the same, they want you to prove it assuming that it's finite. But if you want to handle the infinite case as well, you can't just plug in $|a|=\infty$. You need to assume that $o(a)$ is not finite and then conclude that $o(xax^{-1})$ can't be finite either. See user587192's answer, for example. – stressed out Sep 10 '18 at 03:07
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By induction, one has $$ (xax^{-1})^n=xa^nx^{-1},\tag{1} $$ which implies that the order of the element $y:=xax^{-1}$ must be at most the order of $a$. But the order of $y$ is at least the order of $a$ since if $|y|=n$, then by (1) one has $$ a^n = x^{-1}(xa^nx^{-1})x=x^{-1}y^nx=e.\tag{2} $$

If the order of $a$ is infinite, then, (2) shows that the order of $y$ cannot be finite.

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An abstract proof could be:

$$A_x: a \mapsto xax^{-1}$$ is a group automorphism. So, $ord(a) = ord(xax^{-1})$

A proof "on foot" could be:

It is enough to show $$a^n = e \Leftrightarrow (xax^{-1})^n = e$$. But, that's trivial since

$$(xax^{-1})^n = \underbrace{(xax^{-1}) \cdots (xax^{-1})}_{n} = xa^nx^{-1}$$ So, $$a^n = e \Rightarrow (xax^{-1})^n = xa^nx^{-1} = xex^{-1} = xx^{-1}=e$$. And $$(xax^{-1})^n =e \Rightarrow xa^nx^{-1} = e \Rightarrow a^n = x^{-1}(xa^nx^{-1})x = x^{-1}ex = x^{-1}x = e$$

trancelocation
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