How do two conjugate elements of a group have the same order?

Question

I'm reading group action in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

I have a problem of understanding the last sentence:

Since conjugation is an automorphism, any two conjugate elements have the same order.

Assume $x,y \in G$ are conjugate, then they are equivalent. As such, $gxg^{-1} = y$ for some $g \in G$. This means $gx = yg$. From here, I could not get how $x,y$ have the same order.

Could you please elaborate on this point?

If $x^n=e$ then $y^n=(gxg^{-1})^n=(gxg^{-1})(gxg^{-1})\cdots(gxg^{-1})=gx^ng^{-1}=e$, and vice versa — J. W. Tanner, Jul 01 '20 at 23:32
You might find several posts on the site about order of conjugate elements: For example: Prove that any conjugate of $a$ has the same order as $a$. or Order of conjugate of an element given the order of its conjugate. There is also this question - however, the OP explicitly says that they want to avoid using argument based on automorphisms: Order of conjugate element equals the order of the element. — Martin Sleziak, Jul 02 '20 at 17:28

score 14 · Accepted Answer · answered Jul 01 '20 at 23:41

Mac Lane and Birkhoff are saying that it's not obvious (at least not directly) that $x$ and $gxg^{-1}$ have the same order. But once we know that $x \mapsto gxg^{-1}$ is an automorphism then it becomes obvious, since all automorphisms preserve order.

To see why, let $\varphi : G \to G$ be an automorphism. Then let $x \in G$ have order $n$, and let $\varphi x$ have order $m$. Now

$$(\varphi x)^n = \varphi (x^n) = \varphi e = e$$ So $m$ divides $n$.

Similarly,

$$(\varphi^{-1} \varphi x)^m = \varphi^{-1}((\varphi x)^m) = \varphi^{-1} e = e$$ And $n$ divides $m$ too, so they must be equal.

There is also a direct computational proof for the conjugation isomorphism. It is basically the exact same proof as above, but writing $gxg^{-1}$ everywhere I wrote $\varphi$ above. I encourage you to try to prove it yourself!

I hope this helps ^_^

Thank you for actually explaining what the authors write and not just providing a different proof. — Carsten S, Jul 02 '20 at 08:52

Shaun · Answer 2 · 2020-07-02T00:08:19.690

9

Since $y=gxg^{-1}$, we have

$$\begin{align} y^n&=\underbrace{(gxg^{-1})\dots(gxg^{-1})}_{n\text{ times}}\\ &=\underbrace{g\cdot x\cdot (g^{-1}g)\cdot\dots\cdot (g^{-1}g)\cdot x \cdot g^{-1}}_{n\text{ times }x}\\ &=gx^ng^{-1}, \end{align}$$

so, if $x^n=e$, then $y^n=e$, and vice versa (by the inverse of conjugation).

edited Jul 02 '20 at 00:08

answered Jul 01 '20 at 23:36

Shaun

44,997

Koro · Answer 3 · 2020-07-02T14:26:55.913

Suppose that $x, y \in G$ are conjugate. Hence $\exists g\in G$ such that: $gxg^{-1}=y$
Note that : $y^2=(gxg^{-1})gxg^{-1}=gx^2g^{-1}$ and also $x^2=g^{-1}y^2g$ . Now show using induction that $y^n=gx^ng^{-1}$ and also $x^n=g^{-1}y^ng$ for $n\in \mathbb N$
Let $|x|=m$ and $|y|=p$ and hence $y^m =gx^mg^{-1}=e$, which implies that $p$ divides $m$$ \tag{1}$.
But also, $x^p=g^{-1}y^pg=g^{-1}eg=e$, which implies that $m$ divides $p$. $\tag{2}$

By (1) and (2), $m=p$

score 3 · Answer 4 · answered Jul 01 '20 at 23:40

If $gxg^{-1}=y$ and $x^n=e$, then $y^n=(gxg^{-1})^n=(gxg^{-1})(gxg^{-1})\cdots(gxg^{-1})=gx^ng^{-1}=e$.

Similarly, if $x=g^{-1}yg$ and $y^n=e$, then $x^n=g^{-1}y^ng=e.$

Thus, if $y$ and $x$ are conjugates, then $y^n=e\iff x^n=e$.

So if $r$ is the order of $y$ (the least positive $n$ such that $y^n=e$), it is the order of $x$ too.

How do two conjugate elements of a group have the same order?

4 Answers4

Linked

Related