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

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?

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 2 '20 at 8:52

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).

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), $$m=p$$

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.