I just tried to solve this question on a exam:

Let $G$ be a group with only one element $x$ of order $n$, $n$ natural. Show that $x\in Z(G)$. (I'm using multiplicative notation)

$Z(G)$ is the subset of G such that every element of $Z(G)$ commutes with every element of G (called the Center of G).

I tried to do it like this: Suppose that x does not commute with at least one $g\in G$. Then $xg\neq gx\rightarrow x\neq gxg^{-1}$. Then,

$x^n\neq (g x g^{-1})^n=g x g^{-1}g x g^{-1}...g x g^{-1}=xx...x=gx^ng^{-1}=1$

Notice that $g x g^{-1}$ cannot have an order $n < m$ because its order depends exclusively of the order of $x$, which is $n$.

That means that there are two distinct elements in $G$ with order $n$, which is a contradiction.

Can someone tell me if it was done correctly?

  • $\begingroup$ When you made (gxg-1)^n the result is g(x^n)g-1 and it is equal to 1. $\endgroup$ – alpha.Debi Apr 12 '12 at 14:56
  • $\begingroup$ Yes, that's correct... I hope its just a minor mistake :D, Just edited. $\endgroup$ – Marra Apr 12 '12 at 14:59
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    $\begingroup$ @GustavoMarra — Even if it is not needed, note that if there is exactly one element of order n in G, then n is 1 or 2. $\endgroup$ – Lierre Apr 12 '12 at 15:02
  • $\begingroup$ Some people tried to do it using this fact, but they couldn't conclude that the thesis was true through this. $\endgroup$ – Marra Apr 12 '12 at 15:05
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    $\begingroup$ I agree, this fact is not useful in the proof. But it is interesting to note the false generality of the statement. $\endgroup$ – Lierre Apr 12 '12 at 15:09

you should know that for any $g\in G$ the order of $g^{-1}xg$ equal with order of x so for any $g\in G$ $g^{-1}xg=x$ then we have $x\in Z(G)$

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    $\begingroup$ That's, in essence, what I had to show. $\endgroup$ – Marra Apr 12 '12 at 15:03

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