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If In a group $G$, the element $a\in G$ is the only element of order $n$, i.e., $a^n=e$ for some positive integer $n$. Then we have to show that $a\in Z(G)=\{x\in G : xg=gx, \forall g\in G\}$.

How can I show this?

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  • $\begingroup$ As Andreas Caranti observed, this is only possible for $n=1$ or $2$. $\endgroup$ – Derek Holt Apr 28 '15 at 7:45
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Hint : take $g \in G$. What is the order of $gag^{-1}$ ?

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First note that it is not enough to say $a^{n} = e$ to make sure that $a$ has order $n$. This only shows that $a$ has order dividing $n$. (For instance $a = e$ satisfies the above, for all $n$, and $e$ has order $1$.) You have to add that $a^{k} \ne e$ for all $k \mid n$.

Then note that you must have $n = 2$ here. This is because all the $\varphi(n)$ powers $a^{k}$, for $0 \le k < n$ with $\gcd(k, n) = 1$, have also order $n$. If $a$ is unique of its order, then one must have $\varphi(n) = 1$, and thus $n = 2$.

Finally, as noted in the other answers, $g^{-1} a g$ has the same order as $a$ for all $g \in G$, as $x \mapsto g^{-1} x g$ is an automorphism of $G$, so $g^{-1} a g = a$.

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Look at the conjugates of $a$. They all have the same order. What do you conclude?

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$$\left(g\cdot a\cdot g^{-1}\right)^n=g\cdot a^n\cdot g^{-1}=e$$

And $a$ is the only element of order $n$

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