Let $a$ be the unique element of any order in a group $G$. Is $a$ in the centre of $G?$ I can prove that the result is true if $o(a)$ is 2 by taking another element $z=xax^{-1} $ and then by using uniqueness.
But for in general case I cannot understand how can I proceed further to prove it.
Please help me. Thank you in advance
 A: If you have a valid proof for the case when the order of $a$ is two, $o(a)=2$, then I'm pretty sure that you're done with the question — because the order of $a$ doesn't really matter here. The order of $z=xax^{-1}$ is equal to the order of $a$; so, just as you said, by uniqueness it implies that $xax^{-1}=a$, and you're done.
A: What you want to show is that $\forall x \in G, x.a.x^{-1} = a$. The thing you miss is probably that $(x.a.x^{-1})^k = x.a^k.x^{-1}$ for any integer $k$. If you write the product down, you will see that all the intermediate $x^{-1}.x$ terms cancel each other $(x.a.x^{-1})^n = (x.a.x^{-1})(x.a.x^{-1})\ldots(x.a.x^{-1})$
For a more rigorous proof you can do it by induction.
From here you have $(x.a.x^{-1})^n = x.a^n.x^{-1} = x.x^{-1} =1$. This means that $x.a.x^{-1}$ has an order $m$ that divises $n$.
Now, we just need to show that $m=n$. For that, one can see that $1 = (x.a.x^{-1})^m = x.a^m.x^{-1}$ which implies $a^m=1$. By definition of the order of an element we can deduce $m=n$ since the order is the smallest positive integer that verifies this property.
We proved that $x.a.x^{-1}$ is of order $n$ which implies, by uniqueness of an element of such order, that $x.a.x^{-1} = a$ which is what we wanted.
A: This statement is vacuously true for $o(a) > 2$ however. Put another way: Let $a$ be any element in $G$ where $o(a)$ is at least $3$. Then there is always another element $a' \not = a$ such that $o(a')=a$.
Indeed, write $y=o(a)$, and let $\varphi(y)$ be the number of positive integers $x < y$ that are relatively prime to $y$. Then on the one hand, $\varphi(y) > 1$ iff $y > 2$. Then on the other hand, there are $\varphi(y)$ elements $a'$ in $\langle a \rangle$ that generate $\langle a \rangle$, and each of these elements $a'$ that generate $\langle a \rangle$ satisfy $o(a')=y$ [why is that]. As $\varphi(y)$ is strictly greater than $1$ for $y>2$, there are $\varphi(y)-1 \ge 1$ other elements $a' \not = a$ with $a' \in \langle a \rangle$ that generate $\langle a \rangle$, and thus satisfy $o(a')=y$. So $a$ is not the only element in $G$ that has order $y$ in $G$.
