# A multiple choice question on finite group

Let $G$ be a finite group such that $Z(G)=1$. Let there exist $m$ such that $G$ has a unique element of order $m$. Which of the following statements is true?

(a) $m=1$

(b) $m$ is prime

(c) $m=2p^{n}$ where $p$ is prime.

(d) $m=pq$ where $p$ and $q$ are two distinct prime numbers

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Any thoughts of your own whatsoever on this problem? People are much more willing to help you if you show that you've tried the problem yourself. –  Zev Chonoles Feb 17 '13 at 7:14
Edit: nevermind, missed the condition on the center of the group –  Jim Feb 17 '13 at 7:31
@ZevChonoles People are much more willing to help you if you show that you've tried the problem yourself... I see what you mean, and I agree with what you mean, but is this assertion actually true (without qualifications)? –  Did Feb 17 '13 at 7:57
@Did: The only questionable part, I think, is much. Even I am more willing to help someone who shows some thoughts, though that rarely affects how likely I am to help. –  Brian M. Scott Feb 17 '13 at 11:22
Let $x$ be the unique element of order $m$. Then for all $y \in G$, the conjugate $y^{-1} x y$ has also order $m$, so that $y^{-1} x y = x$, or $xy = yx$, for all $y \in G$, that is $x \in Z(G) = \{ 1 \}$, so that $x = 1$ and $m = 1$.
As noted in the comments, most notably by @CutieKrait, if you drop the requirement on $Z(G)$, then you cannot rule out the cyclic group of order $2$, which has a unique element of order $2$. So (b) would be also a possibility, with $m = 2$.