Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

closed as off-topic by Najib Idrissi, Daniel Fischer, Hakim, Olivier Bégassat, user1729 Jul 31 at 11:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Daniel Fischer, Hakim, Olivier Bégassat, user1729
If this question can be reworded to fit the rules in the help center, please edit the question.

2  
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
1  
@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

1 Answer 1

up vote 2 down vote accepted

The case for (a).

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 $m = 2$.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.