Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We know that the number of elements of order $k$ in a finite group $G$ is equal to $\sum |cl_{G}(x_{i})|$=$\sum|G/C_{G}(x_{i})|$ such that $|x_{i}|=k$. It is clear that for a prime $p$ if $p\mid |cl_{G}(x_{i})|$, then $G$ has an element of order $p$. Now let $p\mid \sum |cl_{G}(x_{i})|$. My question is: Is $G$ has an element of order $p$?($G$ is not $p$-group)

share|improve this question
    
What do you sum over in the last sum? Over all the conjugacy classes? –  labirintas Dec 3 '12 at 18:31
1  
If you are adding the order of all conjugacy classes, then you get the order of $G$, so the result follows from Cauchy's Theorem. If you are only adding conjugacy classes corresponding to elements of a given order, then the answer is no: for the Klein $4$-group, $G$ has 3 elements of order $2$, so $3$ divides the sum of sizes of conjugacy classes of elements of order $2$, but $G$ has no elements of order $3$. –  Arturo Magidin Dec 3 '12 at 18:54
    
@Arturo Magidin: Thanks. Note that $G$ is not $p$-group. –  Mart Dec 3 '12 at 19:39
    
@user123: (i) You have failed to clarify what you mean with yoru ntoation. (ii) Take the direct product of the Klein 4-group and a group of order not divisible by 6. You still get exactly 3 elements of order 2, but no elements of order 3. –  Arturo Magidin Dec 3 '12 at 19:49

1 Answer 1

up vote 3 down vote accepted

As noted, it is unclear what your sums are over.

If the $x_i$ are conjugacy class representatives for all conjugacy classes, then $\sum|\mathrm{cl}_G(x_i)| = |G|$, so the question has an affirmative answer by Cauchy's Theorem.

More likely is that the $x_i$ are representatives from the conjugacy classes of elements of order $k$; in that case, the answer is "no". Although the example I give in the comments is a $p$-group, it is easy to turn it into an example which is not a $p$-group: take $p=3$, $k=2$, and let $G=C_2\times C_2\times A$, where $A$ is any nontrivial group of order not divisible by $6$ (e.g., $A=C_5$); then $G$ is not a $p$-group, but an element $(a,b,c)$ has order $2$ if and only if $c=1$ and $a$ or $b$ are nontrivial; so $G$ has exactly three elements of order $2$, each of which is its own conjugacy class, so the sum equals $3$. However, $|G|$ is not divisible by $3$, so $G$ does not have any elements of order $3$.

share|improve this answer
    
@Arturo Magindin: Thank you so much for your answer. –  Mart Dec 4 '12 at 13:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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