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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)

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What do you sum over in the last sum? Over all the conjugacy classes? – labirintas Dec 3 '12 at 18:31
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
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 $3$.

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

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