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If a prime number $p$ divide the number of elements of order $k$ (for some $k\neq p$) in a finite group $G$, then whether we can say that $p$ divide order of $G$?

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Try adding a bit more detail to this question. It seems that English might not be your first language, so don't be discouraged, but you'll get better answers if you say a bit about where you encountered this problem, what you've tried etc. – Simon Hayward Dec 3 '12 at 15:40
@Simon Hayward: We know that the number of elements of order $k$ is equal to $\sum|cl_{G}(x_{i})|$ such that $|x_{i}|=k$. It is clear that if $p\mid|cl_{G}(x_{i})|$ , then $G$ has an element of order $p$. But when $p\mid\sum |cl_{G}(x_{i})|$, then this is a question that whether group $G$ has an element of order $p$? – Simon Dec 3 '12 at 15:55

No. Consider a group of prime order.

Addition: you can get some more insight into the problem by considering the 3rd Sylow theorem. Take for example a group $G$ of order $p_1p_2\ldots p_r$ where $p_k$ are pairwise different primes. Then the number of elements of order $p_k$ equals $n_k(p_k-1)$, where $n_k$ is the number of $p_k$-Sylow groups. The 3rd Sylow theorem yields that $n_k$ divides $|G|/p_k$. Hence your assertion holds for those primes dividing $n_k$.

The situation becomes more complicated if the primes $p_k$ occur with multiplicities in the group order. Then things depend strongly on the structure of the $p_k$-Sylow groups and their intersections.

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Thanks. If $G$ is not $p$-group, then how? – Simon Dec 3 '12 at 15:40
@Mathematics123: there are groups for which your statement is true, and there are those for which it is false. – akkkk Dec 3 '12 at 15:58
@akkkk: Can you give me an example when it is false? – Simon Dec 3 '12 at 16:02
@Mathematics123: uh... Hagen just did... – akkkk Dec 3 '12 at 16:06

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