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 $G$?
Prime divisor in a finite group
1
$\begingroup$
finite-groups
-
0Try 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. – 2012-12-03
-
0@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 $p$? – 2012-12-03