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In a finite group what is relationship between the number of Sylow $p$-subgroups with the number of elements of order a multiple of $p$?

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One relationship is that the number of $p$-singular elements ( that is, elements whose order is divisible by $p$) is divisible by the number of Sylow $p$-subgroups of $G$. This is a consequence of a theorem Frobenius, together with Sylow's theorem, though I don't recall seeing the fact stated in print. Let $P$ be a Sylow $p$-subgroup of $G.$ Frobenius proved that if $n$ divides the order of finite group $G,$ then the number of solutions of $g^{n}=1$ in $G$ is an integer multiple of $n.$ Hence the number of solutions of $x^{[G:P]} = 1$ in $G$ is divisible by $[G:P].$ This number is also $|G| - \#$ ($p$-singular elements of $G$). Hence the number of $p$-singular elements of $G$ is divisible by $[G:P]$. This is in turn divisible by $[G:N_{G}(P)],$ which is the number of Sylow $p$-subgroups of $G.$

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@Geoff Robinson: This is exactly what I was looking for! Thank you. –  Jiang May 4 '13 at 8:03
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