2
$\begingroup$

Let $p$ be a prime number. Can one find always a finite non-$p$-group $G$, such that the portion of number of $p$-elements of $G$ comes arbitrarily close to the total number of elements of $G$. That is, if $0 < \epsilon < 1$ can one find a $G$, not a $p$-group, such that $\#\{p\text{-elements in } G\}/\#G = 1 - \epsilon$ ?

  • 0
    @Gerry Myerson : That doesn't quite work for this question as phrased: However large $q$ is, the proportion of elements of order $p$ is $1 - \frac{1}{p}.$ Remembering the identity, and counting it (as usual) as a $p$-element, the proportion has limit $1 - \frac{1}{p}$ as $q \to \infty.$2012-02-16

1 Answers 1

10

As mentioned in the comment, Gerry Myerson's example does not quite answer the question, but it is in the right spirit. Choose $\varepsilon >0,$ and for a given prime $p,$ choose an integer $n >0$ so that $p^{-n} < \varepsilon.$ Choose any prime $q$ such that $q \equiv 1$ (mod $p^n$). There is a Frobenius group $G$ of order $qp^n$ with kernel $K$ of order $q$ and complement $H$ of order $p^n.$ Then $G \backslash K$ is the set of non-identity $p$-elements of $G.$ The proportion of elements of $G$ which are non-identity $p$-elements is thus $1 - p^{-n}.$ Allowing for the identity, which makes a positive contribution to the proportion of $p$-elements, the proportion of elements of $G$ which are $p$-elements is greater than $1 - \varepsilon.$

  • 0
    thanks, great example, hadn't thought of Frobenius groups.2012-02-16