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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 $\#\{p\text{-elements in } G\}/\#G = 1 - \epsilon$ ?

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$C_p^n\times C_q$ for $n$ big enough where $p$ is prime with $q=3$ if $p=2$ and $q=2$ otherwise? Unless I am miss-understanding something? – user1729 Feb 16 '12 at 10:06
@user1729: I don't think that works. Unless the $C_q$-component of an element is trivial, the element won't be a $p$-element. Therefore at most half of the elements of your group are $p$-elements irrespective of $n$. – Jyrki Lahtonen Feb 16 '12 at 10:28
@Jyrki, exactly. In the example of user1729 the fraction of p-elements is 1/q (counting the trivial element as a p-element). – Nicky Hekster Feb 16 '12 at 10:35
If $p\lt q$ are primes, and $p$ divides $q-1$, then there's a noncommutative group of order $pq$. All of its elements save one have order $p$ or $q$, and it's probably not too hard to work out how many of each, and see what happens as $q\to\infty$. – Gerry Myerson Feb 16 '12 at 10:44
@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 $q \to \infty.$ – Geoff Robinson Feb 16 '12 at 11:17

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 $1 - \varepsilon.$

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thanks, great example, hadn't thought of Frobenius groups. – Nicky Hekster Feb 16 '12 at 12:02

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