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 $\#\{p\text{-elements in } G\}/\#G = 1 - \epsilon$ ?
The number of p-elements in a finite group
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group-theory
finite-groups
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0$C_p^n\times C_q$ for $n$ big enough where $p$ is prime with $q=3$ if $p=2$ and $q=2$ otherwise? Unless I am miss-understanding something? – 2012-02-16
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0@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 $n$. – 2012-02-16
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0@Jyrki, exactly. In the example of user1729 the fraction of p-elements is 1/q (counting the trivial element as a p-element). – 2012-02-16
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1If $p\lt q$ 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 $q\to\infty$. – 2012-02-16
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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