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Do the set of finite non-abelian groups and the set of finite abelian groups have the same cardinality?

In that case, is it possible to define some sort of density of finite groups and calculate the density of finite abelian/non-abelian groups within the set of all finite groups?

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The collection of all finite abelian groups is not a set –  Amr Feb 7 '13 at 13:08
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I think he means to consider the finite groups "up to isomorphism", so all the permutation groups on the sets {1,2,3,4,...,n} for each n. –  user58512 Feb 7 '13 at 13:14
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The set of all finite groups up to isomorphism is countable. Hence any infinite subset is also countable. In particular there are countably many abelian groups and countably many non-abelian groups. –  Lior B-S Feb 7 '13 at 13:16
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There are several ways to count finite groups, for example by order. –  Lior B-S Feb 7 '13 at 13:17
    
user58512 is right. –  Aqwis Feb 7 '13 at 13:18

1 Answer 1

up vote 4 down vote accepted

Consider the case of finite $p$-groups, $p$ a fixed prime. The number of (isomorphism classes of) abelian $p$-groups of order $p^n$ equals the number $\operatorname{Part}(n)$ of partitions of $n$. For this number we have the asymptotic expression (straight from the Wikipedia) $$ \operatorname{Part}(n) \sim \frac {1} {4n\sqrt3} \exp\left({\pi \sqrt {\frac{2n}{3}}}\right) \mbox { as } n\rightarrow \infty $$

Now the Higman-Sims estimate for the number of (isomorphism classes of) $p$-groups of order $p^n$ is $$ \exp\left(\log(p) \left(\frac{2}{27} + O(n^{-1/3})\right) n^3\right), $$ so unless I'm mistaken, abelian groups are rather scarce here.

On the other hand, the Higman-Sims estimate relies on counting group of nilpotence class two.

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