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Firstly, does the family of all non-isomorphic non-Abelian groups have a well defined cardinality? How about the family of all non-isomorphic Abelian groups? If they are both defined, how do they compare?

Does the statement "there are more non-Abelian groups than Abelian groups" make any mathematical sense at all?

I am aware of a related question but my question should be different.

Zuriel
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    Possibly relevant: https://math.stackexchange.com/questions/2774302/is-there-a-meaningful-measure-of-the-probability-for-a-finite-group-to-be-commut in the comments – twnly Jan 28 '19 at 01:18
  • There are several models of randomness for finitely-presented groups; in the correct regime, a "typical group" is non-abelian. However, what you literally count are not groups but their presentations. See https://en.wikipedia.org/wiki/Random_group – Moishe Kohan Jan 28 '19 at 13:41

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Firstly, does the family of all non-isomorphic non-Abelian groups has a well defined cardinality?

No, being a proper class, it does not have a cardinality.

How about the family of all non-isomorphic Abelian groups?

No: same reason.

If they are both defined, how do they compare?

N/A

Does the statement "there are more non-Abelian groups than Abelian groups" make any mathematical sense at all?

Maybe, but not in the sense of cardinality. You would have to be more specific about the context you are working in.

rschwieb
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If we restrict ourselves to finitely presented abelian or nonabelian groups, there are countably many nonisomorphic groups in each set.

Since there are countably many finite group presentations, we need only find a countable subset of each. Indeed, the free groups, $F_n$ ($n\ge 2)$, and the free abelian groups, $\mathbb{Z}^n$, will do. These are pairwise non-isomorphic (which seems obvious, but all the proofs I know are mildly technical).

We can even find the same result among finite groups: the cyclic groups, $\mathbb{Z}/n\mathbb{Z}$, and the symmetric groups $S_n$ ($n \ge 3$) are pairwise non-isomorphic (they have different orders) and there are countably many.

In the case of finite simple groups, there is a sense in which your statement holds: according to the classification of finite simple groups there is only one infinite family of finite simple abelian groups (cyclic groups of prime order), but two (or more, depending on how you count) families of nonabelian groups: the alternating groups $A_n$ ($n \ge 5$) and the simple groups of Lie type.

Rylee Lyman
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  • Thanks! So (perhaps counter-intuitively) there are no more non-Abelian finitely presented groups than Abelian finitely presented groups. Somehow I heard the saying that "most groups are non-Abelian", which should make no sense. – Zuriel Jan 28 '19 at 01:30
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    Sometimes when people say "most", they're referring to probabilistic methods: consider it this way. If you were to randomly write down a finite group presentation, most ways of doing so would "asymptotically almost surely" yield nonabelian groups (since writing down $ab = ba$ for each generator seems fairly unlikely). I believe there are theorems in this vein, but I don't know enough about random groups to say for sure. – Rylee Lyman Jan 28 '19 at 01:35
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    Derek Holt's comment in the question linked above makes this precise for finite groups! (https://math.stackexchange.com/questions/2774302/is-there-a-meaningful-measure-of-the-probability-for-a-finite-group-to-be-commut) – Rylee Lyman Jan 28 '19 at 01:37