# Thread: Possible Cardinality of non-Abelian Groups

1. ## Possible Cardinality of non-Abelian Groups

Here is a question which I came up with.
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Let $\displaystyle \mathcal{G}$ be a finite non-abelian group. What are all the possible cardinalities of this group?
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For example, it cannot be 1 because that is the trivial group which is abelian. It cannot be 2 or 3, because up to isomorphism there is only 1 namely $\displaystyle \mathbb{Z}_{2,3}$ which is abelian. With 4 you have two possibilities the usual $\displaystyle \mathbb{Z}_4$ and the Klein 4 group which is also abelian. With 5 (prime) there can only be $\displaystyle \mathbb{Z}_5$. Now with 6 we finally have our first non-abelian group, the dihedral group. With 7 since it is prime we have the same thing ...
Which orders are possible?

I don't have anything approaching a complete answer to your question, but I can offer some further hints, some of which you may already know...

Every group of prime order is cyclic (hence abelian), so we can rule out those. Also every group of order p^2, p a prime, is abelian (like the Klein-4 group for instance). We also have:

If p < q are distinct primes, then
1) if p does not divide q-1, then every group of order pq is cyclic (there is only one group of order 15=3*5 for instance)
2) if p divides q-1, then there are exactly two groups of order pq, one cyclic and one non-abelian (6 = 2*3, as you mentioned)

And that is about as far as I can go off the top of my head. This is about as far as Hungerford goes in his beginning graduate text "Algebra". I am sure there are more theorems like these, but i would be very surprised to find that all orders of non-abelian groups have been classified (not that it hasn't, just that I would be surprised ). It's certainly a worthwhile question nonetheless.

3. The first few such orders are listed here.

4. Thank you rgep that partially answers the question.