For which $n$ is it "feasible" to classify groups of order $n$?

Solution 1:

So this is a little cheating, but you asked for an interesting order:

Classify all groups of order $24k+4$. ALL OF THEM.

I find it interesting that there is even an answer!

I'll let you assume someone else has already classified the groups of order $4$ and $n/4 = 6k+1$, though the latter can get pretty tricky [ no one has managed $5^{10} = 6(1627604)+1$ yet ].

Proposition: For each distinct group $N$ of order $6k+1$, find the conjugacy classes of (a) automorphisms of $N$ with order dividing 4, (b) unordered pairs of commuting automorphisms of $N$ with orders dividing 2 [ where $(x,y) \equiv (x,xy)$ and $(x,y) \equiv (y,x)$ as well ]. Then for each class, we get an isomorphism class of group, either (a) $C_4 \ltimes N$ or (b) $K_4 \ltimes N$. These are precisely all isomorphism classes of groups of order $24k+4$.

Proof: Burnside's fusion theorem and checking some details on isomorphisms of semi-directs. $\square$

In general the groups of order 4 are easy: $C_4$ and $K_4$. Since the $6k+1$ can be tricky in general, let's choose a specific number.

Example: For example, $n=316=24\cdot 13+4 = 4\cdot 79$. First we have someone else classify the groups of order $4$ and $n/4 = 79$. We get $C_4$ and $K_4$ for the Sylow $2$-subgroup and then just $C_{79}$ for the Sylow $2$-complement.

Now the automorphism group of $C_{79}$ is cyclic of order 78. It has (a) the trivial automorphism $1 = (x \mapsto x)$ and the order two automorphism $-1 = (x \mapsto x^{-1})$ and that's it, and (b) the pairs $(1,1)$, $(1,-1)\equiv(-1,-1)\equiv(-1,1)$. That gives us the four groups (a) $C_4 \times C_{79}$, $C_4 \ltimes C_{79} = \langle a,b: a^4=b^{79}=1, ba=ab^{-1} \rangle$, and (b) $K_4 \times C_{79}$, $C_2 \times D_{2\cdot 79}$.

Generalization: This also works for $24k+20$. However, $24k+12$ (the last of the $8k+4$ cases) presents some added difficulty, as $A_5$ of order $60 = 24(2) +12$ demonstrates. This difficulty comes from the 3. Without it, the Sylow 2-subgroup cannot be affected by the rest of the group, and so “the rest of the group” becomes a normal subgroup of order $6k\pm1$, so we just get a semi-direct product. The $24k+12$ case has been settled as well, but the proof by Gorenstein–Walters is a few hundred pages long and involves some very interesting mathematics.

Even crazier, is that we can handle some $32k+16$. If the $16$, I mean the Sylow $2$-subgroup, is $C_{16}$ or $C_4 \times C_4$, then handling the 3 is easy (there is no “$A_5$” case), and nearly the same proposition holds! This is a shorter, earlier result of Brauer, and it was one of the earlier uses of modular character theory.

Solution 2:

You should also check out the following papers.

Hans Ulrich Besche, Bettina Eick and E.A. O'Brien, A millenium project: constructing small groups and

John H. Conway, Heiko Dietrich and E.A. O'Brien, Counting groups: gnu's, moas and other exotica

You can google the titles and find a .pdf document. The last paper has a table of the number of groups of order $n \lt 2048$.