# For which $n$ is it “feasible” to classify groups of order $n$?

Classifying groups of small order is a standard exercise in group theory classes, and especially after learning the Sylow Theorems and Direct/Semidirect products, it is possible to classify groups of a relatively large order (i.e., $>100$). For example, in the section on semidirect products, Dummit and Foote have a handful of multipart exercises where you are asked to classify groups of orders which have ~$15$ isomorphism types each. On the other hand, they never (as far as I can tell in what I have read) completely classify groups of order $16=2^4$ (although I think they mention almost all of them at various points). Of course, groups of order $2^n$ are notoriously hard to classify (the most comprehensible article doing so for $n=16$ that I have read was $12$ pages long).

If we only look at odd order groups, does classifying all odd groups, say, with order below $100$, become feasible? And how many powers of $2$ are needed to make a classification too tedious? $2^4$ by itself is already rather tedious, so I guess $4$ would have to be the smallest power dividing the order?

Edit: Does anyone know of some nice group orders, where classifying all groups of that order is nontrivial, but isn't too difficult and gives some interesting groups?

• @AndresCaicedo:en.wikipedia.org/wiki/List_of_small_groups classify up to order 16 and mention library that classify up to 2000 (except 1024) and also there is an article linked at the end The Groups of order Sixteen Made Easy – Gina Jan 8 '14 at 2:31
• Hall–Wales (1968) classified the finite simple groups of order 604800, and only a few years later the classification of all groups of order 604801 was completed in a single page paper. ;-) – Jack Schmidt Jan 8 '14 at 3:39
• Nishant, hehe :-) More seriously, if no prime squared divides the group order, then the calculation is very easy (logarithmic in the group order, so "linear time algorithm"). If no prime cubed divides the group order, then the calculation is fairly easy, but not completely trivial (GAP's package cubefree will do the calculation for you as long as the number is less than a hundred thousand or so). – Jack Schmidt Jan 8 '14 at 3:46
• $p^8$ is done, and I think I heard $2^{11}$ is counted (and an algorithm would let you sample randomly from the isomorphism classes; at any rate you could do that for $2^{10}$ for sure). Once you have something like $p^3 q^3 r^3$ the combinations start to get pretty nasty. (You can do $pqr^2$ by hand without too much trouble, this sylow is normal, blah blah; but $p^3 q^3 r^3$ and you get part of the sylow is normal, then part of that one, then back to the first, blah blah, forever). – Jack Schmidt Jan 8 '14 at 3:49
• A related problem I like better: Sylow's theorem means that the prime factorization of $|G|$ requires there be these specific groups of order $p^n$, $q^m$, etc. Instead of just looking at $|G|$, we look at the isomorphism class of those $p$-groups ($q$-groups, etc.). So instead of saying "Classify the groups of order 24" you say "Classify the groups whose Sylow $p$-subgroups are all quaternion of order 8, cyclic of order 3, or trivial." – Jack Schmidt Jan 8 '14 at 3:52

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.

• That's pretty cool! I guess it boils down to the order being 4 times an odd number (which can be 1, 3, or 5 mod 6), and you do mention all 3 of those cases...I guess if 3 divides the order of the group, then the Sylow 3 may not be normal, which is what makes it more difficult? Hmm, how does this generalize when the small group is of a different order, say, order 9 or 25? Do these sorts of results still apply? – Nishant Jan 8 '14 at 5:22
• Exactly. The groups I mentioned ($C_4,K_4,C_{16},C_4 \times C_4$) have at most one bad prime, $3$. The groups of order $9$ (both!) have $2$ as their only bad prime, and the groups of order $25$ have $p=2$ [ $C_{25}$ ] and $p=2,3$ [ $C_5\times C_5$ ] as bad primes. So the results don't sound as cool when you remove them. – Jack Schmidt Jan 8 '14 at 5:27
• Why doesn't $C_{25}$ have $3$ as a bad prime? – Nishant Jan 8 '14 at 5:35
• The automorphism group of (any subgroup of) $C_{25}$ is not divisible by $3$. The automorphism group of $C_{25}$ is cyclic of order $20$. – Jack Schmidt Jan 8 '14 at 5:40
• (Writing down the exact condition for a prime to be bad is some amazing mathematics in my opinion, but pretty hard too. math.stackexchange.com/questions/629960 is a pre-requisite. However, for abelian $P$ the requirement is simple: a prime $q$ is bad for the abelian $p$-group $P$ iff $q$ divides the order of the automorphism group of $P$. That is basically Burnside's fusion theorem.) – Jack Schmidt Jan 8 '14 at 5:45

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 $n \lt 2048$.