Posted by: Steven Sam | December 21, 2009

Groups of size less than 60

My girlfriend is taking an abstract algebra course this year, and they had the following “longterm project” to think about for the term: Show that every group of size less than 60 is either Abelian or not simple. If we assume Burnside’s theorem that a group of order $p^aq^b$ (p and q are primes) is solvable, this gives every case except for 30 and 42. (A proof of Burnside’s theorem can be found here on p.19, Exercise 6 of one my old solution manuals, but you’ll need Serre’s book to see what the cited theorems are). But since I had never done this exercise, I wanted to write a post about how to do this using only techniques from a first course in algebra (i.e., Sylow theorems).

Let’s recall what the Sylow theorems say. First, let G be a finite group of order n. Let p be a prime and write $n = p^km$ where p does not divide m. A maximal subgroup of G whose order is a power of p is a Sylow p-subgroup. Let $n_p$ be the number of Sylow p-subgroups. Then we have the following theorem.

Theorem. 1. The size of a Sylow p-subgroup is $p^k$.
2. All Sylow p-subgroups are conjugate to one another via inner automorphisms of G.
3. The index of the normalizer of any Sylow p-subgroup is $n_p$. In particular, $n_p$ divides m and is congruent to 1 modulo p.

The third statement is usually enough. If we can show that $n_p = 1$ for some p, then the corresponding Sylow p-subgroup (assuming p divides n) must be normal since any conjugate of it must have the same size. There were three general facts that have relatively short proofs that take care of most of the cases. Then the rest was a (quick) case-by-case analysis. If the reader knows of another quick general fact that makes the problem more efficient, leave a comment!

Fact 1. Any p-group of size bigger than p is not simple.

Proof. The center of a p-group is never the identity subgroup. This is usually a standard fact proved in a first course, so I’ll omit it. Using this, either the group is Abelian (and any nontrivial proper subgroup gives us the desired normal subgroup), or the center gives us the desired normal subgroup.

Fact 2. A group of order pq with p<q primes is not simple.

Proof. The number of Sylow q-subgroups divides p and is congruent to 1 modulo p. Hence there can only be 1 of them.

Fact 3. A group G of order 2k with k odd has a subgroup of index 2 (which is normal), and hence is not simple.

Proof. Let P(G) be the group of permutations of G thought of as a set. We have an injective homomorphism $F \colon G \to P(G)$ defined by sending g to the permutation obtained by left multiplication by g. By the Sylow theorems (or more simply, Cauchy’s theorem), we have a nonidentity element g of G of order 2. The cycles of F(g) are each of size 2, so in disjoint cycle notation, F(g) is a product of k 2-cycles, and hence is an odd permutation. So the kernel of the sign homomorphism $G \to P(G) \to \mathbf{Z}/2$ has index 2 and is the desired subgroup.

This accounts for everything except the following numbers: 12, 20, 24, 28, 36, 40, 44, 45, 48, 56.

Using part 3 of Sylow’s theorems, we can immediately get rid of the following cases:
20: There can only be 1 Sylow 5-subgroup.
28: There can only be 1 Sylow 7-subgroup.
40: There can only be 1 Sylow 5-subgroup.
44: There can only be 1 Sylow 11-subgroup.
45: There can only be 1 Sylow 5-subgroup.

In the cases of 24, 36, and 48, there are either 1, 4, or 16 Sylow 3-subgroups (16 is only a possibility for 48). In the last case, the Sylow 3-subgroups have size 3 and hence pairwise intersect in the identity element. So their union consists of 33 elements. A Sylow 2-subgroup doesn’t intersect any of them except in the identity, so contributes 15 more elements. So we only have one Sylow 2-subgroup in this case. Otherwise, there’s either 1 or 4. We assume that it’s 4. In this case, let X be the set of Sylow 3-subgroups. We get a homomorphism $F \colon G \to P(X)$ (where again P(X) is the permutation group of X) by the conjugation action. By Sylow (2), the kernel is a proper subgroup of G. Since P(X) has size 24, in cases 36 and 48, the map is not injective and hence the kernel gives us our desired normal subgroup. Otherwise, if G has size 24 and the kernel is trivial, then G must be the symmetric group $\mathfrak{S}_4$, which we know is not simple (take the alternating subgroup for a normal subgroup).

All that is left to do is 12 and 56. These can be handled by counting arguments. First we do 12. There is either 1 or 4 Sylow 3-subgroups. In the case of 4, their union consists of 9 elements. A Sylow 2-subgroup has size 4, so adds another 3 elements, giving 12 in total. Hence there would be 1 Sylow 2-subgroup in this case. In a similar way, we can do 56. There is either 1 or 8 Sylow 7-subgroups. In the case of 8, their union gives us 49 elements. A Sylow 2-subgroup has size 8, so it contributes an additional 7 elements, giving 56. So there can only be 1 Sylow 2-subgroup in this case.

So that’s the end of the proof. We can extend the exercise further, i.e., the next non-Abelian simple group has order 168. Courtesy of Wikipedia, I’ll list the first few non-Abelian simple groups and their orders. The notation $\mathbf{F}_q$ denotes the finite field with q elements (q being a prime power), the notation $\mathbf{SL}_2(\mathbf{F}_q)$ denotes the $2 \times 2$ determinant 1 matrices over this field, and the notation $\mathbf{PSL}_2(\mathbf{F}_q)$ denotes the quotient of this group by the center $\{\pm I\}$ where I is the identity matrix (if q is even, then $I = -I$). Also $\mathfrak{A}_n$ denotes the alternating subgroup of the corresponding symmetric group.

Order 60: $\mathfrak{A}_5$, which is isomorphic to $\mathbf{PSL}_2(\mathbf{F}_5)$ and $\mathbf{SL}_2(\mathbf{F}_4)$
Order 168: $\mathbf{PSL}_2(\mathbf{F}_7)$
Order 360: $\mathfrak{A}_6$, which is isomorphic to $\mathbf{PSL}_2(\mathbf{F}_9)$
Order 504: $\mathbf{SL}_2(\mathbf{F}_8)$
Order 660: $\mathbf{PSL}_2(\mathbf{F}_{11})$
Order 1092: $\mathbf{PSL}_2(\mathbf{F}_{13})$
Order 2448: $\mathbf{PSL}_2(\mathbf{F}_{17})$
Order 2520: $\mathfrak{A}_7$
Order 3420: $\mathbf{PSL}_2(\mathbf{F}_{19})$
Order 4080: $\mathbf{SL}_2(\mathbf{F}_{16})$

Have fun!

-Steven

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Responses

1. You can also get some mileage out of the strong Cayley theorem, as shown here.

2. That was one of my favorite homework problems in my first abstract algebra course, very elementary, but still a lot of work. Here’s another good one, which is field theoretic, but also elementary:

Let $k$ be a field that isn’t algebraically closed, and $f_1,\ldots, f_n$ functions into $k$ (can take them as polynomials in some number of variables over $k$, say.) Then there exists a polynomial $g\in k[x_1,\ldots,x_n]$ such that $g(f_1,\ldots,f_n)=0$ if and only if $f_1=\ldots=f_n=0$.

3. What happened to 52? I thought it would be necessary to show (easily) that it only has 1 Sylow-13 group?

4. Yes, I left out 52. Thanks for pointing that out.