## Thursday, November 11, 2010

### Simple Groups of Order 60

We will prove that there is only one simple group of order 60 (up to isomorphism) which is the alternating group $A_5$. We will not prove that $A_5$ is simple here, it will be assumed that this result is known. So all it remains is to show any simple group of order 60 is isomorphic to the alternating group.

The proof is not so trivial. It takes some work. With the right tools it is not too hard. But I think it is a fun proof. The ideas in this proof are actually important, they appear in other kinds of problems too. We will assume knowledge of Sylow theorems in our discussion.

We will also need to know the notion of "group action on a set". This concept was introduced here if you want to learn about it.

There is a one-to-one correspondence between group actions and homomorphisms between groups into symmetry groups. We will explain what this means. This is an important idea in mathematics, it appears in many other kinds of problem too other than determining simple groups of order 60, it is a good time to introduce it.

Let G is a group acting on a set X and let g be a fixed element from G. Then consider the function $$f_g: X \to X$$ defined by $$f_g(x) = gx$$. It is easy to see that this function is a permutation of X i.e. it is a bijection from X to itself - it is a trivial exercise by using the definition of group action. Therefore, we can define the function $$f:G\to S_X$$ by $$f(g) = f_g$$, where $$S_X$$ is the group of symmetries of X. It is easy to see that f is a homomorphism.

Furthermore, if $$f:G\to S_X$$ is a homomorphism then we can define $$gx = f(g)(x)$$, and it is another trivial exercise to see that this gives a group action on X. Thus, we see that any group action induces a "permutation representation", and every permutation representation gives rise to a group action. We can jump between these two equivalent notions whenever it is convenient for us.

Here is an application of what we have been saying.

Cayley's Theorem: If G is a finite group then G is a permutation group i.e. G is a subgroup of the symmetric group.

Proof: Let G act on itself by left-multiplication. That is, let X=G and define gx to be the product gx as in the group. Then clearly, G acts on G by left-multiplication. This action, by what we said above, induces a group homomorphism $$f:G\to S_G$$. Note that $$S_G\simeq S_n$$. The action of G on G is "faithful" i.e. if gx = x for all x then g = 1. Therefore, the homomorphism f is injective. And so we see that G imbeds into $$S_n$$. This completes the proof.

This theorem is not useful for us, we just mentioned it to clarify what we have been discussing. Let us finally get to determing simple groups of order 60.

First, factor $$60 = 2^2 \times 3 \times 5$$. Let $$n_2, n_3,n_5$$ be the number of Sylow 2-subgroups, number of Sylow 3-subgroups, and number of Sylow 5-subgroups, respectively. By Sylow's third theorem we know that $$n_p\equiv 1(\bmod p)$$ and that $$n_p$$ divides 60. Let us list the possibilites for $$n_p$$ where p=2,3,5:

$$n_2 = 1,3,5,15$$
$$n_3 = 1,4,10$$
$$n_5 = 1,6$$

If there is just one Sylow p-subgroup P then $$gPg^{-1}$$ is a Sylow p-subgroup, since it is unique it follows that $$P = gPg^{-1}$$ for all $$g\in G$$. Thus, then P is a normal subgroup. Since it is assumed that G is simple it means that it cannot have a unique Sylow p-subgroup for any p. Which means the possibilities for $$n_p$$ must look as follows:

$$n_2 = 3,5,15$$
$$n_3 = 4,10$$
$$n_5 = 6$$

Let $$X=\{P_1,...,P_n\}$$ be all the Sylow p-subgroups of G. Now we can let G act on X by conjugation, that is, define the group action $$gP_k = gP_kg^{-1}$$. It is an easy exercise to check that this defines a group action. Once we have a group action we immediately get an induced permutation representation $$f:G \to S_n$$. Note that we write $$S_n$$ instead of $$S_X$$ because $$S_n$$ is isomorphic to $$S_X$$ since $$X=n$$. What is the kernel of f? The kernel, whatever it is, must be a normal subgroup of G. Since it is assumed that G is simple it follows that $$\text{ker}(f) = \{ 1\},G$$. It cannot be the case that the kernel is all of G. Because that would imply that every element of g fixes each $$P_k$$, and then that would mean each $$P_k$$ is normal in G, which is impossible. Therefore, the kernel must be trivial. But then this means $$f:G\to S_n$$ is an imbedding of G into $$S_n$$. Which means $$G\leq S_n = n!$$. This shows that the numbers $$n_p$$ must satisfy the inequality $$G \leq (n_p)!$$. But we know that G = 60. Therefore, we can immediately rule out $$n_2=3$$ and $$n_3=4$$. Thus, the possibilities for $$n_p$$ must look as follows:

$$n_2 = 5,15$$
$$n_3 = 10$$
$$n_5 = 6$$

We want to show that $$n_2\not = 15$$ and so it will mean that there are precisely 5 Sylow 2-subgroups. With that knowledge we can apply our work above to conclude that there is an imbedding $$f:G\to S_5$$. Hence it will mean that we can think of G as a subgroup of index 2 in $$S_5$$. From that point on it will be easy to show that $$G=A_5$$. But before we can say that we need to show why $$n_2\not = 15$$.

For that we will use a simple combinatorical argument. If H and K are two Sylow 5-subgroups they have order 5, so they are prime cyclic. Which means if H and K are distinct then $$H\cap K = 1$$. Therefore, in total the number of elements among all Sylow 5-subgroups (excluding the identity) is equal to $$4\cdot 6 = 24$$. This tells us that G has at least 24 elements of order 5. If H and K are two Sylow 3-subgroups they have order 3, so they are prime cyclic. Which means if H and K are distinct then $$H\cap K = 1$$. Therefore, in total the number of elements among all Sylow 3-subgroups (excluding the identity) is equal to $$2\cdot 10 = 20$$. This tells us that G has at least 20 elements of order 3. Now suppose that G had 15 Sylow 2-subgroups. If H and K where two such distinct subgroups the intersection $$H\cap K$$ no longer has to be distinct. Because H and K are not prime cyclics, and they do not even need to be cyclic. But they can be Klein 4-subgroups. So their intersection $$H\cap K \leq 2$$, they intersect in one or two elements. Which means there are at least (excluding the identity) $$2\cdot 15=30$$ elements of order dividing 4. Thus, G has at least $$1+24+20+30 = 75$$ elements - which is a contradiction. Which means G must have precisely 5 Sylow 2-subgroups.

Thus, we see that G imbeds into $$S_5$$ since G = 60 it follows that G is a index two subgroup of $$S_5$$. Which means that G is a normal subgroup of $$S_5$$. We managed to reduce our classification problem into determining all normal index two subgroups of $$S_5$$. It is not so surprising that $$A_5$$ is the only such subgroup. This is true in general. So how about we prove this theorem in general. I found a very easy way to prove this that is not complicated at all.

Theorem: The only index two subgroup of $$S_n$$ is $$A_n$$.

Proof: We know that this subgroup, call it G, is normal. If G was to contain a 2-cycle then it would have to contain all conjugates of the 2-cycle. And hence G would contain all 2-cycles. But the 2-cycles generate $$S_n$$ so that this would force $$G=S_n$$ which is impossible. Therefore, G does not contain any 2-cycle. Form the factor group $$S_n/G$$ this group is just like $$\mathbb{Z}_2$$. If $$\sigma_1,...,\sigma_k$$ are 2-cycles it follows, by the isomorphism to $$\mathbb{Z}_2$$, that $$\sigma_1\sigma_2...\sigma_k\in G$$ if and only if k is even (since modulo $$G$$, the $$\sigma_j$$'s behave like 1 modulo 2). And so we see that G contains all the even permutations and no odd permutations. Thus, G is indeed the alternating group $$A_n$$.

This completes the proof and thereby completes the classification of all simple groups of order 60.