Recall that a group is said to be simple if it contains precisely two normal subgroups, and that a group is solvable if it admits a normal series such that each quotient is abelian. Every subgroup of a solvable group, and every quotient of a solvable group is solvable. It should be clear that if a group is simple nonabelian, it is not solvable. The title of the post, albeit a bit foolish, attemps to convey the following idea: groups are simple because they admit exactly two normal subgroups, but the moral reason behind this varies wildly between each family of simple groups. For example, the most trivial family of simple groups is , and the reason they are simple is no mystery: any subgroup must have index a divisor of , which is either or . It follows every (a cyclic group of prime order) admits exactly two subgroups, hence exactly two normal subgroups. In this post we will look at two other infinite families of finite simple groups: the family of alternating groups and the family of projective special (or unimodular) groups .
The following is based on the exposition in Jacobson’s Basic Algebra I, which I encourage anyone (acquainted or not with undergraduate algebra) to look at.
Simple groups have been objects of interest in mathematics all the way back to Galois, who constructed the Galois group of a polynomial over a field (modernly defined as for a splitting field of over ) as a certain permutation group of its roots. Galois proved his celebrated theorem that the equation is solvable by radicals over if and only if the Galois group is solvable. If is of degree , then one can consider as a subgroup of , in particular if one considers a general equation of degree over (i.e. we’re solving the equation generically in the coefficients )
then it can be shown the corresponding Galois group is . We will show now that is a normal subgroup (for it has index ) which is simple if — we shall call the degree of the alternating group. This nomenclature arises from group actions: a group action is said to be of degree if it is defined on a set of precisely elements, and the alternating group acts canonically on the -set of letters . Since is simple, it follows that is not solvable for (why?) and settles the famous theorem (Galois’ theorem notwithstanding) that the general equation of degree is not salvable by radicals if . Let us prove a lemma, first:
Lemma 0 The alternating groups of degree at least five are generated by the three cycles, and all three cycles are conjugate.
Proof Since is generated by the transpositions it follows is generated by the double transpositions , so it suffices to show every double tranposition is a product of three cycles. The reader can verify that if are distinct and that , which proves the first claim. Since for any permutation , it follows easily that every three cycle is conjugate to in . If is even, we’re done. Else, since we can choose different from and take , for and commute. .
We can now prove the
Theorem (Galois) The alternating groups of degree at least five are all simple.
Proof We will use the previous lemma, of course. Indeed, we will show that every nontrivial normal subgroup of with contains a three cycle, which proves the claim. To this end, we note that if is nontrivial it contains a nontrivial permutation that fixes some letter (show that for any permutation that fixes no letter there exists a permutation such that fixes some letter). We claim that if we pick such nontrivial permutation that is not a three cycle, we may conjugate it to obtain a permutation with more fixed points. It will follow that any permutation with a maximum number of fixed points is a three cycle. Indeed, let be nontrivial with fixed points, but not a three cycle. Using the cycle decomposition in we can write as a permutation of the form or . Let . We then have and and in either case so that . Any letter is fixed by so if it is fixed by it is fixed by . The reader can verify that in the first case fixes , and since moves we obtained more fixed points. In the second case fixed and , and again we have more fixed points. This proves the theorem.
We now move on to a more geometric situation, and consider the projective special groups. Since these are not as popular as the alternating groups, let us take a moment to recall how they are defined and dwell on them a bit further. To do so we’ll have to recall, in turn, a series of definitions and results in basic group theory.
Definition 1 We say that a group is perfect if it is equal to its commutator subgroup , generated by all commutators with . A group is said to act on a set if there is a group homomorphism . The kernel of this homomorphism is said the be the kernel of the action (it is of course a normal subgroup of ). We shall say the action is faithful if the group morphism is injective.
Definition 2 We usually denote by the image of under the automorphism associated to and by the subset and call it the -orbit of . We denote by the subset of elements that fix and call it the stabilizer of the point in . It is not hard to see the collection of distinct orbits of a fixed action partition , and that stabilizers are always subgroups. It is not hard to see, either, that . There is an explicit bijection: map to , and check this is a bijection. It is in fact a -set morphism, but we won’t be caring about this here.
Definition 3 We say a partition of is -stable if for every and , . In words: the action permutes the blocks of the partition. Evidently the trivial partition and the singleton partition are stable, as so is the partition under orbits. We say that an action is primitive if admits only two stable partitions, else we call the action imprimitive. In particular any primitive action on a set with more than one element must be transitive.
We leave it to the reader to verify that a transitive action is imprimitive if and only if there exists a subset of of at least two elements such that for any either or . One direction is easy, for the converse note that and their complement form a partition that is stable.
Definition 4 Given a group acting on a set , we have a canonical action on the product by . We say that a group action is two-fold transitive if acts transitively on under this canonical action. In words: for every pair of distinct pairs , and there exists such that and .
Having all the ingredients to do so, we may as well present our main tool to prove that the family consists also of simple groups:
Theorem (Iwasawa, 1941) Suppose a perfect group acts primitively on a set , and that for some the stabilizer contains a normal abelian subgroup whose conjugates generate . Let be the kernel of the action. Then the quotient group is simple.
To prove this, we aid ourselves in the following.
(1) Any two-fold transitive action is primitive.
(2) If a group acts primitively on a set and is not contained in the kernel of the action, acts transitively on .
(3) If a subgroup acts transitively on a set , then for any .
(1) Let be a proper subset of at least two elements . By two-fold transitivity there is some such that and , so that is nonempty and .
(2) Our set is partitioned into -orbits, and since is normal we have for any , which implies the partition is -stable. Since the partition is not that by singletons by hypothesis, we must have only one -orbit.
(3) This is rather easy, and is left as an exercise.
We now prove Iwasawa’s criterion. Let be normal. By , acts transitively on . Let be the stabilizer in the statement of the theorem. By above, we know that . Now is normal in and since the conjugates of generate , it must be that . It follows that is abelian, so that . .
Given a field , let denote the collection of one dimensional subspaces of , which we call the -dimensional projective space over . The linear group of (invertible) matrices of size over acts on the projective space by for a one dimensional subspace. We will content ourselves with finite dimensional vector spaces, as above. In the following series of lemmas we will gather some information on the special linear groups that will allow us to apply Iwasawa’s criterion to the previously defined action of on the projective space. Some minor details and calculations are omitted. The reader is strongly encouraged to carry them out in detail!
Lema 1. The kernel of the action just defined (for any !) is the subgroup of scalar matrices with which is the center of . The quotient group thus obtained is called the projective linear group , and the quotient associated to the restriction of this action to one of the special linear group is called the special projective group .
Proof It is clear any scalar matrix is in the kernel of the action. Suppose now that fixes every one dimensional subspace. For each we can write and , hence is diagonal if . If fix a basis and note that so that with . We have simply proved that every invertible linear transformation which has every vector as an eigenvector must be a scalar multiple of the identity.
Lema 2 The special linear group is generated by the elementary matrices .
Proof We can prove a stronger result, namely that if is a PID then is generated by the elementary matrices. This follows from the Smith Normal Form for matrices over principal ideal domains and the Whitehead lemma. Indeed, given unimodular we may find invertible products of elementary matrices plus permutation matrices such that is diagonal. If is the permutation matrix that interchanges row (or column) and , , so we can assume we have only elementary matrices and matrices of the form . But and , and commutes with every other elementary matrix. It follows we may gather the and using they are involutions obtain a diagonal matrix with products of elementary matrices, and hence simply show diagonal unimodular matrices are product of elementary matrices. The Whitehead lemma now says that is elementary, so we may multiply to the right to replace the first nonzero diagonal element by , and iterate to obtain for .
Lemma 3 Any special linear group is perfect unless it is or .
Proof If , we can use the Steinberg relations to write , so we can assume that . In this case , and then gives some such that . Taking transposes and recalling the Whitehead lemma gives what we wanted.
Lemma 4 The special linear group acts doubly transitively on the projective space if .
Proof We want to show that given two pairs of different lines, equivalently two pairs , of linearly independent vectors, there is a unimodular matrix that takes to . If we can always extend the to a basis , write and subsequently extend the matrix obtain one of determinant . If then the matrix has nonzero determinant , and we may then consider the matrix associated to the linear transformation that sends and in our basis, which has determinant .
Lemma 5 The stabilizer of in the special linear group contains a normal abelian subgroup whose conjugates generate the whole special group.
Proof The stabilizer of consists of matrices of the form
We can consider the group morphism that sends such matrix to the submatrix of . This has kernel the matrices of the form
which the reader can verify is abelian. Evidently any matrix is contained in , and the fact that with extended with the identity if necessary shows the conjugates of contain any matrix . Finally, the relation allows us to obtain the remaining elementary matrices from these two.
The reader is encouraged to check that and that are not simple. In fact there is a solvable series : this is intimately related to the fact we can solve any quartic equation by radicals and that there is a canonical procedure using what is called a resolvent cubic of (related to the Klein four group ) and the discriminant of the polynomial (which allows us to figure out the Galois group by knowing that of ).
In an upcoming post we will hopefully dwell into the Mathieu groups, and , which were studied by Émile L. Mathieu.