Section III.15. Factor-Group Computations and Simple Groups

Transcription

1 III.15 Factor-Group Computations 1 Section III.15. Factor-Group Computations and Simple Groups Note. In this section, we try to extract information about a group G by considering properties of the factor group (or quotient group ) G/N. We also introduce a new class of groups, which have received much attention. Example. Let G be a group with identity e. Consider the normal subgroup N = {e}. Since N has one element, then all cosets of N have one element in fact, the coset gn = {ge} = {g}. So the factor group G/N is isomorphic to G. Example. For any group G, the factor group G/G is isomorphic to {e}. This is because there is only one coset of G (treating G as a normal subgroup of G) and so G/G is a group with one element. Note. The above two examples are extreme cases of collapse of the cosets of G down to elements of G/N. If G is a finite group and N {e} is a normal subgroup, then G/N is a smaller group than G and so may have a more simple structure than G (using quotation marks when referring to the text s wording). There is a correspondence between the multiplication of cosets in G/N and the multiplication of their representatives in G (as given in Corollary 14.5).

2 III.15 Factor-Group Computations 2 Lemma. If G is a finite group and N is a subgroup of G where N = G /2, then N is a normal subgroup of G. Proof. Since all cosets of N must be the same size and the cosets partition G, then there are only two cosets of N, namely N and an where a G \ N. Now for any g G, (1) if g N then gn = Ng = N, and (2) if g G \ N then gn = Ng since this is the only coset of N other than N itself. So gn = Ng for all g G, and N is a normal subgroup. Example Consider S n /A n. Recall that the symmetric group S n has order n! and the alternating group A n has order n!/2. So, by Lemma, A n is a normal subgroup and hence S n /A n is defined. So there are two cosets in S n /A n. As in the proof of Lemma, for any σ S n, (1) if σ A n (i.e., σ is an even permutation) then σa n = A n, and (2) if σ S n \A n (i.e., σ is an odd permutation) then σa n = S n \A n. So the two cosets correspond to the even permutations and the odd permutations. If we denote cosets as E n and O n, we get the multiplication table for S n /A n as: E n O n E n E n O n O n O n E n As the text comments, the factor group S n /A n does not give us details about products of specific elements of S n, but it does give us information about products of types of elements of S n (even and odd, in this case).

3 III.15 Factor-Group Computations 3 Example Falsity of the Converse of the Theorem of Lagrange. We have claimed in the past that the alternating group A 4 (of order 4!/2 = 12) does not have a subgroup of order 6. Recall that Lagrange s Theorem states that the order of a subgroup divides the order of its group. This example shows that there may be divisors of the order of a group which may not be the order of a subgroup. Suppose, to the contrary, that H is a subgroup of A 4 of order 6. By Lemma, H must be a normal subgroup of G. Then A 4 /H has only two elements, H and σh where σ A n \ H. Since A 4 /H is a group of order 2, then it is isomorphic to Z 2 and the square of each element (coset) is the identity (H). So H H = H and (σh) (σh) = σ 2 H = H. So if α H then α 2 H and if β / H (then β σh) then β 2 H. So, the square of every element of A 4 is in H. But in A 4 we have (1,2,3) = (1,3,2) 2 and (1,3,2) = (1,2,3) 2 (1,2,4) = (1,4,2) 2 and (1,4,2) = (1,2,4) 2 (1,3,4) = (1,4,3) 2 and (1,4,3) = (1,3,4) 2 (2,3,4) = (2,4,3) 2 and (2,4,3) = (2,3,4) 2. So all 8 of the above (distinct) permutations are in H. This is a contradiction, since we assumed H = 6. Therefore no such H exists.

4 III.15 Factor-Group Computations 4 Exercise Classify Z Z/ (0, 1) according to the Fundamental Theorem of Finitely Generated Abelian Groups. Solution. Notice (0,1) = {0} Z = H and the cosets are (x,y) + {0} Z = {x} (Z + y) = {x} Z for all (x,y) Z Z. Now {x} Z is isomorphic to Z (define φ({x} Z) = x and φ is an isomorphism). So Z Z/ (0, 1) is isomorphic to Z. Note. The previous exercise foreshadows the following result. Theorem Let G = H K be the direct product of groups G and H. Then H = {(h,e) h H} is a normal subgroup of G. Also, G/H is isomorphic to K. similarly G/K is isomorphic to H where K = {(e, k) k K}. Note. The humble Theorem 15.8 reveals why we have been studying factor groups because they allow us to FACTOR GROUPS (sometimes, at least)! It follows that if H is a normal subgroup of G = H K then G = (G/H) H since H = H. So the reason we have addressed factor groups is so that we can take a given group and factor (or better yet, decompose ) it into smaller groups. We know from the statements at the beginning of this section that H 1 = {e} and H 2 = G are normal subgroups of G, but they do not yield interesting factor groups because G/H 1 = G and G/H2 = {e}. So very soon our attention will turn to normal subgroups which are neither trivial nor improper. Theorem A factor group of a cyclic group is cyclic.

5 III.15 Factor-Group Computations 5 Exercise Classify (Z 4 Z 8 )/ (1,2). Solution. Notice (1,2) = {(1,2),(2,4),(3,6),(0,0)} = H and so Z 4 Z 8 / (1,2) = 4 8/4 = 8. All groups are abelian, including the factor group. The abelian groups of order 8 are Z 2 Z 2 Z 2, Z 2 Z 4, and Z 8 (we don t distinguish between Z 2 Z 4 and Z 4 Z 2 ). We try to determine which of these three choices is correct by considering orders of elements of the factor group. Consider the coset (0, 1) + H. The order of this coset is 8 since: ((0,1) + H) + ((0,1) + H) + + ((0,1) + H) = (0,k) + H }{{} k times and the smallest value of k for which this yields the identity is k = 8. Since the only choice from the three groups which has elements of order 8 is Z 8, then the factor group must be isomorphic to Z 8. Note. We now introduce a new class of groups, which have a rich history. Details are given in the supplement to this section. Definition A group is simple if it is nontrivial and has no proper nontrivial normal subgroups. Example. For any prime p, the group Z p is simple because it has no proper nontrivial subgroups by Lagrange s Theorem (and therefore has no proper nontrivial normal subgroups).

6 III.15 Factor-Group Computations 6 Note. We now see that simple groups play a role similar to that which prime numbers play in number theory! That is, if G is simple then the only factor groups of G are G/G = {e} and G/{e} = G. Notice that we cannot say that simple groups are equivalent to indecomposable groups. For example, Z p n, where p is prime, is indecomposable but it is not simple when n > 1. We also cannot say that if H is a normal subgroup of G, then G = (G/H) H (as illustrated by considering indecomposable Z 4 and normal subgroup Z 2 = {0,2}). The nature by which group G relates to its normal subgroups will be spelled out in more detail in the Jordan-Hölder Theorem (Theorem 35.15). Theorem The alternating group A n is simple for n 5. Note. The proof we give of Theorem is not elegant! It requires a clunky argument based on several cases. We defer it until the end of this section and give it as a supplement to these notes. Note. We are interested in when G/N is simple for normal subgroup N of G. When this occurs, we have sort of found a maximal factor of G. Note. Theorem showed us that a group homomorphism sort of maps subgroups back and forth. The following result shows that the subgroup property of normality is also mapped back and forth by homomorphisms.

7 III.15 Factor-Group Computations 7 Theorem Let φ : G G be a group homomorphism. If N is a normal subgroup of G, then φ[n] is a normal subgroup of φ[g]. Also, if N is a normal subgroup of φ[g], then φ 1 [N ] is a normal subgroup of G. Note. The proofs of the two parts of Theorem are Exercises and Definition A maximal normal subgroup of a group G is a normal subgroup M not equal to G such that there is no proper normal subgroup N of G properly containing M. Theorem M is a maximal normal subgroup of G if and only if G/M is simple. Note. Personally, I find the proof in the text hard to follow (it may just be me, since the previous editions of Fraleigh present the same proof as given in the 7th edition). The proof depends on the use of the homomorphism of Theorem 14.9 which maps G to G/M. In addressing both M is maximal and M/G is simple we must be concerned with nontrivial and proper subgroups. Since a homomorphism is not one to one, this requires special attention to the details. We now present a proof with many more details than the proof given in Fraleigh (though still based on Fraleigh s general argument s). Notice that both claims in the theorem are established by contradiction.

8 III.15 Factor-Group Computations 8 Proof of Theorem Let M be a maximal normal subgroup of G. Define γ : G G/M as γ(g) = gm. Then by Theorem 14.9, γ is a homomorphism with Ker(γ) = M. Suppose G/M is not simple and that N is a proper nontrivial normal subgroup of G/M. Then by Theorem 15.16, γ 1 [N ] is a normal subgroup of G. Since N is nontrivial then N {e} = {M} (remember M is the identity in G/M). Also, we have Ker(γ) = M, so Ker(γ) N and φ 1 [N ] M. But {e} = {M} N and so M φ 1 [N ]. So φ 1 [N ] properly contains M. Since N is a proper subgroup of G/M, then it contains some but not all of the cosets of M. Since the cosets of M partition G (Section II.10), then φ 1 [N ] contains some but not all elements of G. That is, φ 1 [N ] is a proper subgroup of G. Therefore, φ 1 [N ] is a nontrivial proper normal subgroup of G which properly contains M. But this contradicts the maximality M. So no such N exists and G/M is simple. Now let G/M be simple. Suppose that M is not a maximal normal subgroup and that there is a proper normal subgroup N of G properly containing M (i.e., M N G). Then by Theorem 15.16, γ[n] is a normal subgroup of γ[g] = G/M where γ is as defined above. Now Ker(γ) = {M} (remember that M is the identity in G/M). So γ[n] is a nontrivial normal subgroup of G/M. Now we show that γ[n] is a proper subgroup of G/M. Recall that γ maps elements of G to cosets of M. So the only way that γ[n] = G/M is if N contains an element of each coset of M. But we assumed M N, and so if N contains an element of each coset of M, then (since N is a group and so is closed under the binary operation) N must contain all cosets of M that is, M = G. But this contradicts the choice of N as a proper subgroup of G. Hence N does not contain an element from each coset of M and γ[n] G/M. That is, γ[n] G/M. Therefore, γ[n] is a proper nontrivial

9 III.15 Factor-Group Computations 9 subgroup of G/M. But this contradicts the fact that G/M is simple. Therefore, no such N exists and M is a maximal normal subgroup of G. Definition. For group G, define the center of G as Z(G) = {z g zg = gz for all g G}. Note. The letter Z in the notation above is from the German zentrum for center. In Exercise 5.52, it is shown that Z(G) is an abelian subgroup of G. Of course, if G itself is abelian, then Z(G) = G. Since Z(G) is abelian, then it is a normal subgroup of G. Definition. For group G, consider the set C = {aba 1 b 1 a,b G}. C is the commutator subgroup of G. (C is shown to in fact be a group below.) Theorem Let G be a group. Then the set C = {aba 1 b 1 a,b G} is a subgroup of G. Additionally, C is a normal subgroup of G. Furthermore, if N is a normal subgroup of G then G/N is abelian if and only if C N. Revised: 10/25/2014