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Math 412. Simple Groups

DEFINITION: A G is simple if its only normal are {e} and G. Simple groups are rare among all groups in the same way that prime numbers are rare among all . The smallest

non- is A5, which has 60.

THEOREM 8.25: A abelian group is simple if and only if it is finite of prime order.

THEOREM: The Alternating Groups An where n ≥ 5 are simple. The simple groups are the building blocks of all groups, in a sense similar to how all integers are built from the prime numbers. One of the greatest mathematical achievements of the Twentieth Century was a classification of all the finite simple groups. These are recorded in the Atlas of Simple Groups. The mathematician who discovered the last-to-be-discovered finite is right here in our own department: Professor Bob Greiss. This simple group is called the because its order is so big—approximately 8 × 1053. Because we have classified all the finite simple groups, and we know how to put them together to form arbitrary groups, we essentially understand the structure of every finite group. It is difficult, in general, to tell whether a given group G is simple or not. Just like determining whether a given (large) is prime, there is an algorithm to check but it may take an unreasonable amount of time to run.

A.WARM UP. Find proper non-trivial normal subgroups of the following groups: Z, Z35, GL5(Q), S17, D100. Are these groups simple?

Because Z is abelian, all subgroups are normal. So the even integers form non-trivial normal of Z. The subgroup h7i = {0, 7, 14, 21, 28} is normal. The subgroup SL5(Q) is normal in GL5(Q) because it is a . It is the kernel of the determi- nant map. The subgroup A17 is index two, hence normal, in S17. The subgroup R100 is index two, hence normal, in D100

B.EASY PROOFS. Let G be an arbitrary group. (1) Prove that if G is simple, then every non-trivial1 G → H is injective. (2) Prove that if G is simple, then every surjective homomorphism G → H is an . (3) If G = H × K, where neither H nor K is trivial, then G is not simple.

(1) To check injective, we can check the kernel is trivial. The kernel is a of the source G. Since G is simple, ker = {e} or G. Since the map is non-trivial, ker 6= G. So the kernel is trivial, and the map is injective. (2) Follows immediately, since the map in injective and surjective. (3) The subgroup H × {eK } is easily seen to be normal.

C.PROOFOF THEOREM 8.25 Let G be an arbitrary simple abelian group. (1) Take any non-identity element g ∈ G. Prove that g generates G. [Hint: Consider the subgroup hgi.]

1By non-trivial, I mean the map is not sending every element to e. (2) Deduce that our simple abelian group G is cyclic. (3) Show that an infinite hhi is isomorphic to Z. [Hint: Define Z → hhi sending n 7→ hn]. (4) Conclude that our simple abelian group G must be finite. (5) Prove that a cyclic of order nm has a normal subgroup of order n. [Hint: if g generates, look at gm.] (6) Prove that a finite cyclic group is simple if and only if it has prime order. [Hint: Langrange] (7) Prove Theorem 8.25.

(1) Because G is abelian, every subgroup is normal. So hgi is a normal subgroup, and not trivial, since g 6= e. So hgi = G. (2) Immediate. (1) shows that G is cyclic, and in fact, generated by any non-trivial element. (3) Suppose G = hgi is infinite. Define φ : Z → G sending n 7→ gn. Clearly φ is a homomor- phism, since φ(a + b) = ga+b = gagb = φ(a)φ(b). It is surjective, because every element of hgi is of the form gn for some n ∈ Z. To check it is injective, we show the kernel is trivial. If n ∈ ker φ, then φ(n) = gn = e, which means that G is finite, a contradition. QED. (4) We know that Z is not simple (for example the even integers are a normal subgroup). So neither is any isomorphic group. Hence an infinite cyclic group is never simple, and our simple abelian group G (which we know is cyclic) must be finite. (5) Suppose G is cyclic of order mn. Then if g generates G, then gm generates a group with n elements, namely {gm, g2m, . . . , gnm = e}. (6) This follows immediately. If the order of G is not prime, then we can |G| = mn where m, n 6= 1, |G|. As above, G has a proper non-trivial normal subgroup of order n, so is not simple. Conversely, if the order of G is prime, then by Lagrange’s theorem, every non- identity element of G has order p, so generates all of G. So any non-trivial subgroup of G has an element of order p in it, and hence all elements of G are in it. (7) This follows from the steps above: if G is simple and abelian, we saw in (2) that is it cyclic, then in (4) that is finite, then in (6) that its order is p.

D.CONJUGACYCLASSES: Recall that G acts on itself by conjugation: g · h = ghg−1. The orbits of this action are called conjugacy classes, so that O(h) denotes the set of all elements of G conjugate to h.

(1) Let H be an arbitrary subgroup of a group G. Show that H ⊂ ∪h∈H O(h). (2) Show that if N is a normal subgroup and n ∈ N, then O(n) ⊂ N. (3) Show that H is normal if and only if H = ∪h∈H O(h). By throwing away repeated orbits, we can say that a subgroup is normal if and only if it is a disjoint union of conjugacy classes. (4) Discuss and review with your workmates what was proved about conjugacy classes in Sn on Problem Set 11: each in Sn consists of all of the same “cycle type.” Why? (5) There are three disjoint conjugacy classes in S3. List out the elements of each. (6) We know An is normal in Sn because it has index two. Write A3 explicitly as a union of disjoint conjugacy classes, as guaranteed by (3). (7) Write the normal subgroup A4 explicitly as a union of disjoint conjugacy classes, as guaranteed by (3). This is easiest if you remember the definition of An.

−1 (1) Take arbitrary h ∈ H. Note that e · h = ehe = h ∈ O(h). So H ⊂ ∪h∈H O(h). (2) Take n ∈ N. For all g, we have gng−1 ∈ N, by definition of normal. So O(n) ⊂ N for all n ∈ N. (3) This follows from (1) and (2). −1 (4) In Sn, take any cycle (i1 i2 . . . it). We have seen that σ · (i1 i2 . . . it) = σ(i1 i2 . . . it)σ = −1 (σ(i1) σ(i2) . . . σ(it)). The same is true for products of cycles: if τ1, . . . τr are cycles, then σ(τ1, . . . τr)σ = −1 −1 −1 σ(τ1)σ σ(τ2)σ ··· σ(τr)σ s So, when we act by conjugation on a product of disjoint cycles, we get back a product of the same type of disjoint cycles. Also, we can take any product of disjoint cycles to any other of the same type. (5) {e}, {(1 2 3), (1 3 2)}, {(1 2), (1 3), (2 3)}. (6) A3 = {e} ∪ {(1 2 3), (1 3 2)}. (7) A4 = {e} ∪ O((1 2 3)) ∪ O((1 2)(3 4)).

E.TECHNIQUETO FIND NORMAL SUBGROUPS: Let (G, ◦) be an arbitrary finite group. Consider the decomposition of G into a disjoint union of distinct conjugacy classes G = O(g1) ∪ O() ∪ · · · ∪ O(gt). (1) Explain why finding a normal subgroup is the same as finding a (disjoint) collection of conjugacy classes which is closed under ◦. (2) There are five disjoint conjugacy classes in S4. Describe them. Also count the cardinality of each. (3) Show that the conjugacy class in S4 consisting of all 2-cycles is not closed under multiplication. (4) Every subgroup must contain the identity. Given that, what unions of conjugacy classes in S4 contain e and have the right size to possibly be a subgroup? (5) Show that the union of the two conjugacy classes O(e) and O((12)(34)) in S4 is closed under multiplication. Use this to find a normal subgroup of S4 which is a Klein 4-group. (6) By considering which unions of conjugacy classes in S4 are closed under composition, show that A4 and the Klein 4-group you found in (5) are the only normal subgroups of S4.

(1) A N is a subgroup if it is closed under composition and taking inverses. Because G is finite, we don’t have to check explicitly for inverses, because g−1 = gn−1 where n is the order of g, so that being closed under multiplication gives us inverses for free. So any subset closed under multiplication will be a subgroup. To be a normal subgroup, it must be a union of conjugacy classes, by D(3). (2) The conjugacy classes in S4 are these: {e},O((12)),O((123)),O((1234)),O((12)(34)), where, for example, O((12)) is the conjugacy class of (12), which is the set of all transpositions. Their cardi- nalities are 1, 6, 8, 6, 3, respectively. (3) The set O((1 2)) is not closed under multiplication because (1 2)(2 3) = (1 2 3), which is not of the same cycle-type, hence not in O((1 2)). (4) We can consider conjugacy classes of sizes 1 + 3 or 1 + 3 + 8. No other sum of the integers 1, 6, 8, 6, 3 (including 1) will divide 24 (other than 1+ 6+8+6+3 and 1 alone, but these give the whole group S4 and {e} respectively). So the only possibilities for proper, non-trivial subgroups of S4 are {e} ∪ O((1 2)(3 4)) and {e} ∪ O((1 2)(3 4)) ∪ O((1 2 3)). Thus there are most two proper non-trivial normal subgroups of S4. To check whether or not these are normal subgroups, we need to check whether or not each is closed under composition. (5) The set {e} ∪ O((1 2)(3 4)) = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is closed under multiplication, hence a normal subgroup. It is a Klein 4-group, because every element is order 2. (6) Similarly, the union {e} ∪ O((1 2)(3 4)) ∪ O((1 2 3)) consists of all even permutations, so is closed under composition. This is A4. There are no further normal subgroups.

F. NORMAL SUBGROUPSOF S5:

(1) List out and describe all conjugacy classes of S5. Count the number of elements in each. (2) Every normal subgroup must contain the identity. Given that, what unions of conjugacy classes have the right size to possibly be a subgroup? (3) Verify for each of the possibilities in the last part whether your union of classes is closed under composition. (4) Prove that A5 is the only proper, non-trival normal subgroup of S5.

(1) The homomorphism is the sign map and its kernel is An, which is a normal subgroup, like all kernels. (2) The conjugacy classes sort out by cycle type. The are (a) {e}, or cardinality 1. (b) O((12)), the set of all 2-cycles, of cardinality 10. (c) O((123)), the set of all 3-cycles, of cardinality 20. (d) O((1234)), the set of all 4-cycles, of cardinality 30. (e) O((12345)), the set of all 5-cycles, of cardinality 24. (f) O((12)(34)), the set of all products of disjoint transpositions, of cardinality 15. (g) O((123)(45)), of cardinality 20. Sanity check: 1 + 10 + 20 + 30 + 24 + 15 + 20 = 120 = 5!. (3) We need to add the numbers 1, 10, 15, 20, 20, 24, 30 to see if we can achieve a divisor of 120 = 23 · 3 · 5, while have 1 be one of the summands. Since the only odd factors of 120 are 1, 3, 5, 15, none of which can be achieved in this way, we know any such sum will be even. Since 1 is a summand, also 15 must be. We must consider only 16 + 10, 16 + 20, 16 + 40, 16 + 24, 12 + 10 + 24, 16 + 20 + 24 (since all other sums are greater than 60, so not proper divisors of 120. Of these, only 16 + 24 = 40 and 16 + 20 + 24 = 60 divide 120. This gives three possible sets that are unions of conjugacy classes and have cardinality dividing 120: (a) {e} ∪ O((12)(34)) ∪ O((12345)) (b) {e} ∪ O((12)(34)) ∪ ((12345)) ∪ O((123)(45)) (c) {e} ∪ O((12)(34)) ∪ O((12345)) ∪ O((123)) We need to check each to see if any is closed under composition. (a) The set {e} ∪ O((12)(34)) ∪ O((12345)) is not closed under composition, since (12)(34) ◦ (45)(12) = (12)(345)(12) is 3-cycle, hence not in these conjugacy classes. (b) The set {e} ∪ O((12)(34)) ∪ ((12345)) ∪ O((123)(45)) is not closed under composition for the same reason. (c) {e} ∪ O((12)(34)) ∪ O((12345)) ∪ O((123)) consists of all even transpositions, so is closed under com- position. We know anyway it has to be, because we know A5 is a normal subgroup, so this has to be A5. (4) Done.

BONUS: Prove that A5 is simple by using the techniques above. Be cautious! The conjugacy classes in A5 are not the same as those in S5. Do you see why, for σ ∈ A5, the conjugacy class of σ in A5 is typically smaller than the conjugacy class in S5?