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Home , Inner automorphism

MATH 412 PROBLEM SET 11, SOLUTIONS

Reading: • Hungerford 8.4, 9.1, 9.2

Practice Problems: These are “easier” problems critical for your understanding, but not to be turned in. Be sure you can do these! You may see them or something very similar on Quizzes and Exams. Section 8.4: 9, 19, 21, Section 9.1: 1, 2, 3, 4, 5 Section 9.2: 1, 3

Problems due Friday. Write out carefully, staple solutions to A, B and C in one pack, and stable solutions to D and E in another. Write your name and section number on each.

(A) Let F be a field. Consider the SL2(F) of 2 × 2 matrices with entries in F and determinant one. 1 (1) Show that the of SL2(F) is Z = {±I}, where I is the 2 × 2 identity matrix. (2) Let PSL2(F) be the SL2(F)/Z. Compute the order of PSL2(Z7) as a group. 2 (3) For each of the following, compute its order as an element of PSL2(Z7). 0 2 • A = ± 3 0 1 1 • B = ± 6 0 1 3 • C = ± 0 1

Note: PSL2(Z7) is the smallest simple group that is neither cyclic nor isomorphic to An for some n.

Solution. a b 1 1 1 0 (1) Let A = be an arbitrary element of SL ( ), B = , and C = . We c d 2 F 0 1 1 1 −c a − d  b 0  compute AB − BA = and AC − CA = . If A ∈ Z, then AB − BA = 0 c d − a −b AC − CA = 0, so b = c = a − d = 0. Since A has determinant one, this forces A = ±I. Conversely, ±I both clearly commute with any matrix. 2 2 (2) We have computed the order of GL2(Z7) as (7 − 1)(7 − 7) = 2016 based on a previous homework. SL2(Z7) is a subgroup of index 7 − 1 = 6, so its order is 2016/6 = 336 by Lagrange’s theorem. Then, PSL2(Z7) = SL2(F)/Z, where Z has order 2, so the order of PSL2(Z7) is equal to the index of Z, which is 336/2 = 168, by Lagrange’s theorem again.

1Hint: First show that the center consists of scalar matrices aI by computing AB − BA when A is arbitrary and 1 1 1 0 B = or B = . 0 1 1 1 2That is, do not compute its size as a coset. 1 MATH 412 PROBLEM SET 11, SOLUTIONS 2

(3) We note that the order of Ng in G/N is the least n (if one exists) such that (Ng)n = Ngn = N, which is the least integer n (if one exists) such that gn ∈ N. • A2 = −I, so the order of NA is 2. • B2 6= ±I, and B3 = −I, so the order of NB is 3. • C7 = I, and Cn 6= ±I for n < 7, so the order of NC is 7.

(B) Let G be an , not necessarily finite. (1) Let T be the set of elements of G of finite order. Prove that T is a subgroup of G. (2) Show that every element of G/T has infinite order.

Solution. (1) We note first that e ∈ T . If t, t0 ∈ T , then there are m, n such that tm = t0n = e. Then (tt0)mn = tmnt0mn, using the abelian property, and tmn = (tm)n = en = e and t0mn = (tn)m = em = e, so (tt0)mn = e. Finally, the order of an element is equal to the order of its inverse, so T is closed under inverses. It follows that T is a subgroup. (2) Let T g ∈ G/T . If g has finite order in G then g ∈ T , so T g = T , which is the identity element. If T g 6= T , then g∈ / T . Suppose that (T g)n = T . Then gn ∈ T , so gnm = (gn)m = e for some m, so g ∈ T . Thus, the only element of finte order is the identity.

(C) Give an example of a group G where the set of elements of finite order does not form a subgroup of G.

−1 1 0 1 Solution. Let G = GL ( ). Let M = and N = . Note that M 2 = N 2 = I. However, 2 R 0 1 1 0 1 −1 MN = , which has characteristic polynomial t2 − t + 1. This characteristic polynomial has 1 0 distict roots, neither of which is a root of unity, so no power of MN is the identity.

(D) Let P be the partition function: P(n) is the number of ways to write the positive integer n as sum of positive integers. For example, P(2) = 2, since 2 = 2 and 2 = 1 + 1, and P(4) = 5: 4 = 3 + 1 = 2 + 2 = 2+1+1=1+1+1+1. Express the number of abelian groups of order n, up to isomorphism, in terms of values of the partition function. Compute the number of abelian groups, up to isomorphism, of order 1000.

a1 at Solution. From the structure theorem, we know that the abelian groups of order n = p1 ··· pt are in bjiection with the number of ways to write n as a product of prime powers, which in turn is in a1 at with the number of ways to write p1 as a product of powers of p1,..., pt as a product of powers of pt simultaneously. It follows that the number is P(a1) ···P(at).

(E) An of a group G is a homomorphism φ : G → G that is invertible. An automorphism φ is an inner automorphism if there exists some h ∈ G such that φ(g) = hgh−1 for all g ∈ G. (1) Prove that the of G form a group Aut(G) under composition. (2) Prove that the inner automorphisms of G form a subgroup Inn(G) ≤ Aut(G). (3) Use the first isomorphism theorem to show that Inn(G) ∼= G/Z(G). MATH 412 PROBLEM SET 11, SOLUTIONS 3

(4) Suppose G is a generated by g ∈ G of order n. Show that there is a homomorphism φ : G → H for every choice of h ∈ H whose order divides n. (5) Let p be a prime. What is Aut(Zp)? (6) Let p be a prime. What is Aut(Zp × Zp)?

Solution. (1) Let φ and ψ be automorphisms of G. We need to check that φ ◦ ψ is also an automorphism. But the composition of group homomorphisms is a , and a composition of is a bijection, so φ ◦ ψ is an isomorphism from G to G. We then need to verify that it satisfies the axioms of a group. It is associative, because composition of any kind of functions always is (Proposition from the Appendix). It has an identity, namely the identity map. And by definition, being an automorphism, it has an inverse. So AutG is a group. (2) To check the inner automorphisms are a subgroup, first note that since the identity map is conjugation by e, Inn(G) is nonempty. Conjugation by g followed by conjugation by h is the same as conjugation by gh; we proved this when we showed that conjugation gives a . Similarly, conjugation by g−1 is the to conjugation by g, so Inn(G) is a subgroup. (3) Consider the map ψ : G → Aut(G) given by ψ(g) is conjugation by g. Since conjugation by g followed by conjugation by h is the same as conjugation by gh, this map is a homomorphism. By definition of Inn(G), it is surjective. Conjugation by g is trivial if and only if g ∈ Z(G), since ghg−1 = h for all h ∈ G is equivalent to gh = hg for all h ∈ G. Thus, the map ψ : G → Aut(G) has Z(G) and image Inn(G). The desired isomorphism is given by the first isomorphism theorem. (4) Let h have order dividing n. Then the map φ : G → H given by φ(ga) = ha is a homomorphism. Conversely, if φ : G → H is a homomorphism, φ(g)n = φ(gn) = e so the order of φ(g) must divide n. Consequently, there is a bijection between the homomorphisms from G to H and elements of H with order dividing n. (5) Let φ : Zp → Zp be an automorphism. Then φ([a]p) = aφ([1]p), so φ is determined by the image of [1]p. Thus, each homomorphism is one of {φ[0], φ[1], . . . , φ[p−1]}, where φi([a]p) = [ia]p for all a. Note that φ[j] ◦ φ[i]([a]p) = [ija]p = φ[ij]([a]p) for each a, so φ[j] ◦ φ[i] = φ[ij]. Thus, if × [i] ∈ Zp , then φ[i] is invertible, hence an automorphism. Clearly φ[0] is not injective, so it is × not an automorphism. Now, the map Zp → Aut(Zp) given by [i] 7→ φ[i] is clearly a bijection, and by the computation above, it is a homomorphism, hence is an automorphism. (6) Let φ : Zp × Zp → Zp × Zp be a homomorphism. Write elements of Zp × Zp as column vectors a . Using the definition of homomorphism, we see that b a  1 0 1 0 φ = φ a + b = aφ + bφ b 0 1 0 1 so each homomorphism φ is a linear transformation, hence agrees with multiplication by a matrix φ(v) = Mv for some 2 × 2 matrix M. Let φM correspond to the homomorphism φM (v) = Mv. From linear algebra, we know that there is an inverse function to φM if and only if there is an inverse matrix to M, if and only if M ∈ GL2(Zp). Thus, φM ∈ Aut(Zp ×Zp) if and only if M ∈ GL2(Zp). Thus, we have a bijective map ψ : GL2(Zp) → Aut(Zp × Zp) given by ψ(M) = φM . It remains only to see that ψ is a homomorphism.

ψ(MN)(v) = φMN (v) = (MN)v = M(Nv) = φM (φN (v)) = (φM ◦ φN )(v) = (ψ(M) ◦ ψ(N))(v) for all v, so ψ(MN) = ψ(M) ◦ ψ(N). Thus, ψ is an isomorphism.