(c)Karen E. Smith 2017 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. Classification of Finite Abelian Groups THEOREM 9.7. FUNDAMENTAL STRUCTURE THEOREM FOR FINITE ABELIAN GROUPS: Let G be a finite abelian group. Then G is isomorphic to a group of the form a1 × a2 × a3 × · · · × an Zp1 Zp2 Zp3 Zpn where p1; p2; : : : pn are (not necessarily distinct!) prime numbers. Moreover, the list of prime powers a1 a2 an appearing fp1 ; p2 ; : : : ; pn g is unique, up to re-ordering. A. WARM UP. (1) Let G be isomorphic to a product of cyclic groups of prime power order as in the Theorem above. What is the order of G? (2) Pretend you know nothing about groups of order 4 except that they are all abelian. What does the structure the theorem above tell us about the possible isomorphism types of (abelian) groups of order 4? Does this jibe with your knowledge about groups of order 4? (3) What does the structure theorem say about the group Z6? Do you know another proof? (4) What does the structure theorem say about an abelian group of prime order? Do you know another proof? ∼ (5) Prove that for arbitrary groups K; H, there is an isomorphism K × H −!= H × K. (6) Explain why the group a1 × a2 × a3 × · · · × an is isomorphic to any group obtained by Zp1 Zp2 Zp3 Zpn reordering the factors. a1 a2 an (1) jGj = p1 p2 : : : pn . (2) The only groups of order four, up to isomorphism, would be Z4 and Z2 × Z2. ∼ (3) Z6 = Z2 × Z3. This also follows from the Chinese remainder theorem. ∼ (4) Every abelian group of prime order must be = Zp. We also know this from Lagrange’s theorem, since the elements would have order p, hence generate the whole group, making it cyclic (and thus abelian). (5) The map (h; k) 7! (k; h) is easily seen to be invertible, and preserves the group structure. (6) Follows from previous: reordering the factors is invertible and preserves structure. B. ABELIAN GROUPS OF ORDER 8. (1) Suppose that G is abelian and has order 8. Use the Structure Theorem to show that up to isomor- phism, G must be isomorphic to one of three possible groups, each a product of cyclic groups of prime power order. (2) The Structure Theorem guarantees that the three possibilities for groups of order eight you found are non-isomorphic. How so? Verify, without using the theorem, that the three groups are non- isomorphic. [Hint: consider orders of elements.] (3) Explain why the parenthetical remark (not necessarily distinct) in the statement of the structure theorem is an important point to remember. (4) The abelian group U24 is order eight. To which of the three groups in (1) is it isomorphic. (5) Name two non-abelian groups of order eight. (1) The abelian groups of order 8 are (up to isomorphism): Z8, Z4 × Z2 and Z2 × Z2 × Z2: (2) We see that Z8 is the only group with an element of order 8, Z4 × Z2 is the only group with an element of order 4 but not 8. Also Z2 × Z2 × Z2 is the only group on the list with all non-zero elements of order 2. (3) Note that the primes are not distinct in this example! All are 2. (4) U24 = {±1; ±5; ±7; ±11g. All elements are order 2, except 1, so this group must be isomorphic to Z2×Z2×Z2: (5) D4 is non-abelian of order 8. So is the quaternion group Q, which we can describe as the 3 × 3 upper triangular matrices with 1’s on the diagonal and elements in Z2, under muliplication. C. PRACTICE USING THE STRUCTURE THEOREM (1) Determine the number of abelian groups of order 12, up to isomorphism. (2) Find a list of groups such that every abelian group of order 36 is isomorphic to exactly one group on your list. (3) For p prime, how many isomorphism types of abelian groups of order p5? (4) Decompose G = Z2 × Z12 × Z36 as (isomorphic to) a product of cyclic groups of prime power order. (5) Prove that an abelian group of order 100 with no element of order 4 must contain a Klein 4-group. (6) Prove that every abelian group of order 210 is cyclic. Hint: What does the structure theorem say about the number of isomorphism types of abelian groups of order 210?] (7) Using your experience with Problem C(4) as a guide, conjecture and prove a statement about abelian groups of order n, where n is square-free, meaning that no square integer divides n. (1) We can factor 12 into prime powers in two ways: 2 × 2 × 3 or 22 × 3. So there are two abelian groups of order 12, up to isomorphism, Z2 × Z2 × Z3 and Z4 × Z3: (2) We can factor 36 into prime powers in four ways: 2 × 2 × 3 × 3; 22 × 3 × 3, 2 × 2 × 32, and 22 × 32. So every abelian group of order 36 is isomorphic to one of the following four: Z2 × Z2 × Z3 × Z3; Z2 × Z2 × Z9; Z4 × Z3 × Z3; Z4 × Z9: (3) We have seven ways to write p5 as a product of prime powers: p5; p4 × p; p3 × p2; p3 × p × p; p2 × p2 × p; p2 × p × p × p; p × p × p × p × p: So there are 7 abelian groups of order p5 (up to isomorphism). ∼ ∼ (4) We can use the Chinese Remainder Theorem: Z2 ×Z12 ×Z36 = Z2 ×Z4 ×Z3 ×Z36 = Z2 ×Z4 ×Z3 ×Z4 ×9: (5) An abelian group of order 100 that does not contain an element of order 4 must be isomorphic to either Z2 × Z2 × Z25; or Z2 × Z2 × Z5 × Z5: The first contains the Klein 4-group f(a; b; 0) j a; b 2 Z2g, and the second contains the Klein 4-group f(a; b; 0; 0) j a; b 2 Z2g: (6) Since 210 = 2 × 3 × 5 × 7, any abelian group of order 210 is isomorphic to Z2 × Z3 × Z5 × Z7. In particular, the cyclic group Z210 is isomorphic to this one, and so every abelian group of order 210 is isomorphic to both. (7) If p1; : : : ; pt are distnct primes, then any group of order n = p1 ··· pt is cyciic. Proof: there is only one way to ∼ t decompose: G = Zp1 × ::: Zp . This means all abelian groups of order n are isomorphic to this one. But also Zn is abelian of order n, so all groups are isomorphic to it as well. D. IDENTIFYING PRODUCT GROUPS. Fix an abelian group G. Suppose that H and K are subgroups of G such that H \ K = feGg. (1) Prove that there is an injective homomorphism φ H × K ! G (h; k) 7! hk: (2) Prove that furthermore, if G is finite with jGj = jKjjHj, then G ∼= H × K. −1 (1) It suffices to show ker φ = feH×K g. Take arbitrary (h; k) 2 ker φ. So hk = eG. This means that k = h 2 H since H is closed under inverse (being a subgroup). So k 2 K \ H = feGg, which means that k = eG = eK , and also h = eG = eH . So the kernel is f(eH ; eK )g and φ is injective. (2) This follows from the pigeon hole principal. E. THE CHINESE REMAINDER THEOREM: If m and n are relatively prime, then there is a ring isomor- phism Zmn ! Zn × Zm. (1) Use the Chinese Remainder theorem to prove that any finite cyclic group G can be written as a product of cyclic groups of prime power order as in the Structure Theorem. (2) Suppose a finite abelian group G is isomorphic to a product of cyclic groups. Prove the existence part of the fundamental structure theorem for abelian groups. F. p-GROUPS. Let G be a finite abelian group of order n. Suppose that a prime p divides n. Let G(p) be the subset of G consisting of elements whose orders are a power of p (note that 1 = p0). (1) Prove that G(p) is a subgroup of G. (2) Suppose that p and q are distinct primes dividing n. Prove that G(p) \ G(q) = feGg. (3) In the case G = U21, compute jGj, and find G(p) for each p dividing jGj. Verify that their intersection is trivial. e1 et (4) Prove that if n = q1 : : : qt is the factorization of jGj into distinct prime powers, then there is an injective homomorphism G(q1) × G(q2) × · · · × G(qt) ,! G. [Hint: Generalize D.] (5) PROPOSITION: Suppose that pe is the largest power of p dividing the order of a finite abelian group G. Show that G(p) has order pe. Assuming this Proposition, show that the map in (4) is an isomorphism. e1 e2 et (6) Prove that if jGj = p1 p2 ··· pt is the factorization of jGj into distinct prime powers, then ∼ G = G(p1) × G(p2) × · · · × G(pt): (7) Prove the Proposition in (5) by contradiction. [Hint: If not, show that the quotient group G=G(p) contains a (non-identity) element whose order is a positive power of p. Show that any g 2 G representing this coset has order a power of p, so must be in G(p).] (1) Let G(p) be the subset of G consisting of elements whose orders are a power of p. Note that G(p) is non-empty a+b a+b a+b since jej = p0. Take g; h 2 G(p), and suppose jgj = pa and jhj = pb.