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Math. 5363, exam 1, solutions 1. Prove that every finitely generated abelian group is the homomorphic image of a free finitely generated abelian group. Let G be a finitely generated abelian group and let {g1 , g2 , ..., gn } ⊆ G be a set of generators. Let {x1 , x2 , .., xn } be a Z-basis of Zn . Then f : Zn 3 a1 x1 + a2 x2 + ... + an xn → a1 g1 + a2 g2 + ... + an gn ∈ G is a well defined surjective group homomorphism and G = f (Zn ). 2. Describe all the abelian groups of order 44, up to an isomorphism. The classification of the finitely generated abelian groups implies that a group of order 44 = 4 × 11 is isomorphic either to C4 × C11 or to C2 × C2 × C11 . 3. Show that the group of the characters of the additive Z is isomorphic to the multiplicative group T of the complex numbers of absolute value 1. Define the following map (3.1) Char(Z) 3 χ → χ(1) ∈ T. Since, for two characters χ and χ0 (χχ0 )(1) = χ(1)χ0 (1) the map (3.1) is a group homomorphism. Suppose χ is in the kernel of the homomorphism (3.1). Then χ(1) = 1 Then χ(n) = χ(1)n = z n for all n ∈ Z. Thus χ is a trivial character. Therefore the map (3.1) is injective. 1 2 Given z ∈ T define χ(n) = z n (n ∈ Z). Then, (3.1) maps this character to z. Thus (3.1) is surjective. 4. State Sylow’s Theorem. Look it up in the book. 5. Are the Sylow 2-groups in S4 abelian or not. Why? Let K ⊆ S4 be the subgroup generated by the cycle (1234). This group has 4 elements. Since |S4 | = 3 × 8, there is a Sylow 2-subgroup H ⊆ S4 containing K. Notice that for any permutation σ ∈ S4 , σ(1234)σ −1 = (σ(1)σ(2)σ(3)σ(4)). Hence, σ commutes with the cycle iff (σ(1)σ(2)σ(3)σ(4)) = (1234), which means that σ is a power of (1234). Thus there is no commutative extension of K in S4 . In particular H is not commutative. 6. Suppose G is a finite group, H is a subgroup, and a ∈ G is an element with an ∈ H for some integer n with GCD(n, |G|) = 1. Prove that a ∈ H. Since GCD(n, |G|) = 1, there are integers x, y such that xn + y|G| = 1. Hence, a = a1 = axn ay|G| = axn (a|G| )y = axn = (an )x ∈ H. 7. Let G be a group of order 2pq, where p and q are primes with 2 < p < q. Prove that if q + 1 6= 2p then the Sylow q-subgroup is normal. 3 The number of Sylow q subgroups of G is equal to qk + 1, where k is a non-negative integer. Furthermore, qk +1 divides 2p. Since 2 < p < q, we must have k = 0. Thus there is only one Sylow q subgroup. Since all Sylow q subgroups are conjugate, this unique subgroup is normal. 8. Let G be a non-abelian group of order 6. Prove that G is isomorphic to S3 . Let G be a non-abelian group of order 6. Since G is not abelian, it does not contain any element of order 6. Also, it can’t happen that every element other than 1 is of order 2. Therefore, there is element a ∈ G of order 3. This element generates the subgroup H = {1, a, a2 } ⊆ G of index 2. In particular, H is a normal subgroup. Since |G| = 2 × 3, there is a Sylow subgroup of G of order 2. In other words, there is an element b ∈ G of order 2. Since H does not contain any elements of order 2, we see that b ∈ G and b ∈ / H. The conjugation by b preserves H, but does not commute with a. Hence, bab−1 = a2 . Thus G = {1, a, a2 , b, ab, a2 b} and the following relation holds (8.1) ba = a2 b. 4 The relation (8.1) determines the multiplication table for G. (Explicitly, a · a = a2 , a · a2 = a3 = 1, a · b = ab, a · ab = a2 b, a · a2 b = b, a2 · a = 1, a2 · a2 = a, a2 · b = a2 b, a2 · ab = b, a2 · a2 b = ab, b · a = a2 b, b · a2 = a2 ba = a4 b = ab, b · b = 1, b · ab = a2 b2 = a2 , b · a2 b = a2 bab = a4 b2 = a, ab · a = aa2 b = b, ab · a2 = aa2 ba = ba = a2 b, ab · b = a, ab · ab = aa2 bb = 1, ab · a2 b = abba = a2 , a2 b · a = ab, a2 b · a2 = b, a2 b · b = a2 , a2 b · ab = a, a2 b · a2 b = 1.) Thus up to a group isomorphism, there is only one non-abelian group of order 6. In particular G is isomorphic to S3 .