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2 Math 411: Abstract Algebra Midterm - Page 2 of 8 10/29/18 1. (20 points) What are the definitions of the following terms? 1. Group. Solution: A group is a set G together with a binary operation : G G G satisfying: (a) (associativity) For all g 1, g 2, g 3 G, g 1 (g 2 g 3 ) = (g 1 g 2 ) g 3. (b) (existence of an identity element) There is an element e G such that for all g G, g e = e = e g. (c) (existence of inverses) For each g G there is an element h G such that 2. Normal Subgroup. g h = e = h g. Solution: A subset H G is called a normal subgroup of G if the following conditions are satisfied: (a) H is closed under the group law on G, i.e. for all h 1, h 2 H the product h 1 h 2 H. (b) H is a group under the restriction of the group law from G. (c) For all g G and h H the conjugate ghg 1 H.

3 Math 411: Abstract Algebra Midterm - Page 3 of 8 10/29/18 2. (18 points) Let G and H be groups. Let f : G H be a group homomorphism. Show that ker(f) is a normal subgroup of G. Show that Im(f) is not necessarily a normal subgroup of H. Solution: Let h 1, h 2 ker(f), then f(h 1 h 2 ) = f(h 1 )f(h 2 ) = ee = e; thus h 1 h 2 ker(f). Since f(e) = e, we have e ker(f). Finally since f(h 1 1 ) = f(h 1) 1 = e 1 = e, we have h 1 1 ker(f). We conclude that ker(f) is a subgroup of G. If g G, then f(gh 1 g 1 ) = f(g)f(h 1 )f(g) 1 = f(g)ef(g) 1 = f(g)f(g) 1 = e. Hence, gh 1 g 1 ker(f). We conclude ker(f) is a normal subgroup. It is not necessarily the case that Im(f) is a normal subgroup. Consider the case where G = Z/2Z and H = S 3 and f : G H is the homomorphism mapping 1 to the 2-cycle (1 2). The image of f is not normal, as conjugation by H = S 3 transitively permutes the three cyclic subgroups generated by a 2-cycle.

4 Math 411: Abstract Algebra Midterm - Page 4 of 8 10/29/18 3. (18 points) Let G be a group and H be a subgroup. Show that if [G : H] = 2, then H is a normal subgroup of G. Solution: Let ϕ : G S G/H = S2 be the homomorphism given by left multiplication on cosets. We claim ker(ϕ) = H. We know that ker(ϕ) = Core(H) H. Conversely, observe that if g G, then the image ϕ(g) is either a 2-cycle (and g fixes no point of G/H) or g ker(ϕ). Since the elements h H fix the coset H, it must be the case that H ker(ϕ). We conclude H = ker(ϕ) is normal.

5 Math 411: Abstract Algebra Midterm - Page 5 of 8 10/29/18 4. (45 points) Determine if the following statements are true or false. No justification is needed. F All elements of order n in S n are conjugate. Explaination: The group S 6 contains two conjugacy classes of elements of order 6. The 6-cycles and the elements of cycle type {2,3}. Explaination: Explaination: T All finite groups of order n are isomorphic to a subgroup of S n. G acts on G by left multiplication. F Two finite groups are isomorphic if they have the same number of elements. Z/4Z and Z/2Z Z/2Z have the same order but are not isomorphic. T For any pair of groups G and H there is a homomorphism ϕ : G H. Explaination: Between any pair of groups G and H there is the trivial homomorphism defined by mapping all elements of G to e. F A homomorphism ϕ : S n S m is either injective or has cyclic image of order at most 2. Explaination: There is a surjective homomorphism from S 4 S 3. If n 4, the claim is true. T Every cyclic group is abelian. Explaination: Powers of a generator commute. F If G and H are non-trivial cyclic groups, then G H is non-cyclic. Explaination: The group Z/2Z Z/3Z is cyclic. It is generated by (1, 1). Explaination: F The automorphism group of a cyclic group is cyclic. Aut(Z/8Z) = Z/2Z Z/2Z. F If G is a group and σ, τ G. The order of the product στ lcm( σ, τ ). Explaination: Every element of S n is a product of elements of order 2 (2-cycles.) F A subset S G of a finite group which is closed under the group operation is a subgroup. Explaination: The set S may be empty. True if S is non-empty. T If G is a finite group, the order H of a subgroup H G divides G. Explaination: Lagrange s Theorem F A subgroup of an infinite group has order 1 or. Explaination: The group R contains {1, 1}. The group GL n (R) contains S n. F The order of a permutation σ S n divides n. Explaination: A 2-cycle in S 3 is a counterexample. Explaination: F The centralizer C G (x) of an element x G is an abelian subgroup of G. C G (e) = G F A subgroup G S n is transitive if and only if n divides G. Explaination: There is a non-transitive embedding of Z/2Z Z/2Z in S 4 generated by a pair of disjoint 2- cycles.

6 Math 411: Abstract Algebra Midterm - Page 6 of 8 10/29/18 5. (24 points) Let x := (1 2)(3 4)(5 6). 1. What is the size of the conjugacy class of x in S 6? Solution: The conjugacy class of x in S 6 is the set of elements with cycle type {2, 2, 2}. There are 6! 3!2 3 = 15 elements with this cycle type, as there are 6! ways to assign distinct values to a 1,..., a 6 in (a 1 a 2 )... (a 5 a 6 ) from the set {1,... 6}, and for each assignment of a 1,..., a 6 reordering the cycles (3! ways) and reordering the values contained in the cycles (2 3 ways) define the same element of S What the order of the centralizer of x in S 6? Solution: The centralizer of x is the stablizer of x under the conjugation action. Hence by the orbit stabilizer theorem: C S6 (x) = S 6 / {conjugates of x in S 6 } = 3!2 3 = Find an explicit element of order 3 in C S6 (x). Solution: We find an element that cyclically permutes the cycles appearing in x. Recall that if g S 6, then gxg 1 = (g(1) g(2))(g(3) g(4))(g(5) g(6)). Let g := (1 3 5)(2 4 6). Then gxg 1 = (3 4)(5 6)(1 2) = x and g has order 3.

7 Math 411: Abstract Algebra Midterm - Page 7 of 8 10/29/18 6. (25 points) 1. Let A be a finite, abelian group. Show that S n contains a transitive subgroup isomorphic to A if and only if n = A. Here are two proofs. Proof 2 differs from Proof 1 as it essentially contains a proof of the orbit-stabilizer theorem. In proof 1 that work is hidden, as we appeal to the general theory of G-sets. Solution 1: A transitive A-set X is isomorphic (as an A-set) to A/H for some subgroup H A. It follows that a transitive A-set must have size X = A/H A. The kernel of the associated map φ : A S X is Core(H), the largest normal subgroup contained in H. When A is abelian every subgroup is normal. Hence, Core(H) = H. Therefore, if A is abelian, then A is isomorphic to its image under φ if and only if H = {e}, or equivalently if and only if X = A (as an A-set). Applying this to the special case that X = {1, 2, 3,..., n} and A S n acts through the restriction of the action of S n to A, we obtain A is isomorphic to a transitive subgroup of S n only if n = A. Conversely, we have that if n = A, then A is isomorphic to a transitive subgroup of S A = Sn. Solution 2: Assume that A S n is a transitive subgroup. Consider the map ev 1 : A {1, 2,... n} defined by a a(1). We claim ev 1 is a bijection. The map ev 1 is surjective because A is transitive. To see that ev 1 is injective consider elements a, a A such that ev 1 (a) = a(1) = a (1) = ev 1 (a ), or equivalently such that a 1 a (1) = 1. We claim a = a, which is equivalent to showing a 1 a (k) = k for all k {1,... n}. Let x A such that x(1) = k. Then a 1 a (k) = a 1 a (x(1)) = x(a 1 a (1)) = x(1) = k, where the second equality holds because A is abelian. We conclude a = a and ev 1 is a bijection. Hence, A = n. Conversely if A = n, then the transitive action of A on A by left multiplication yields an injective homomophism A S A. Identifying A with {1,..., n} as a set, gives an isomorphism S A Sn. The image of A under the composition A S A Sn is a transitive subgroup of S n isomorphic to A.

8 Math 411: Abstract Algebra Midterm - Page 8 of 8 10/29/18 2. Do there exist non-abelian groups with this property? Yes. A group G will have the property that G is isomorphic to a transitive subgroup of S n if and only if n = G exactly when it has the property that for subgroups H G, the group Core(H) = {e} if and only if H = {e}. The group Q 8 has this property as Core(H) = 1 for any non-trivial subgroup H Q 8, so Q 8 is isomorphic to a transitive subgroup of S n if and only if n = 8.

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