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Math 417 Exam 2 (Solutions) Prof. I.Kapovich April 13, 2009 Problem 1.[20 points] For each of the following statements indicate whether it is true or false. You DO NOT need to provide explanations for your answers in this problem. (1) If G is a group and H / G is a normal subgroup then for every g ∈ G and every h ∈ H we have gh = hg. (2) For every n ≥ 2 there exists an onto homomorphism f : Dn → Z2 . (3) For every n ≥ 2 and every σ ∈ Sn we have σ (10n)! = . (4) The subset GL(2, R) ⊆ M2 (R) is a subring of M2 (R). (5) If G is a group and H / G is a normal subgroup then for every g ∈ G the order of the element g ∈ G is equal to the order of the element gH in the factor group G/H. Answers: (1) False. The correct conclusion is that gH = Hg for every g ∈ G. However, if H is a normal subgroup of G, this does not necessarily imply that H is contained in the center of G. For example, this is not the case for AN / An (for n ≥ 3) and for SL(n, R) / GL(n, R) (for n ≥ 2). (2) True. The subgroup H = hai ≤ Dn has order n and index [Dn : An ] = = 2n n = 2 in Dn . Therefore H is normal in Dn and the factor group Dn /H is a cyclic group of order 2, which is therefore isomorphic to Z2 . The quotient homomorphism Dn → Dn /H is onto, and when composed with an isomorphism between Dn /H and Z2 , it yields an onto homomorphism Dn /H → Z2 . |Dn | |H| (3) True. Since |Sn | = n!, we have σ n! = for every σ ∈ Sn . However, (10n)! = 10n · (10n − 1) · · · · · (n + 1) ·n · (n − 1) · · · · · 2 · 1 = m · n! | {z } m is divisible by n!. Therefore every σ ∈ Sn we have σ (10n)! = (σ n! )m = . 0 0 (4) False. The zero matrix does not belong to GL(2, R) and hence 0 0 GL(2, R) is not a subgroup of the additive group of M2 (R). (5) False. For example, if G is an arbitrary group and H = G then H / G and G/H is a trivial group, so that every element of G/H has order 1. However, if |G| > 1, then not every element of G has order 1. Problem 2.[20 points] (1) Determine whether the groups Z2 × Z8 and Z16 are isomorphic. Justify your answer. (2) Let R∗ = {x ∈ R : x 6= 0} and R+ = {x ∈ R : x > 0}. Determine whether the groups (R∗ , ·) and (R+ , ·) are isomorphic. Justify your answer. Solution. (1) The groups Z2 × Z8 and Z16 are not isomorphic. Indeed, the group Z16 has an element of order 16 (namely 1) but Z2 × Z8 has no elements of order 16. To see the latter, recall that for every a ∈ Z2 , b ∈ Z8 and g = (a, b) ∈ Z2 × Z8 we have |g| = lcm(|a|, |b|). The possible values of |a| are 1 and 2 and the possible values 1 2 of |b| are 1, 2, 4, 8. Hence the maximal possible value of lcm(|a|, |b|) is 8, so that Z2 × Z8 indeed has no elements of order 16. (2) The groups (R∗ , ·) and (R+ , ·) are not isomorphic. The group (R+ , ·) has no elements if order 2 since if x > 0 then x2 = 1 if and only if x = 1. However, (R∗ , ·) has an element of order 2, namely the element −1 ∈ R∗ . Problem 3.[20 points] (1) Prove that if G is a group and H / G, K / G are normal subgroups in G then H ∩ K is again a normal subgroup of G. Show the details of your work. (2) Give an example of a group G and a normal subgroup H /G such that |H| > 1 and that the group G/H is non-abelian. Justify that your example has the required properties. Solution. (1) Recall first that by the Normality Test (Theorem 3 in Ch 2.8) a subgroup Y ≤ G is normal in G if and only if for every g ∈ G we have g −1 Y g ⊆ Y , that is, for every g ∈ G and every y ∈ Y we have g −1 yg ∈ Y . There was a theorem prooved in class (see Theorem 3 in Ch 2.3) that the intersection of two subgroups is always a subgroup, so that H ∩ K ≤ G.Thus we only need to verify that the subgroup H ∩ K is normal in G. Let g ∈ G and a ∈ H ∩ K be arbitrary. We need to show that g −1 ag ∈ H ∩ K. Since H / G and a ∈ H, we have g −1 ag ∈ H. Since K / G and a ∈ K, we have −1 g ag ∈ K. Hence a ∈ H ∩ K, and therefore H ∩ K / G by the Normality Test, as required. (2) Let K be any nonabelian group (e.g. K = S3 ) and let G = K × K. Consider H = {(k, 1) ∈ K × K : k ∈ K}. It was verified in one of the homework problems that H / G and that G/H ∼ = K. Since K was chosen to be nonabelian, this example has all the required properties. We recall the argument why G/H ∼ = K. Consider the map f : G → K defined by f (k, k 0 ) = k for any k, k 0 ∈ K. Then it is easy to check that f is an onto homomorphism and that Ker(f ) = H. Hence H / G and by the Isomorphism Theorem we have G/H ∼ = K. Problem 4.[20 points] Let R 6= {0} be a ring such that r2 = r for every r ∈ R. Prove that R is commutative and that char(R) = 2. Solution. Note: This was one of the homework problems. For every r ∈ R we have r + 1 = (r + 1)2 = r2 + 2r + 1 = r + 2r + 1 = 3r + 1 =⇒ 2r = 0. Thus for every r ∈ R we have 2r = 0. Since R 6= 0, this implies that char(R) = 2. Note that since char(R) = 2, we have 2r = 0, that is r = −r for every r ∈ R. Let a, b ∈ R be arbitrary. Then 3 a + b = (a + b)2 = a2 + ab + ba + b2 = a + ab + ba + b =⇒ ab + ba = 0. Hence ba = −ab for every a, b ∈ R. Since r = −r for every r ∈ R, we have −ab = ab and hence ba = ab for every a, b ∈ R. Thus R is commutative, as required. Problem 5.[20 points] a b Let G = { : a, b, c ∈ R, a 6= 0, c 6= 0}, so that G is a subgroup of GL(2, R) 0 c and thus is a group withrespect to matrix multiplication (you do not have to verify 1 b this). Let K = { : b ∈ R}. Prove that K is a normal subgroup of G and 0 1 ∗ ∗ ∼ that G/K = R × R , where R∗ = R − {0} is the multiplicative group of nonzero real numbers. Solution. a b Consider the function f : G → R × R defined as f = (a, c). 0 c We claim that f is an onto homomorphism and that Ker(f ) = K. First, check that f is a homomorphism: aa1 ab1 + bc1 a b a1 b1 = (aa1 , cc1 ) = =f f 0 cc1 0 c 0 c1 a b a1 b1 (a, c)(a1 , c1 ) = f f 0 c 0 c1 ∗ ∗ Thus indeed f is a homomorphism. x 0 The map f is onto since for any x, y ∈ R we have f = (x, y). 0 y The kernel of f is: ∗ a b a b ∈G:f = (1, 1) = Ker(f ) = 0 c 0 c 1 b a b ∈ G : (a, c) = (1, 1) = : b ∈ R = K. 0 c 0 1 Thus indeed K = Ker(f ) and hence K / G is a normal subgroup. We have verified that f : G → R∗ ×R∗ is an onto homomorphism with Ker(f ) = K. Therefore by the Isomorphism Theorem G/K ∼ = R∗ × R∗ , as required.