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Solutions: Homework 1 Solutions: Homework 1 January 17, 2020 Problem 1. Let ? be the binary operation on Z defined by a ? b = a − b: Is ? commutative? Is it associative? Proof. It is not commutative because 1 ? 2 = −1 while 2 ? 1 = 1; hence 1 ? 2 6= 2 ? 1: It is not associative because (a ? b) ? c = (a − b) − c = a − b − c while a ? (b ? c) = a − (b − c) = a − b + c and hence (a ? b) ? c 6= a ? (b ? c) unless c = 0: Problem 2. Let ? be the binary operation on Z+ defined by a ? b = 2ab: Is ? commutative? Is it associative? Proof. It is commutative because a ? b = 2ab = 2ba = b ? a; because ab = ba: To check for associativity, let us look at a ? (b ? c) = a ? 2bc = 2a2bc and (a ? b) ? c = 2ab ? c = 2c2ba : If a = b = 1 and c = 4; we have 1 ? (1 ? 4) = 224 = 216; while (1 ? 1) ? 4 = 28; and they are not equal. So ? is not associative. Problem 3. Let H be the subset of M2(R) consisting of matrices of the form a −b H = ja; b 2 : b a R (a) Is H closed under matrix addition? (b) Is H closed under matrix multiplication? Proof. (a) We want to prove that if A and B are two matrices in H; then A + B is also in a1 −b1 a2 −b2 H: Let A = and B = with a1; b1; a2; b2 2 R: Then b1 a1 b2 a2 a + a −(b + b ) A + B = 1 2 1 2 b1 + b2 a1 + a2 which clearly lies in H: So H is closed under matrix addition. (b) We want to prove that if A and B are two matrices in H; then AB is also in H: Let a1 −b1 a2 −b2 A = and B = with a1; b1; a2; b2 2 R: Then b1 a1 b2 a2 a a − b b −(a b + b a ) AB = 1 2 1 2 1 2 1 2 a1b2 + a2b1 a1a2 − b1b2 which clearly lies in H: So H is closed under matrix multiplication. 1 Problem 4. Either prove the statement or give a counterexample. Every commutative binary operation on a set having just two elements is associative. Proof. This statement is false. Consider the following table: ? a b a b a b a a This binary operation is obviously commutative, as a ? b = b ? a; but not associative as (a ? a) ? b = b ? b = a; while a ? (a ? b) = a ? a = b: In problems 5, 6 and 7, determine whether or not φ is an isomorphism of binary structures. If it is not an isomorphism, why not? Problem 5. (Z; +) with (Z; +) where φ(n) = −n for all n 2 Z: Proof. φ(n) = φ(m) implies that −n = −m and hence n = m: So φ is one-to-one. For any n 2 Z; φ(−n) = n; hence it is also onto. So φ is a bijection. For n; m 2 (Z; +); φ(n+m) = −(n+m) = −n−m = φ(n)+φ(m): So it is an isomorphism. Problem 6. (Z; +) with (Z; +) where φ(n) = 2n for all n 2 Z: Proof. φ is not onto as there does not exist n 2 Z such that φ(n) = 1: So it is not bijective, and hence not an isomorphism. Problem 7. (M2(R);:) with (R;:) where φ(A) is the determinant of the matrix A: 0 0 0 0 1 0 1 0 Proof. φ = det = 0 = det = φ : So φ is 0 0 0 0 0 0 0 0 not injective, hence not bijective, and hence not an isomorphism. Problem 8. Consider the binary operation ? on Z defined by a ? b = ab: Decide whether (Z;?) is a group. If not, which axioms fail? Proof. Integer multiplication is clearly associative, hence G1 axiom holds. For any a 2 Z; 1 ? a = a ? 1 = a; hence 1 clearly acts as the identity in (Z;?) and hence G2 axiom also holds. G3 axiom fails because not every element has an inverse, for example there is no a 2 Z such that 0 ? a = 1; so 0 has no inverse. So (Z;?) is not a group. Problem 9. Let n be a positive integer and let nZ = fnmjm 2 Zg: (a) Show that (nZ; +) is a group. (b) Show that (nZ; +) is isomorphic to (Z; +): Proof. (a) + is associative. So G1 axiom holds. 0 + nm = nm + 0 = nm for all nm 2 nZ: So 0 is the identity, hence G2 axiom holds. For any nm 2 nZ; nm + n(−m) = n(−m) + nm = 0; so G3 axiom also holds. So (nZ; +) is a group. a (b) Define the map φ :(nZ; +) ! (Z; +) by φ(a) = n : Note that this makes sense because a a b as a is in nZ; n divides a and hence n is an integer. φ(a) = φ(b) implies that n = n ; hence a = b and so φ is one-to-one. For any k 2 Z; φ(nk) = k; hence φ is also onto. So φ is a a+b a b bijection. Now, φ(a + b) = n = n + n = φ(a) + φ(b): So φ is an isomorphism. 2 Problem 10. Give a table for a binary operation on the set fe; a; bg of three elements satisfying axioms G2 and G3 but not axiom G1: Proof. We shall choose e as our identity and a and b to be the inverses of each other, i.e. a ? b = b ? a = e: So ? satsifies G2 and G3: ? e a b e e a b a a a e b b e b Now, (a ? a) ? b = a ? b = e; while a ? (a ? b) = a ? e = a: So ? is not associative, hence G1 fails. Problem 11. Show that if G is a group with identity e and with an even number of elements, then there is a 6= e in G such that a ? a = e: Proof. This is equivalent to proving that if G has an even number of elements, then there exists an a 6= e in G such that the inverse of a is itself. We prove this by contradiction. Suppose that for all a 6= e in G; the inverse of a is not a: Then, pairing together all elements other than e with its inverse gives us an even number of elements. Counting e; this implies that G has an odd number of elements. This contradicts our hypothesis that G has an even number of elements. Problem 12. Show that every group G with identity e such that x ? x = e for all x 2 G is abelian. Proof. Let a; b be any two elements in G: Then (a ? b) ? (a ? b) = e by our hypothesis on G: By associativity, we have ((a ? b) ? a) ? b = e Then (((a ? b) ? a) ? b) ? b = e ? b = b Again, by associativity, we have ((a ? b) ? a) ? (b ? b) = b Since b ? b = e; we have (a ? b) ? a = b Then ((a ? b) ? a) ? a = b ? a Again, by associativity, we have (a ? b) ? (a ? a) = b ? a Since a ? a = e; we have a ? b = b ? a So G is abelian. 3.
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