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 = |a, b ∈ . 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 ∈ 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 ∈ 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 ∈ Z. Proof. φ(n) = φ(m) implies that −n = −m and hence n = m. So φ is one-to-one. For any n ∈ Z, φ(−n) = n, hence it is also onto. So φ is a bijection. For n, m ∈ (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 ∈ Z. Proof. φ is not onto as there does not exist n ∈ 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 ∈ 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 ∈ 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 = {nm|m ∈ Z}. (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 ∈ nZ. So 0 is the identity, hence G2 axiom holds. For any nm ∈ 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 ∈ 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 {e, a, b} 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 ∈ 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.