Solutions: Homework 1

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.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    3 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us