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Unformatted text preview: Math 541 Solutions to HW #3 1. Gallian Chapter 2: 3,5 • #3 Show that (a) { 1 , 2 , 3 } under multiplication modulo 4 is not a group, but that (b) { 1 , 2 , 3 , 4 } under multiplication modulo 5 is a group. (a) This is not a group, since it is not closed. Consider that 2 · 2 ≡ 0 (mod 4), and that 0 is not in the set. (b) This is a group. A quick multiplication table shows that the operation is binary. By associa tivity of multiplication in the integers, ( a · b ) · c = a · ( b · c ), so the operation is associative. Consider any element a ∈ Z 5 { } . Then 1 · a = a · 1 = a , so there is an identity, namely 1. Consider any element a ∈ Z 5 { } . Then a has an inverse. The justification follows: 1 · 1 = 1 ≡ 1 (mod 5); 2 · 3 = 6 ≡ 1 (mod 5); 3 · 2 = 6 ≡ 1 (mod 5); 4 · 4 = 16 ≡ 1 (mod 5). • #5 Find the inverse of the matrix 2 6 3 5 in GL (2 , Z 11 ). We have 2 6 3 5 1 = 1 2 · 5 3 · 6 5 6 3 2 = 1 3 5 5 8 2 = 4 5 5 8 2 = 9 9 10 8 . 2. Compute 2 2047423023 (mod 11). • Fermat’s Little Theorem tells us that a p ≡ a (mod p), where p is a prime. Or, if a and p are coprime, then a p 1 ≡ 1 (mod p). Since 2 and 11 are coprime, we make use of the second part of Fermat’s Little Theorem, which tells us that 2 10 ≡ 1 (mod 11). This also means that 2 20 = 2 10 · 2 10 ≡ 1 · 1 = 1 (mod 11), and in general that 2 10 b ≡ 1 (mod 11), where b is any positive integer. Thus, 2 2047423023 = 2 2047423020 · 2 3 ≡ 1 · 2 3 = 8 (mod 11)....
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This note was uploaded on 11/03/2010 for the course MATH 541 taught by Professor Pollack during the Fall '09 term at BU.
 Fall '09
 Pollack
 Algebra, Multiplication

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