Download (f g)(h(x)) = f(g(h(x))) = f((g h)(x))

MATH 103A: MODERN ALGEBRA I (WINTER 2014)
HW2 SOLUTIONS
Problem 1. a) A parallelogram that is neither a rectangle nor a rhombus only has two rotation
symmetries, R0 and R180 .
b) A rhombus that is not a rectangle has two rotation symmetries, R0 and R180 , and two reflection
symmetries, D and D 0 (flipping about the two diagonals).
Problem 2. The symmetries of the infinitely long strip of equally spaced H’s are:
• Two rotation symmetries R0 and R180 .
• One horizontal reflection symmetry through the center of all the H’s.
• Infinitely many vertical reflection symmetries through the center of any H or through any
vertical line exactly halfway between any two adjacent H’s.
• Infinitely many translation symmetries to the left or right by any number of H’s.
The group of symmetries is not Abelian. To see this, fix an H on the infinite strip and consider
the symmetries R1 and R2 given by letting R1 be translation to the right by one H and R2 be
horizontal reflection through the center of the chosen H. Then the composition R1 R2 moves
the chosen H one unit to the right whereas R2 R1 moves the chosen H one unit to the left. We
conclude that R1 R2 6= R2 R1 .
Problem 3. a) Multiplication mod n, for any positive n, is associative.
Proof: Since multiplication is associative, ( ab)c = a(bc). Thus, ( ab)c mod n = a(bc) mod n.
Now apply the result from problem 0.9 from HW1 to get what we need to show.
b) Division of nonzero rational numbers is not associative.
Counter example: (81 ÷ 9) ÷ 3 = 9 ÷ 3 = 3 but 81 ÷ (9 ÷ 3) = 81 ÷ 3 = 27.
c) Function composition of polynomials with real coefficients is associative: Suppose f , g, h are
such polynomials then for any x in the domain,
[( f ◦ g) ◦ h]( x ) = ( f ◦ g)(h( x )) = f ( g(h( x ))) = f (( g ◦ h)( x )) = [ f ◦ ( g ◦ h)]( x )
d) Multiplication of 2 × 2 matrices with integer entries is associative. This follows from a result
in Math 20F.
Problem 4. a) {0, 4, 8, 12} is closed under addition mod 16 (Just compute the addition table).
b) {0, 4, 8, 12} is not closed under addition mod 15. Counter example: under addition mod 15,
4 + 12 = 1 which doesn’t belong to the set.
c) {1, 4, 7, 13} is closed under multiplication mod 15 (Just compute the multiplication table).
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d) {1, 4, 5, 7} is not closed under multiplication mod 9. Counter example: under multiplication
mod 9, 4.5 = 2 which doesn’t belong to the set.
Problem
5. Let GL(2,R) be the group of 2 × 2 invertible matrices with real entries. Let A =
1 0
2 1
and B =
be matrices in GL(2, R). Then one can easily check that AB 6= BA. Thus,
0 2
0 1
GL(2, R) is not Abelian.
Problem 6. Let GL(3, R)+ be the set of all 3 × 3 real matrices A such that det( A) > 0. Then
GL(3, R)+ is a group under matrix multiplication, as followed:
• Closure: ∀ A, B ∈ GL(3, R)+ , AB is a 3 × 3 real matrix with det( AB) = det( A)det( B) > 0
so AB ∈ GL(3, R)+
• Associative: ∀ A, B, C ∈ GL(3, R)+ , we have ( AB)C = A( BC ) (This follows from results in
Math 20F).
• Identity: The 3 × 3 identity matrix I3 satisfies det( I3 ) = 1 > 0, so I3 ∈ GL(3, R)+ , and
AI3 = I3 A = A for all A ∈ GL(3, R)+ .
• Inverse: ∀ A ∈ GL(3, R)+ , the inverse matrix A−1 is also in GL(3, R)+ because A−1 is a
3 × 3 real matrix with det( A−1 ) = det1( A) > 0.
Problem 7. Let G be a group. Let a ∈ G and let a−1 be the inverse of a in G. Then:
( a−1 )−1 = ( a−1 )−1 e = ( a−1 )−1 ( a−1 a) = (( a−1 )−1 a−1 ) a = ea = a
Thus, ( a−1 )−1 = a
Problem 8. Let G be a group.
• (⇒) Suppose G is Abelian. Then for all a, b ∈ G, ab = ba. Multiplying a to the left and b to
the right of both sides gives a( ab)b = a(ba)b. Thus, a2 b2 = ( ab)2 , for all a, b ∈ G.
• (⇐) Now suppose that a2 b2 = ( ab)2 for all a, b ∈ G. Since G is a group, the inverses
a−1 and b−1 exist and are in G. Multiplying a−1 to the left and b−1 to the right of both
sides yields a−1 ( a2 b2 )b−1 = a−1 ( ab)2 b−1 . Then, ( a−1 a) ab(bb−1 ) = ( a−1 a)ba(bb−1 ) and,
therefore, ab = ba for all a, b ∈ G. Hence, G is Abelian.
Hence, G is Abelian ⇔ a2 b2 = ( ab)2 for all a, b ∈ G.
Problem 9. Let G be a group. Let a, b ∈ G and let n be a positive integer. Then:
( aba−1 )n = ( aba−1 )( aba−1 )...( aba−1 ) = ab( a−1 a)b( a−1 a)b...( a−1 a)ba−1 = abn a−1
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