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Exam 2009-2015, questions and answers
Abstract Algebra (King's College London)
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CM121A, Abstract Algebra May 2009 Exam Solutions
A1. a) Find all x ∈ Z satisfying the congruence 19x ≡ 401 mod 10.
Solution: This is the same as −x ≡ 1 mod 10, so the solution is
x ≡ −1 mod 10, or x = −1 + 10k with k ∈ Z.
b) Find all solutions with x, y ∈ Z to the linear Diophantine equation
19x + 10y = 401.
Solution 1: By part a), 19x − 401 is divisible by 10 if and only if
x = −1 + 10k for some k ∈ Z, so these are the possible values of x.
Solving for the corresponding value of y gives 10y = 401 − 19(−1 +
10k) = 420 − 190k, so y = 42 − 19k.
Solution 2: The Euclidean Algorithm gives 19 = 10 + 9, 10 =
9 + 1, so 1 = 10 − (19 − 10) = 2 · 10 − 19. Multiplying by 401 gives
−401 · 19 + 802 · 10 = 401, so x = −401, y = 802 is a solution. The
general solution is therefore
x = −401 + 10k, y = 802 − 19k.
c) Find all solutions to this linear Diophantine equation with x and y
positive integers.
Solution: The inequalities −1+10k > 0, 42−19k > 0 are equivalent
to k > 1/10, k < 42/19, so the values of k giving positive solutions
are k = 1, 2 (or k = 41, 42 for the parametrization is Solution 2).
The corresponding solutions are
x = 9, y = 23 and x = 19, y = 4.
A2. Consider the following permutations σ, τ ∈ S4 defined by
µ ¶ µ ¶
1 2 3 4 1 2 3 4
σ= , τ= .
4 3 2 1 4 1 3 2
(Recall that S4 is the group of permutations of the set {1, 2, 3, 4}, and
that the above notation means that σ(1) = 4, σ(2) = 3, etc.)
a) Express σ and τ using cycle notation.
Solution: σ = (14)(23), τ = (142).
b) Find the orders of σ and τ .
Solution: σ has order 2, τ has order 3.
c) Find στ .
Solution: στ = (14)(23)(142) = (243).
1
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Exam 2009-2015, questions and answers
Abstract Algebra (King's College London)
Scan to open on Studocu
Studocu is not sponsored or endorsed by any college or university
Downloaded by Sudo Expert ()
, lOMoARcPSD|51280880
CM121A, Abstract Algebra May 2009 Exam Solutions
A1. a) Find all x ∈ Z satisfying the congruence 19x ≡ 401 mod 10.
Solution: This is the same as −x ≡ 1 mod 10, so the solution is
x ≡ −1 mod 10, or x = −1 + 10k with k ∈ Z.
b) Find all solutions with x, y ∈ Z to the linear Diophantine equation
19x + 10y = 401.
Solution 1: By part a), 19x − 401 is divisible by 10 if and only if
x = −1 + 10k for some k ∈ Z, so these are the possible values of x.
Solving for the corresponding value of y gives 10y = 401 − 19(−1 +
10k) = 420 − 190k, so y = 42 − 19k.
Solution 2: The Euclidean Algorithm gives 19 = 10 + 9, 10 =
9 + 1, so 1 = 10 − (19 − 10) = 2 · 10 − 19. Multiplying by 401 gives
−401 · 19 + 802 · 10 = 401, so x = −401, y = 802 is a solution. The
general solution is therefore
x = −401 + 10k, y = 802 − 19k.
c) Find all solutions to this linear Diophantine equation with x and y
positive integers.
Solution: The inequalities −1+10k > 0, 42−19k > 0 are equivalent
to k > 1/10, k < 42/19, so the values of k giving positive solutions
are k = 1, 2 (or k = 41, 42 for the parametrization is Solution 2).
The corresponding solutions are
x = 9, y = 23 and x = 19, y = 4.
A2. Consider the following permutations σ, τ ∈ S4 defined by
µ ¶ µ ¶
1 2 3 4 1 2 3 4
σ= , τ= .
4 3 2 1 4 1 3 2
(Recall that S4 is the group of permutations of the set {1, 2, 3, 4}, and
that the above notation means that σ(1) = 4, σ(2) = 3, etc.)
a) Express σ and τ using cycle notation.
Solution: σ = (14)(23), τ = (142).
b) Find the orders of σ and τ .
Solution: σ has order 2, τ has order 3.
c) Find στ .
Solution: στ = (14)(23)(142) = (243).
1
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