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MAT3702 Abstract Algebra – Complete Study Guide

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Description: This study guide covers essential abstract algebra concepts, including groups, subgroups, cyclic groups, rings, fields, homomorphisms, and mathematical proofs. It is designed to help students strengthen their understanding of algebraic structures and prepare for mathematics assignments and examinations.

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MAT3702 ABSTRACT ALGEBRA — REVISION
PACK EXAM 4 2026 PAST EXAM
QUESTIONS & ANSWERS WITH WORKED
EXPLANATIONS



SECTION A: SETS, FUNCTIONS & RELATIONS

Question 1: Set Operations
Let A={1,2,3,4}A={1,2,3,4} and B={3,4,5,6}B={3,4,5,6}. Determine
A∪BA∪B.
A. {3,4}{3,4}
B. {1,2,5,6}{1,2,5,6}
C. {1,2,3,4,5,6}{1,2,3,4,5,6}
D. ∅∅

Answer: C. {1,2,3,4,5,6}{1,2,3,4,5,6}.
Solution: The union contains every element belonging to either
set, without repetition.




Question 2: Set Intersection
Using the sets in Question 1, determine A∩BA∩B.
A. {1,2}{1,2}
B. {3,4}{3,4}

,C. {5,6}{5,6}
D. {1,2,3,4,5,6}{1,2,3,4,5,6}

Answer: B. {3,4}{3,4}.
Solution: The intersection contains only elements common to
both sets.




Question 3: Injective Functions
A function f:A→Bf:A→B is injective if:
A. Every element of BB has a preimage.
B. Every element of AA maps to itself.
C. Different elements of AA always have different images.
D. The domain and codomain are identical.

Answer: C. Different elements of AA always have different
images.
Solution: Injectivity means f(x)=f(y)⇒x=yf(x)=f(y)⇒x=y.
Equivalently, distinct domain elements cannot map to the same
codomain element.




Question 4: Surjective Functions
A function f:A→Bf:A→B is surjective when:
A. Every element of BB is the image of at least one element of AA.
B. Every element of AA has two images.

, C. ff is necessarily constant.
D. The domain is empty.

Answer: A. Every element of BB is the image of at least one
element of AA.
Solution: Surjectivity means that for every b∈Bb∈B, there exists
a∈Aa∈A such that f(a)=bf(a)=b.




Question 5: Equivalence Relation
Define a relation on ZZ by a∼ba∼b if and only if 5∣(a−b)5∣(a−b).
Show that this is an equivalence relation.
Solution:

 Reflexive: a−a=0a−a=0 is divisible by 55.
 Symmetric: If 5∣(a−b)5∣(a−b), then 5∣(b−a)5∣(b−a).
 Transitive: If 5∣(a−b)5∣(a−b) and 5∣(b−c)5∣(b−c), then 5∣(a−c)5∣(a−c).
Thus ∼∼ is an equivalence relation.




Question 6: De Morgan's Laws
Which identity is correct for sets AA and BB in a universal set UU?
A. (A∪B)c=Ac∩Bc(A∪B)c=Ac∩Bc
B. (A∪B)c=Ac∪Bc(A∪B)c=Ac∪Bc

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