MAT3702 EXAM PACK 2026 – QUESTION & ANSWERS

lOMoARcPSD|47389193




2 MAT3702
October/November 2025




QUESTION 1


(a) Let A, B, C be sets such that A ⊆ B and prove that A ∪ C ⊆ B ∪ C. (5)

(b) Let G be a group of nonzero real numbers under multiplication with

H = {x ∈ G | x = 1 or x is irrational}

K = {y ∈ G | y ≥ 1}

(i) and check as to whether H is a subgroup of G or not. (5)
[Hint: check as to whether H is contained in G and the identity of G is in H or not]
(ii) and check as to whether K is a subgroup of G or not. (5)
[Hint: check as to whether K is contained in G and the identity of G is in K or not]

(c) Prove that a group of order 3 is cyclic. (5)

(d) Let G be a group and H a subgroup of G. For a fixed x ∈ G, define

xHx−1 = {xhx−1 | h ∈ H}

and show that if H is abelian, then so is xHx−1 . (5)
[25]


QUESTION 2


(a) Let G be a group, H a subgroup of G and g ∈ G be a fixed element of G. Then show that
NG (gHg −1 ) = gNG (H)g −1 . (10)

(b) Let G be the group of real numbers under addition and G be the group of positive real numbers under
multiplication. Define ϕ : G −→ G by ϕ(x) = 2x and check as to whether ϕ is an isomporphism or
not. (7)
[Hint: first check as to whether ϕ is well-defined or not]

(c) Let f : G1 −→ G2 be a homomorphism of groups.

(i) Define the kernel of f . (2)
(ii) Prove that the kernel of f is a normal subgroup of G1 . (6)
[Hint: first show that the kernel of f is a subgroup of G1 ]

[25]


[TURN OVER]




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3 MAT3702
October/November 2025




QUESTION 3


(a) Given a = 4, p = 7, show that ap−1 ≡ 1(mod p). (4)
√ √
(b) Let ZZ [ 2] = {r + s 2 | r, s ∈ ZZ } and show that
√
(i) ZZ [ 2] is a subring of |R . (6)
√ √
(ii) 1 + 2 is a unit in ZZ [ 2]. (5)

(c) Let R be a commutative ring with identity and a, b, c ∈ R such that a ∈ R is a unit. Then prove
that b ∈ R divides c ∈ R if and only if ab ∈ R divides c ∈ R. (6)

(d) Prove that the zero element in a ring is unique. (4)


[25]


QUESTION 4


(a) Define ZZ 3 [i] = {a + bi | a, b ∈ ZZ 3 } and list all the elements of ZZ 3 [i]. (3)

(b) Let R, S be rings and z : R −→ S be given by z(r) = 0S for every r ∈ R and 0S ∈ S is the zero
element in S.

(i) Check as to whether z is a ring homomorphism or not (first check as to whether z is well-defined
or not). (5)
(ii) If both R, S contain nonzero elements, then what can be said about z? (4)

(c) Units are preserved by isomorphisms and using this fact, check as to whether the rings ZZ 8 and
ZZ 4 × ZZ 2 are isomorphic or not . (7)
[Hint: find units in each ring and compare how many they are]

(d) Let R be a finite commutative ring with identity. Prove that an ideal in R is prime if and only if
it is maximal . (6)


[25]


TOTAL MARKS: 100

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UNISA 2025




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, Question 1



QUESTION 1(a)
Let A, B, C be sets such that A ⊆ B. Prove that

A ∪ C ⊆ B ∪ C.

Proof
Let x ∈ A ∪ C.
By definition of union, this means:

x∈A or x ∈ C.

If x ∈ A, since A ⊆ B, it follows that x ∈ B.
If x ∈ C , then clearly x ∈ C ⊆ B ∪ C .

In either case, x ∈ B ∪ C.

Hence,

A ∪ C ⊆ B ∪ C.

■



QUESTION 1(b)
Let G be the group of nonzero real numbers under multiplication.

, (i) Check whether

H = {x ∈ G ∣ x = 1 or x is irrational}

is a subgroup of G.

Solution
To test whether H is a subgroup of G, we use the subgroup test.
1. Closure
Take x, y ∈ H .
If both are irrational, their product may be rational (e.g. 2 ⋅ 2 = 2).
Since 2 is rational and not equal to 1, it does not belong to H.

Hence, H is not closed under multiplication.
2. Identity
The identity element of G is 1, and 1 ∈ H.
3. Inverses
Even though the inverse of an irrational number is irrational, failure of closure already
disqualifies H .

Conclusion
Since H is not closed, it is not a subgroup of G.




(ii) Check whether

K = {y ∈ G ∣ y ≥ 1}

is a subgroup of G.

Solution
1. Identity
The identity of G is 1, and 1 ≥ 1, so 1 ∈ K.
2. Closure
∈ K , then x ≥ 1 and y ≥ 1.
If x, y
Hence, xy ≥ 1, so xy ∈ K .
3. Inverses
Let x ∈ K with x > 1.