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MAT 3702 Assignment 1 Solutions Semester 1 2020

1. Suppose that A ⊆ X − B so that for all x ∈ A, x ∈ X − B giving that

x ∈ A, x 6∈ B making A

T

B = ∅. Conversely suppose that A

T

B = ∅ so that for all

y ∈ A, y 6∈ B giving that y ∈ X − B making A ⊆ X − B.

2. By the surjectivity of f, for every y ∈ Y , there exists x ∈ X such that f(x) = y.

For every x ∈ X, we have that f(x) = f(x) so that x ∼ x thus making ∼ reflexive.

If x1 ∼ x2, we have that f(x1) = f(x2) giving that f(x2) = f(x1) so that x2 ∼ x1

making ∼ symmetric. If x1 ∼ x2 and x2 ∼ x3, we then obtain that f(x1) = f(x2) and

f(x2) = f(x3) which gives that f(x1) = f(x3) so that x1 ∼ x3 making ∼ transitive.

Hence ∼ is an equivalence relation on X.

3. Suppose f is a one-to-one homomorphism, we have that ker(f) = {1G} so

that G/ker(f) = G/{1G} ∼= G making G the image of f thus rendering f onto.

Conversely suppose that f is an onto homomorphism. Thus G/ker(f) ∼= G making

ker(f) = {1G} so that f becomes one-to-one.

4. Let K = {g ∈ G | f(g) = y} and suppose that K is a subgroup of G. Thus

eG ∈ K, where eG is the identity element in G. Since eG ∈ K, we must have that

f(eG) = y. However f is a homomorphism so that f(eG) = eH . Thus f(eG) = y = eH

so that y = eH. Moreover for all a, b ∈ K, we have that f(a) = y = f(b). Since K

is a subgroup, we have that ab−1 ∈ K so that f(ab−1

) = y. However f(ab−1

) = y =

f(a)f(b

−1

) = f(a)(f(b))−1 = yy−1 = eH. Hence y = eH.

Again let K = {g ∈ G | f(g) = y} and suppose that y = eH . Thus we obtain

that K = {g ∈ G | f(g) = y = eH }. Observe that K ⊆ G and since f is a

homomorphism, we obtain that f(eG) = eH where eG is the identity element in G

so that eG ∈ K. Thus K 6= ∅. Let m, n ∈ K so that f(m) = eH = f(n). Then

f(mn−1

) = f(m)f(n

−1

) = f(m)(f(n))−1 = eHeH = eH so that mn−1 ∈ K. Hence K

is a subgroup of G.

5. Since G is finite, we have that x, y ∈ G will have finite orders. Since x ∈ G has

order 6, we have that x 6= eG such that x

6 = eG, where eG is the identity element in

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