MAT3702/201/2/2018
Tutorial letter 201/2/2018
ABSTRACT ALGEBRA
MAT3702
Semester 2
Department of Mathematical Sciences
IMPORTANT INFORMATION:
This tutorial letter contains solutions to assignment 01
BARCODE
university
Define tomorrow. of south africa
, 1
MAT 3702 Assignment 1 Solutions Semester 2 2018
1. (i) Suppose that g ◦ f (x) = g ◦ f (y) giving g(f (x)) = g(f (y)). Since g is
injective, we get that f (x) = f (y) and the injectivity of f gives that x = y making
g ◦ f injective.
(iii) The surjectivity of g gives that for all d ∈ D, there exists c ∈ C such that
g(c) = d. Also the surjectivity of f gives that there exists b ∈ B such that f (b) = c.
Thus (g ◦ f )(b) = g(f (b)) = g(c) = d making g ◦ f surjective.
2. (i) For all f ∈ T , we have that f (2) = f (2) making f ∼ f and so ∼ is reflexive
on T .
(ii) If f ∼ g, then f (2) = g(2) so that g(2) = f (2) giving that g ∼ f and thus
making ∼ symmetric
(iii) If f ∼ g and g ∼ h, then we get that f (2) = g(2) and g(2) = h(2) giving
that f (2) = h(2) so that f ∼ h making ∼ transitive. Hence ∼ defines an equivalence
relation on T .
3.
1 2 3 1 2 3 1 2 3
ab = =
3 1 2 1 3 2 3 2 1
1 2 3
(ab)−1 =
3 2 1
1 2 3 1 2 3 1 2 3
a−1 b−1 = =
2 3 1 1 3 2 2 1 3
Hence (ab)−1 6= a−1 b−1
4. Observe from the definition that S ⊆ G. We have that en = e so that e ∈ S
and thus making S 6= ∅. Let x, y ∈ S so that xn = e = y n . Since y n = e, we have
that (y n )−1 = (y −1)n = e so that y −1 ∈ S. Thus (xy −1 )n = xn (y −1 )n = ee = e so that
xy −1 ∈ S. Hence S is a subgroup of G.
,2
5. We have that g((a, b)(x, y)) = g((ax, by)) = ax = g((a, b))g((x, y)) making g a
homomorphism. For all x ∈ G, there exists (x, y) ∈ G × H such that g((x, y)) = x.
Hence g is surjective.
6. From the definition of L, we observe that L ⊆ G. Let eG , eH be the identity
elements of G, H respectively. Since K is a subgroup of H, we have that eH ∈ K
and since f is a homomorphism, we have that f (eG ) = eH ∈ K so that eG ∈ L
making L 6= ∅. Let x, y ∈ L so that f (x), f (y) ∈ K. Since K is a subgroup of H and
f (y) ∈ K, we get that (f (y))−1 = f (y −1) ∈ K. Thus f (xy −1) = f (x)f (y −1) ∈ K so
that xy −1 ∈ L and thus making L a subgroup of G.
7. Since f is surjective, all elements in H are of the form f (x) where x ∈ G. So
for all f (x), f (y) ∈ H, we get that f (x)f (y) = f (xy) = f (yx) = f (y)f (x) making H
abelian. Observe that f (xy) = f (yx) since x, y ∈ G and G is abelian and f (x)f (y) =
f (xy) since f is a homomorphism.
Furthermore observe that a homomorphic image of an abelian group is itself
abelian. Thus f (G) is a homomorphic image of an abelian G making f (G) abelian.
However by the surjectivity of f , we get that f (G) = H and therefore making H
abelian as well.
,
Tutorial letter 201/2/2018
ABSTRACT ALGEBRA
MAT3702
Semester 2
Department of Mathematical Sciences
IMPORTANT INFORMATION:
This tutorial letter contains solutions to assignment 01
BARCODE
university
Define tomorrow. of south africa
, 1
MAT 3702 Assignment 1 Solutions Semester 2 2018
1. (i) Suppose that g ◦ f (x) = g ◦ f (y) giving g(f (x)) = g(f (y)). Since g is
injective, we get that f (x) = f (y) and the injectivity of f gives that x = y making
g ◦ f injective.
(iii) The surjectivity of g gives that for all d ∈ D, there exists c ∈ C such that
g(c) = d. Also the surjectivity of f gives that there exists b ∈ B such that f (b) = c.
Thus (g ◦ f )(b) = g(f (b)) = g(c) = d making g ◦ f surjective.
2. (i) For all f ∈ T , we have that f (2) = f (2) making f ∼ f and so ∼ is reflexive
on T .
(ii) If f ∼ g, then f (2) = g(2) so that g(2) = f (2) giving that g ∼ f and thus
making ∼ symmetric
(iii) If f ∼ g and g ∼ h, then we get that f (2) = g(2) and g(2) = h(2) giving
that f (2) = h(2) so that f ∼ h making ∼ transitive. Hence ∼ defines an equivalence
relation on T .
3.
1 2 3 1 2 3 1 2 3
ab = =
3 1 2 1 3 2 3 2 1
1 2 3
(ab)−1 =
3 2 1
1 2 3 1 2 3 1 2 3
a−1 b−1 = =
2 3 1 1 3 2 2 1 3
Hence (ab)−1 6= a−1 b−1
4. Observe from the definition that S ⊆ G. We have that en = e so that e ∈ S
and thus making S 6= ∅. Let x, y ∈ S so that xn = e = y n . Since y n = e, we have
that (y n )−1 = (y −1)n = e so that y −1 ∈ S. Thus (xy −1 )n = xn (y −1 )n = ee = e so that
xy −1 ∈ S. Hence S is a subgroup of G.
,2
5. We have that g((a, b)(x, y)) = g((ax, by)) = ax = g((a, b))g((x, y)) making g a
homomorphism. For all x ∈ G, there exists (x, y) ∈ G × H such that g((x, y)) = x.
Hence g is surjective.
6. From the definition of L, we observe that L ⊆ G. Let eG , eH be the identity
elements of G, H respectively. Since K is a subgroup of H, we have that eH ∈ K
and since f is a homomorphism, we have that f (eG ) = eH ∈ K so that eG ∈ L
making L 6= ∅. Let x, y ∈ L so that f (x), f (y) ∈ K. Since K is a subgroup of H and
f (y) ∈ K, we get that (f (y))−1 = f (y −1) ∈ K. Thus f (xy −1) = f (x)f (y −1) ∈ K so
that xy −1 ∈ L and thus making L a subgroup of G.
7. Since f is surjective, all elements in H are of the form f (x) where x ∈ G. So
for all f (x), f (y) ∈ H, we get that f (x)f (y) = f (xy) = f (yx) = f (y)f (x) making H
abelian. Observe that f (xy) = f (yx) since x, y ∈ G and G is abelian and f (x)f (y) =
f (xy) since f is a homomorphism.
Furthermore observe that a homomorphic image of an abelian group is itself
abelian. Thus f (G) is a homomorphic image of an abelian G making f (G) abelian.
However by the surjectivity of f , we get that f (G) = H and therefore making H
abelian as well.
,
