8.

3 Assignments
SEMESTER 1
ASSIGNMENT 01
DUE DATE: 20 March 2019
UNIQUE ASSIGNMENT NO: 753326

1. Let f, g ∈ S3 be given by
   
1 2 3 1 2 3
f= ; g=
3 2 1 2 1 3

Thus compute (f ◦ g)(1), (f ◦ g)(2), (f ◦ g)(3), (g ◦ f )(1), (g ◦ f )(2), (g ◦ f )(3) and hence write
down f ◦ g, g ◦ f

2. (i) By finding the order of each element in Z4 , check as to whether or not Z4 is cyclic

(ii) If G is a finite group and a, b ∈ G, show that aba−1 and b have the same order

3. Let G be a group and x ∈ G. Define

CG (x) = {g ∈ G | gx = xg}

and show that CG (x) is a subgroup of G.

4. Let G be a finite group and define

Z(G) = {z ∈ G | zy = yz ∀ y ∈ G}

(i) Prove that Z(G) is a normal subgroup of G.

NB: First show that Z(G) is a subgroup of G.

T
(ii) Prove that Z(G) = x∈G CG (x)

(iii) Prove that x ∈ Z(G) if and only if CG (x) = G

(iv) If x ∈ G is the only element of order 2, then show that x ∈ Z(G)

5. Let G and H be finite groups and f : G −→ H be a homomorphism.

(i) Prove that for each x ∈ G, n ∈ Z, f (xn ) = (f (x))n

(ii) If a ∈ G has finite order k, then prove that (f (a))k = e, where e ∈ G is the identity element

6. Let n ∈ Z+ be fixed and define a relation ≡ on Z by x ≡ y (mod n) if x − y is a multiple of n.

Then check as to whether or not ≡ defines an equivalence relation on Z.

10
 1

MAT 3702 Assignment 1 Solutions Semester 1 2019

1. (f ◦ g)(1) = f (g(1)) = f (2) = 2, (f ◦ g)(2) = f (g(2)) = f (1) = 3, (f ◦ g)(3) =

f (g(3)) = f (3) = 1, (g ◦ f )(1) = g(f (1)) = g(3) = 3, (g ◦ f )(2) = g(f (2)) = g(2) =
1, (g ◦ f )(3) = g(f (3)) = g(1) = 2
    
1 2 3  1 2 3   1 2 3 
f ◦g = =
3 2 1 2 1 3 2 3 1
    
1 2 3  1 2 3   1 2 3 
g◦f = =
2 1 3 3 2 1 3 1 2

2. (i) ZZ 4 = {0, 1, 2, 3} and thus has order 4, denoted by | ZZ 4 | = 4. We have

that 1 + 1 + 1 + 1 = 4 = 0, 2 + 2 = 4 = 0, 3 + 3 + 3 + 3 = 12 = 0 giving that
o(0) = 1, o(1) = 4, o(2) = 2, o(3) = 4, where o(x) means the order of x. Since
o(1) = 4 = | ZZ 4 | = o(3), we get that 1, 3 are generators of ZZ 4 thus making ZZ 4 cyclic.
(ii) Since G is finite, all its elements will also have finite order. Suppose that
o(b) = k and o(aba−1 ) = n. Thus we get that bk = e where e ∈ G is the identity
element in G. We get that (aba−1 )k = (aba−1 )(aba−1 ) . . . (aba−1 ) = abk a−1 = aea−1 =
| {z }
k−times
e making n | k. Since we have that o(aba−1 ) = n, we get that (aba−1 )n = abn a−1 = e
which yields that abn a−1 = e = abk a−1 giving that a−1 abn a−1 a = a−1 ea = a−1 abk a−1 a
so that bn = e = bk giving that k | n. Since we have that n | k and k | n, we thus
conclude that k = n making b and aba−1 have the same order.

3. It follows immediately from the definition that CG (x) ⊆ G. We have that

e ∈ G is such that ex = x = xe so that e ∈ CG (x) and so making CG (x) 6= ∅. Let
g, h ∈ CG (x) so that gx = xg, hx = xh. From hx = xh, we get that hxh−1 = x so
that xh−1 = h−1 x and so making h−1 ∈ CG (x). Thus gh−1 x = gxh−1 = xgh−1 so
that gh−1 ∈ CG (x) making CG (x) a subgroup of G.

4. (i) Observe from the definition that Z(G) ⊆ G. For e ∈ G, ex = x = xe ∀ x ∈ G

so that e ∈ Z(G) making Z(G) 6= ∅. Let x, y ∈ Z(G) so that xg = gx, yg = gy ∀ g ∈
2

G. From yg = gy, we obtain for all g ∈ G that ygy −1 = g so that gy −1 = y −1g making
y −1 ∈ Z(G). So xy −1 g = xgy −1 = gxy −1 giving that xy −1 ∈ Z(G) making Z(G) a
subgroup of G. For all g ∈ G, x ∈ Z(G), we get that gxg −1 = xgg −1 = x ∈ Z(G)
making Z(G) a normal subgroup of G.
(ii) Let g ∈ Z(G) so that gx = xg ∀ x ∈ G. Thus g ∈ CG (x) ∀ x ∈ G. Thus
T T
g ∈ x∈G CG (x) making Z(G) ⊆ x∈G CG (x).
T
Conversely suppose that y ∈ x∈G CG (x) giving that y ∈ CG (x) ∀x ∈ G so
T
that yx = xy ∀ x ∈ G. Thus y ∈ Z(G) so that x∈G CG (x) ⊆ Z(G). Hence
T
Z(G) = x∈G CG (x)
(iii) Suppose that CG (x) = G. Thus xg = gx ∀ g ∈ G making x ∈ Z(G).
Conversely suppose that x ∈ Z(G). Thus xy = yx ∀ y ∈ G so that for all
y ∈ G, y ∈ CG (x) making G ⊆ CG (x). However, it is always true that CG (x) ⊆ G.
Thus combining G ⊆ CG (x) and CG (x) ⊆ G yields that CG (x) = G.
(iv) Since x ∈ G has order 2, we have that for all g ∈ G, gxg −1 also has order 2
(see Question 2(ii) above). Since x ∈ G is the only element in G of order 2, we must
have for all g ∈ G that gxg −1 = x giving that gx = xg ∀ g ∈ G and hence making
x ∈ Z(G).

5. (i) For positive exponents, we obtain that f (xn ) = f (xx . . . x}) = f (x)f (x) . . . f (x)) =
| {z | {z }
n−times n−times
(f (x))n since f is a homomorphism. The result is obviously true for n = 0. For nega-
tive exponents, we get that f (x−n ) = f (x−1 )n = f (x−1 −1 −1 −1 −1 −1
| x {z. . . x }) = f (x )f (x ) . . . f (x )) =
| {z }
n−times n−times
(f (x)−1 )n = (f (x))−n . Hence f (xn ) = (f (x))n ∀ n ∈ ZZ.
(ii) Let eG , eH be the identity elements of G, H respectively. Since a ∈ G has order
k, we have that ak = eG so that f (ak ) = f (eG ) = eH since f is a homomorphism.
However f (ak ) = (f (a))k so that (f (a))k = f (ak ) = f (eG ) = eH .

6. For all x ∈ ZZ , x − x = 0 = 0n giving that x ≡ x (mod n) making ≡ reflexive.

If x ≡ y (mod n), then x − y = kn. However we get that y − x = −(x − y) = −kn so
that y ≡ x (mod n) making ≡ symmetric. If x ≡ y (mod n) and y ≡ z (mod n), then
we obtain that x − y = kn and y − z = tn so that x − y + y − z = kn + tn = (k + t)n,
 3

where x − y + y − z = x − z so that x − z = (k + t)n giving that x ≡ z (mod n)

making ≡ transitive. Hence ≡ defines an equivalence relation on ZZ .
 MAT3702/101/3/2019

SEMESTER 1
ASSIGNMENT 02
DUE DATE: 12 April 2019
UNIQUE ASSIGNMENT NO: 687509

1. Let R be a ring and a ∈ R. Define

AR = {r ∈ R | ar = 0R },

where 0R is the zero element in R and show that AR is a subring of R.

2. If S is a subring of a ring R, then show that 0S = 0R , where 0S , 0R are the zero elements of S
and R respectively

3. A ring R in which x2 = x for every x ∈ R is called a Boolean ring. If R is a Boolean ring, prove
that a + a = 0R for every a ∈ R, where 0R is the zero element in R.

4. Let R1 be a commutative ring and f : R1 −→ R2 be a ring isomorphism. Prove that R2 is also

commutative

5. Let I be a two-sided ideal of a ring R and J be a two-sided ideal of a ring H. Prove that I × J
is a two-sided ideal of the ring R × H
NB: First show that I × J is a subring of R × H

6. Let I, J be ideals in a ring R and f : R −→ R/I × R/J be the function defined by f (x) =
(x + I, x + J) and check as to whether or not f is a ring homomorphism.

7. Let I, K be two-sided ideals in a ring R with K ⊆ I. Prove that

I/K = {a + K | a ∈ I}

is a two-sided ideal of the quotient ring R/K

NB: First show that I/K is a subring of R/K

11
 1

MAT 3702 Assignment 2 Solutions Semester 1 2019

1. We have that a0R = 0R so that 0R ∈ AR . For all x, y ∈ AR , we have that

ax = 0R = ay so that ax − ay = 0R − 0R = 0R . But ax − ay = a(x − y) = 0R making
x − y ∈ AR . Also for all x, y ∈ AR , axy = (ax)y = 0R y = 0R making xy ∈ AR . Hence
AR is a subring of R.

2. Given that 0S , 0R are the zero elements in S and R respectively and since S
is a subring of R, 0R ∈ S so that 0S , 0R ∈ S. Thus for all x ∈ S, x − x ∈ S, where
x − x = 0S = 0R since x ∈ R as well. Hence 0R = 0S .

3. For all a ∈ R, we have that (a+a)2 = a+a. However (a+a)2 = a2 +a2 +a2 +a2 =
a + a + a + a = a + a so that a + a = a + a − a − a = 0R

4. Since f is onto, all r2 ∈ R2 are of the form f (r1 ) = r2 , where r1 ∈ R1 . Thus For
all x2 , y2 ∈ R2 , such that x2 = f (x1 ), y2 = f (y1), where y1 , y2 ∈ R1 , we obtain that
x2 y2 = f (x1 )f (y1) = f (x1 y1 ) = f (y1 x1 ) = f (y1)f (x1 ) = y2 x2 making R2 commutative
as well.

5. Firstly observe that I × J ⊆ R × H. Since I, J are ideals of R, H respectively,

we have that 0 ∈ I, 0 ∈ J so that (0, 0) ∈ I × J. Let (x1 , y1 ), (x2 , y2 ) ∈ I × J
so that (x1 , y1) − (x2 , y2 ) = (x1 − x2 , y1 − y2 ). However x1 − x2 ∈ I, y1 − y2 ∈ J
since I, J are ideals of R and H respectively and thus making (x1 , y1 ) − (x2 , y2) =
(x1 − x2 , y1 − y2 ) ∈ I × J. Also we get that (x1 , y1 )(x2 , y2 ) = (x1 x2 , y1 y2 ). However
x1 x2 ∈ I, y1y2 ∈ J since I, J are ideals of R and H respectively and thus making
(x1 , y1 )(x2 , y2 ) = (x1 x2 , y1 y2 ) ∈ I × J. Thus I × J is a subring of R × H. Let
(x, y) ∈ I × J and (r, h) ∈ R × H so that (x, y)(r, h) = (xr, yh), where xr ∈ I, yh ∈ J
since I, J are ideals making (x, y)(r, h) = (xr, yh) ∈ I ×J. Also (r, h)(x, y) = (rx, hy),
where rx ∈ I, hy ∈ J as I, J are two-sided ideals and thus making (r, h)(x, y) =
(rx, hy) ∈ I × J. Hence I × J is a two-sided ideal of R × H.

6. Since we are told that f is a function, thus f is well-defined. We have that

f (x + y) = (x + y + I, x + y + J) = ((x + I) + (y + I), (x + J) + (y + J)) =
2

((x + I, x + J)) + ((y + I, y + J)) = f (x) + f (y). Also we obtain that f (xy) =
(xy+I, xy+J) = ((x+I)(y+I), (x+J)(y+J)) = (x+I, x+J)(y+I, y+J) = f (x)f (y).
Hence f is a ring homomorphism.

7. Observe that I/K ⊆ R/K. Since I is an ideal in R, we have that 0 ∈ I so that

0 + K = K ∈ I/K. Let x + K, y + K ∈ I/K, where x, y ∈ I. Thus x + K − y + K =
x−y +K ∈ I/K since x−y ∈ I as I is an ideal. Also (x+K)(y +K) = xy +K ∈ I/K
since xy ∈ I as I is an ideal. Thus I/K is a subring of R/K. For all x + K ∈ I/K
and r + K ∈ R/K, we have that (x + K)(r + K) = xr + K ∈ I/K as xr ∈ I as I is a
two-sided ideal. Also (r + K)(x + K) = rx + K ∈ I/K as rx ∈ I as I is a two-sided
ideal. Hence I/K is a two-sided ideal of R/K.
 1

MAT 3702 Assignment 1 Solutions Semester 2 2019

1. Suppose that x = y so that x−1 = y −1 so that f (x) = f (y) making f well-

defined. If f (x) = f (y), then x−1 = y −1 so that x = y making f one-to-one. For all
x ∈ G, we have that x−1 ∈ G such that f (x−1 ) = (x−1 )−1 = x making f onto. Hence
f is a bijection.

2.      
1 2 3   1 2 3   1 2 3 
f ◦g = ◦ =
3 1 2 2 1 3 1 2 3
     
1 2 3   1 2 3   1 2 3 
g◦f = ◦ =
2 3 1 3 1 2 1 2 3

3. (i) It follows immediately from the definition that NG (H) ⊆ G. We have that
e ∈ G is such that eHe−1 = H so that e ∈ NG (H) and so making NG (H) 6= ∅.
Let x, y ∈ NG (H) so that xHx−1 = H = yHy −1. From yHy −1 = H, we obtain
that yH = Hy so that H = y −1Hy = y −1 H(y −1)−1 making y −1 ∈ NG (H). Thus
xy −1 H(xy −1)−1 = xy −1 Hyx−1 = x(y −1 Hy)x−1 = xHx−1 = H so that xy −1 ∈ NG (H)
making NG (H) a subgroup of G. For all x ∈ H, we obtain that xHx−1 = H so that
x ∈ NG (H) making H ⊆ NG (H).
(ii) We have from (i) above that NG (H) ⊆ G. If H is normal in G, then for all
g ∈ G, we have that gHg −1 = H so that g ∈ NG (H) making G ⊆ NG (H). Thus from
NG (H) ⊆ G and G ⊆ NG (H), we obtain that NG (H) = G

4. Let x, y ∈ G such that x = y. Thus c−1 xc = c−1 yc so that f (x) = f (y)

making f well-defined. For x, y ∈ G, we obtain that f (xy) = c−1 xyc = c−1 xcc−1 yc =
f (x)f (y) making f a homomorphism. If x, y ∈ G such that f (x) = f (y), then
c−1 xc = c−1 yc =⇒ xc = yc =⇒ x = y making f one-to-one. For all x ∈ G and for
c ∈ G a fixed element, there exits cxc−1 ∈ G such that f (cxc−1 ) = c−1 (cxc−1 )c = x
making f onto. Hence f is an isomorphism.

5. For all x ∈ ZZ, we have that x − x = 0 = 0 × 3 which is a multiple of 3. Thus

2

x ∼ x making ∼ reflexive. If x, y ∈ ZZ such that x ∼ y, then x − y = 3k so that

−y + x = 3k making y − x = −3k = 3(−k) so that y ∼ x making ∼ symmetric. If
x, y, z ∈ ZZ such that x ∼ y and y ∼ z, then x − y = 3k and y − z = 3k ′ so that
x − y + y − z = x − z = 3k + 3k ′ = 3(k + k ′ ) so that x ∼ z making ∼ transitive.
Hence ∼ defines an equivalence relation on ZZ .
 SEMESTER 2
ASSIGNMENT 01
DUE DATE: 16 August 2019
UNIQUE ASSIGNMENT NO: 818352

1. Let G be a group with identity element e ∈ G and f : G −→ G be defined by f (x) = x−1 . Show
that f is well-defined and a bijection

2. Let f, g ∈ S3 be given by
   
1 2 3 1 2 3
f= ; g=
3 1 2 2 3 1

and compute f ◦ g, g ◦ f

3. Let G be a group and H be a subgroup of G. Define

NG (H) = {g ∈ G | gHg −1 = H}

(i) Show that NG (H) is a subgroup of G that contains H.

(ii) If H is normal in G, then show that NG (H) = G

4. Let G be a group and c ∈ G be a fixed element. Define f : G −→ G by f (g) = c−1 gc and show
that f is well-defined and an isomorphism.

5. Define a relation ∼ on Z by x ∼ y if x − y is a multiple of 3. Then check as to whether or not

∼ defines an equivalence relation on Z.

12
 MAT3702/101/3/2019

SEMESTER 2
ASSIGNMENT 02
DUE DATE: 13 September 2019
UNIQUE ASSIGNMENT NO: 835604

1. Let R be a ring and b ∈ R be a fixed element. Define

T = {rb | r ∈ R}

and prove that T is a subring of R.

2. Let R and S be rings and define

R × S = {(r, s) | r ∈ R, s ∈ S}

with addition and multiplication defined componentwise on R × S which thus makes R × S a

ring.

(i) If R and S are both commutative, then prove that R × S is also commutative

(ii) If R and S both have identity, then prove that R × S also has an identity

3. Let R be a ring with identity and u ∈ R be a unit.

(i) Prove that u ∈ R is NOT a zero divisor

(ii) If a, b ∈ R are units, then prove that ab is also a unit and that (ab)−1 = b−1 a−1

√ √ √ √ √
√ 2) = {r + s 2 | r, s ∈ Q} and prove that f : Q( 2) −→ Q( 2) given by f (a + b 2) =
4. Let Q(
a − b 2 is well-defined and an isomorphism.

5. Let φ : R −→ S be a ring homomorphism. Define

K = {r ∈ R | φ(r) = 0S },

where 0S ∈ S is the zero element in S. Prove that

(i) K is a subring of R

(ii) K is a two-sided ideal of R

6. Using contradiction, prove that the rings Z4 and Z2 × Z2 are NOT isomorphic

13
 1

MAT 3702 Assignment 2 Solutions Semester 2 2019

1. We have that 0R b = 0R so that 0R ∈ T . For all xb, yb ∈ T , we have that

xb − yb = (x − y)b ∈ T since x − y ∈ R as x, y ∈ R. Also for all xb, yb ∈ T, xbyb =
(xby)b ∈ T since x, b, y ∈ R so that xby ∈ R. Hence T is a subring of R.

2. (i) Since R and S are both commutative, we obtain that r1 r2 = r2 r1 and

s1 s2 = s2 s1 for all r1 , r2 ∈ R and s1 , s2 ∈ S. Thus for all (r1 , s1 ), (r2 , s2 ) ∈ R × S, we
obtain that (r1 , s1 )(r2 , s2 ) = (r1 r2 , s1 s2 ) = (r2 r1 , s2 s1 ) = (r2 , s2 )(r1 , s1 ) thus making
R × S commutative.
(ii) Since R and S both have identity elements, write 1R , 1S for the identities of
R and S respectively so that (1R , 1S ) ∈ R × S. So for all (r, s) ∈ R × S, we get that
(r, s)(1R , 1S ) = (r1R , s1S ) = (r, s). Also we obtain that (1R , 1S )(r, s) = (1R r, 1S s) =
(r, s). Thus (1R , 1S ) ∈ R × S is the identity.

3. (i) Since u ∈ R is a unit, we have that u 6= 0 and that there exists v ∈ R

such that uv = 1R = vu, where 1R ∈ R is the identity. Suppose that u ∈ R is a zero
divisor. So there exits x 6= 0 ∈ R such that ux = 0 so that vux = v0 = 0 =⇒ (vu)x =
0 =⇒ 1R x = 0 =⇒ x = 0 which is a contradiction since x 6= 0 ∈ R. Hence u ∈ R is
not a zero divisor.
(ii) Since a, b ∈ R are units, a−1 , b−1 ∈ R exist such that aa−1 = 1R = a−1 a
and bb−1 = 1R = b−1 b. Since a−1 , b−1 ∈ R, b−1 a−1 ∈ R such that (ab)(b−1 a−1 ) =
a(bb−1 )a−1 = a1R a−1 = 1R . Also we obtain that (b−1 a−1 )(ab) = b−1 (a−1 a)b =
b−1 1R b = 1R . Thus ab ∈ R is also a unit and actually (ab)−1 = b−1 a−1 .

4. Let a, b, c, d ∈ Q
| such that (a, b) = (c, d) from which we obtain that a = c, b = d
√ √ √ √
giving that a−b 2 = c−d 2 so that f (a+b 2) = f (c+d 2) making f well-defined.
√ √ √ √
We obtain that f (a + b 2 + c + d 2) = f (a + c + (b + d) 2) = a + c − (b + d) 2 =
√ √ √ √ √ √
a − b 2 + c − d 2 = f (a + b 2) + f (c + d 2). Also f (a + b 2)(c + d 2) =
√ √ √ √
f (ac + ad 2 + bc 2 + 2bd) = f (ac + 2bd + (ad + bc) 2) = ac + 2bd − (ad + bc) 2
√ √ √ √ √ √
and f (a + b 2)f (c + d 2) = (a − b 2)(c − d 2) = ac − ad 2 − bc 2 + 2bd =
√ √ √ √ √
ac + 2bd − (ad + bc) 2. Thus f (a + b 2)(c + d 2) = f (a + b 2)f (c + d 2) making
√ √ √ √
f a ring homomorphism. If f (a + b 2) = f (c + d 2), then a − b 2 = c − d 2
2

√ √
so that a = c, b = d giving that a + b 2 = c + d 2 making f one-to-one. For all
√ √ √ √ √ √
a + b 2 ∈ Q(
| 2), there exists a − b 2 ∈ Q(
| 2) such that f (a − b 2) = a + b 2
making f onto. Hence f is an isomorphism.

5. (i) Firstly observe from the definition that K ⊆ R. Since φ is a homomorphism,

we get that φ(0R ) = 0S , where 0R ∈ R is the zero element in R, thus making 0R ∈ K.
Let x, y ∈ K so that φ(x) = 0S = φ(y). Thus φ(x − y) = φ(x) − φ(y) = 0S − 0S = 0S
so that x − y ∈ K. Also φ(xy) = φ(x)φ(y) = 0S so that xy ∈ K. Hence K is a
subring of R.
(ii) From (i) above, we have that K is a subring of R. For all r ∈ R, x ∈ K we have
that φ(rx) = φ(r)φ(x) = φ(r)0S = 0S so that rx ∈ K. Also for all r ∈ R, x ∈ K we
have that φ(xr) = φ(x)φ(r) = 0S φ(r) = 0S so that xr ∈ K. Hence K is a two-sided
ideal of R.

6. Suppose that f : ZZ 4 −→ ZZ 2 × ZZ 2 is an isomorphism. Since f is a ho-

momorphism, we thus obtain that f (0) = (0, 0) and f (1) = (1, 1). Furthermore
we obtain that f (2) = f (1 + 1) = f (1) + f (1) since f is a homomorphism and so
we obtain that f (2) = f (1) + f (1) = (1, 1) + (1, 1) = (0, 0). Also we obtain that
f (3) = f (1 + 1 + 1) = f (1) + f (1) + f (1) since f is a homomorphism giving that
f (3) = (1, 1) + (1, 1) + (1, 1) = (1, 1). However we have that (0, 1), (1, 0) have not
been used, thus contradicting the surjectivity of f since it (f ) was supposed to be an
isomorphism. Hence ZZ 4 and ZZ 2 × ZZ 2 are NOT isomorphic.