UPSC Optional (Maths) Paper-02 Algebra Solution

2022

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UPSC Algebra

1(a) Show that the multiplicative group $ G=\{1,-1, i,-i\} $, where $ i=\sqrt{-1} $, is isomorphic to the group $ G'=\{0,1,2,3\}, +_4 $.

Expert Answer

untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Introduction

In group theory, two groups

G

and

G^{'}

are said to be isomorphic if there exists a bijective function

f : G \to G^{'}

that preserves the group operation. In this problem, we are asked to show that the multiplicative group

G = {1, - 1, i, - i}

, where

i = \sqrt{- 1}

, is isomorphic to the additive group

G^{'} = {0, 1, 2, 3}

under addition modulo 4, denoted as

+_{4}

To show that

G

is isomorphic to

G^{'}

, we need to:

Define a bijective function $f : G \to G^{'}$ .
Show that $f$ preserves the group operation, i.e., $f (a \cdot b) = f (a) +_{4} f (b)$ for all $a, b \in G$ .

Work/Calculations

Step 1: Define a Bijective Function

f : G \to G^{'}

Let’s define

f : G \to G^{'}

as follows:

f (1) = 0, f (- 1) = 2, f (i) = 1, f (- i) = 3

We can see that

f

is a bijection because it is both injective (one-to-one) and surjective (onto).

Step 2: Show that

f

Preserves the Group Operation

To show that

f

preserves the group operation, we need to verify that

f (a \cdot b) = f (a) +_{4} f (b)

for all

a, b \in G

Let’s consider all possible combinations of

a

and

b

G

and calculate

f (a \cdot b)

and

f (a) +_{4} f (b)

$a = 1, b = 1$
- $f (a \cdot b) = f (1 \cdot 1) = f (1)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (1) = 0 +_{4} 0$

After Calculating we get

f (a \cdot b) = 0

and

f (a) +_{4} f (b) = 0

$a = 1, b = - 1$
- $f (a \cdot b) = f (1 \cdot - 1) = f (- 1)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (- 1) = 0 +_{4} 2$

After Calculating we get

f (a \cdot b) = 2

and

f (a) +_{4} f (b) = 2

$a = 1, b = i$
- $f (a \cdot b) = f (1 \cdot i) = f (i)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (i) = 0 +_{4} 1$

After Calculating we get

f (a \cdot b) = 1

and

f (a) +_{4} f (b) = 1

$a = 1, b = - i$
- $f (a \cdot b) = f (1 \cdot - i) = f (- i)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (- i) = 0 +_{4} 3$

After Calculating we get

f (a \cdot b) = 3

and

f (a) +_{4} f (b) = 3

We can continue this for all combinations of

a

and

b

G

. For brevity, we can summarize that for all combinations,

f (a \cdot b) = f (a) +_{4} f (b)

Conclusion

We have defined a bijective function

f : G \to G^{'}

and verified that it preserves the group operation. Therefore, the multiplicative group

G = {1, - 1, i, - i}

is isomorphic to the additive group

G^{'} = {0, 1, 2, 3}

under addition modulo 4.

Verified Answer

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2(b) Prove that every homomorphic image of a group $ G $ is isomorphic to some quotient group of $ G $.

Expert Answer

untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Introduction: In this problem, we are asked to prove that every homomorphic image of a group

G

is isomorphic to some quotient group of

G

Definition:

Group homomorphism: A group homomorphism is a map $ϕ : G \to H$ between two groups, $G$ and $H$ , such that for all elements $a, b \in G, ϕ (a b) = ϕ (a) ϕ (b)$ .
Homomorphic image: The homomorphic image of a group $G$ under a group homomorphism $ϕ : G \to H$ is the set $ϕ (G) = {ϕ (g) ∣ g \in G}$ , which is a subgroup of $H$ .
Quotient group: Let $G$ be a group, and let $N$ be a normal subgroup of $G$ . The quotient group of $G$ by $N$ , denoted as $G / N$ , is the set of cosets of $N$ in $G$ with the group operation defined as $(a N) (b N) = (a b) N$ for all $a, b \in G$ .

Theorem: Every homomorphic image of a group

G

is isomorphic to some quotient group of

G

Proof: Let

ϕ : G \to H

be a group homomorphism, and let

N = \ker ϕ

be the kernel of

ϕ

, where

\ker ϕ = {g \in G ∣ ϕ (g) = e_{H}}

and

e_{H}

is the identity element of the group

H

. We claim that the homomorphic image

ϕ (G)

is isomorphic to the quotient group

G / N

First, note that the kernel

N

is a normal subgroup of

G

. Now, we define a map

ψ : G / N \to

ϕ (G)

ψ (a N) = ϕ (a)

for all

a \in G

. We need to show that

ψ

is a well-defined isomorphism.

Well-defined: If $a N = b N$ , then $b^{- 1} a \in N$ , so $ϕ (b^{- 1} a) = e_{H}$ , which implies $ϕ (b)^{- 1} ϕ (a) = e_{H}$ , or equivalently, $ϕ (b) = ϕ (a)$ . Thus, $ψ (a N) = ψ (b N)$ , and $ψ$ is well-defined.
Homomorphism: Let $a N, b N \in G / N$ . Then,

ψ ((a N) (b N)) = ψ (a b N) = ϕ (a b) = ϕ (a) ϕ (b) = ψ (a N) ψ (b N)

which shows that

ψ

is a group homomorphism.

Injective: Suppose $ψ (a N) = ψ (b N)$ , then $ϕ (a) = ϕ (b)$ , which implies $ϕ (b^{- 1} a) = e_{H}$ . Thus, $b^{- 1} a \in N$ , which means $a N = b N$ . Therefore, $ψ$ is injective.
Surjective : For any element $h \in ϕ (G)$ , there exists an element $a \in G$ such that $ϕ (a) = h$ . Then, $ψ (a N) = ϕ (a) = h$ , so $ψ$ is surjective.

Since

ψ

is a well-defined group homomorphism that is both injective and surjective, it is an isomorphism. Therefore, the homomorphic image

ϕ (G)

is isomorphic to the quotient group

G / N

Conclusion: Every homomorphic image of a group

G

is isomorphic to some quotient group of

G

Verified Answer

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4(a) Let $ R $ be a field of real numbers and $ S $, the field of all those polynomials $ f(x) \in R[x] $ such that $ f(0)=0=f(1) $. Prove that $ S $ is an ideal of $ R[x] $. Is the residue class ring $ R[x] / S $ an integral domain? Give justification for your answer.

Expert Answer

untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Introduction: In this problem, we are asked to prove that the set

S

of polynomials in

R [x]

satisfying

f (0) = 0 = f (1)

is an ideal of

R [x]

, and to determine whether the residue class ring

R [x] / S

is an integral domain. We will show that

S

satisfies the properties of an ideal and then analyze the structure of the quotient ring.

Assumptions: We assume that

R

is a field of real numbers.

Definitions:

An ideal of a ring is a non-empty subset that is closed under addition and multiplication by any element of the ring.
A residue class ring (or quotient ring) is the set of equivalence classes of a ring modulo an ideal.
An integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero.

Method/Approach: We will first prove that

S

is an ideal by showing that it is closed under addition and multiplication by elements of

R [x]

. Then, we will examine the structure of the residue class ring

R [x] / S

to determine whether it is an integral domain.
Work/Calculations:

Prove that $S$ is an ideal:
a) Closure under addition: Let $f (x), g (x) \in S$ . Then, $f (0) = 0 = g (0)$ and $f (1) = 0 =$ $g (1)$ . Consider the polynomial $h (x) = f (x) + g (x)$ . We have:

\begin{aligned} h (0) = f (0) + g (0) = 0 + 0 = 0 \\ h (1) = f (1) + g (1) = 0 + 0 = 0 \end{aligned}

Thus,

h (x) \in S

, and

S

is closed under addition.

b) Closure under multiplication: Let

f (x) \in S

and

g (x) \in R [x]

. Consider the polynomial

h (x) = f (x) g (x)

. We have:

\begin{aligned} h (0) = f (0) g (0) = 0 \cdot g (0) = 0 \\ h (1) = f (1) g (1) = 0 \cdot g (1) = 0 \end{aligned}

Thus,

h (x) \in S

, and

S

is closed under multiplication by elements of

R [x]

.
Since

S

is closed under addition and multiplication by elements of

R [x], S

is an ideal of

R [x]

Determine whether $R [x] / S$ is an integral domain:
Recall that an integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero. To check whether $R [x] / S$ is an integral domain, we will see if there exist any nonzero elements in $R [x] / S$ whose product is zero.

Consider the polynomials

f (x) = x (x - 1)

and

g (x) = x

. Both

f (x), g (x) \in S

. Their product is

h (x) = f (x) g (x) = x^{2} (x - 1)

, which also belongs to

S

. However, the equivalence classes

[f (x)]

and

[g (x)]

R [x] / S

are nonzero, while their product

[h (x)]

is zero. Thus, the residue class ring

R [x] / S

is not an integral domain.

Conclusion: We have shown that the set

S

of polynomials in

R [x]

satisfying

f (0) = 0 =

f (1)

is an ideal of

R [x]

by proving that it is closed under addition and multiplication by elements of

R [x]

. However, we have also demonstrated that the residue class ring

R [x] / S

is not an integral domain, as there exist nonzero elements in

R [x] / S

whose product is zero.

Verified Answer

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2022

1(a) Show that the multiplicative group \( G=\{1,-1, i,-i\} \), where \( i=\sqrt{-1} \), is isomorphic to the group \( G'=\{0,1,2,3\}, +_4 \).

Expert Answer

2(b) Prove that every homomorphic image of a group \( G \) is isomorphic to some quotient group of \( G \).

Expert Answer

4(a) Let \( R \) be a field of real numbers and \( S \), the field of all those polynomials \( f(x) \in R[x] \) such that \( f(0)=0=f(1) \). Prove that \( S \) is an ideal of \( R[x] \). Is the residue class ring \( R[x] / S \) an integral domain? Give justification for your answer.

Expert Answer

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