UPSC Optional (Maths) Paper-02 Algebra Solution

2015

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UPSC Algebra

1(a) (i) How many generators are there of the cyclic group $G$ of order 8? Explain.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

To find the number of generators of a cyclic group

G

of order 8, we need to understand the properties of cyclic groups and their generators.

Step 1: Definition of a Cyclic Group

A cyclic group

G

of order

n

is a group that can be generated by a single element

a

such that every element in

G

can be written as

a^{k}

for

k = 0, 1, 2, \dots, n - 1

Step 2: Generators of a Cyclic Group

In a cyclic group

G

of order

n

, an element

a

is a generator if its order is

n

. The order of an element

a

is the smallest positive integer

m

such that

a^{m} = e

, where

e

is the identity element of

G

Step 3: Criteria for Generators in Cyclic Groups

In a cyclic group of order

n

, the elements that are generators are precisely those elements

a

for which

gcd (a, n) = 1

. In other words,

a

is a generator if and only if

a

is relatively prime to

n

Step 4: Counting the Generators for

G

of Order 8

The group

G

has order 8, so we need to find the number of elements in

G

that are relatively prime to 8. The elements of

G

can be represented as

{0, 1, 2, 3, 4, 5, 6, 7}

Let’s find the elements that are relatively prime to 8:

$gcd (1, 8) = 1$
$gcd (2, 8) = 2$
$gcd (3, 8) = 1$
$gcd (4, 8) = 4$
$gcd (5, 8) = 1$
$gcd (6, 8) = 2$
$gcd (7, 8) = 1$

The elements

1, 3, 5, 7

are relatively prime to 8.

Summary

There are 4 generators of the cyclic group

G

of order 8, and they are

1, 3, 5, 7

Verified Answer

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1(a) (ii) Taking a group $\{e, a, b, c\}$ of order 4, where $e$ is the identity, construct composition tables showing that one is cyclic while the other is not.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

To construct composition tables for groups of order 4, we need to consider the group axioms:

Closure: The operation is closed under the set.
Associativity: The operation is associative.
Identity: There exists an identity element.
Inverses: Every element has an inverse.

Composition Table for a Cyclic Group

Let’s consider a cyclic group

G

with elements

{e, a, b, c}

and

e

as the identity element. In a cyclic group, one element (other than the identity) generates the entire group. Let’s assume

a

is the generator. Then:

$a^{0} = e$
$a^{1} = a$
$a^{2} = b$
$a^{3} = c$
$a^{4} = e$ (since it’s of order 4)

The composition table would look like:

\begin{array}{ccccc} \circ & e & a & b & c \\ e & e & a & b & c \\ a & a & b & c & e \\ b & b & c & e & a \\ c & c & e & a & b \end{array}

Here,

a

generates the entire group, making it cyclic.

Composition Table for a Non-Cyclic Group

Now let’s consider a non-cyclic group

H

with elements

{e, a, b, c}

and

e

as the identity element. In a non-cyclic group, no single element generates the entire group. We can define the operation as follows:

$e$ is the identity: $e \circ x = x \circ e = x$ for all $x$
$a \circ a = b$ , $a \circ b = e$ , $a \circ c = c$
$b \circ a = e$ , $b \circ b = a$ , $b \circ c = c$
$c \circ a = c$ , $c \circ b = c$ , $c \circ c = c$

The composition table would look like:

\begin{array}{ccccc} \circ & e & a & b & c \\ e & e & a & b & c \\ a & a & b & e & c \\ b & b & e & a & c \\ c & c & c & c & c \end{array}

Here, no single element generates the entire group, making it non-cyclic.

Summary

We have constructed composition tables for a cyclic and a non-cyclic group of order 4. The cyclic group is generated by a single element, while the non-cyclic group is not generated by any single element.

Verified Answer

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2(a) If $R$ is a ring with unit element 1 and $\phi$ is a homomorphism of $R$ onto $R'$, prove that $\phi(1)$ is the unit element of $R'$.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

To prove that

ϕ (1)

is the unit element of

R^{'}

, we need to show that for any element

a^{'}

R^{'}

ϕ (1) \cdot a^{'} = a^{'} \cdot ϕ (1) = a^{'}

Step 1: Homomorphism Property

Recall that

ϕ

is a ring homomorphism from

R

onto

R^{'}

. This means that

ϕ

preserves both addition and multiplication:

$ϕ (a + b) = ϕ (a) + ϕ (b)$
$ϕ (a \cdot b) = ϕ (a) \cdot ϕ (b)$

Step 2:

ϕ

is Onto

Since

ϕ

is onto

R^{'}

, for every element

a^{'}

R^{'}

, there exists an element

a

R

such that

ϕ (a) = a^{'}

Step 3: Prove

ϕ (1)

is the Unit Element in

R^{'}

Let

a^{'}

be any element in

R^{'}

. Then there exists

a

R

such that

ϕ (a) = a^{'}

We want to show that

ϕ (1) \cdot a^{'} = a^{'}

and

a^{'} \cdot ϕ (1) = a^{'}

Using the homomorphism property, we have:

ϕ (1) \cdot a^{'} = ϕ (1) \cdot ϕ (a) = ϕ (1 \cdot a) = ϕ (a) = a^{'}

Similarly,

a^{'} \cdot ϕ (1) = ϕ (a) \cdot ϕ (1) = ϕ (a \cdot 1) = ϕ (a) = a^{'}

Summary

We have shown that for any element

a^{'}

R^{'}

ϕ (1) \cdot a^{'} = a^{'}

and

a^{'} \cdot ϕ (1) = a^{'}

. Therefore,

ϕ (1)

is the unit element of

R^{'}

Verified Answer

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4(a) Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields: (i) The set of numbers of the form $b \sqrt{2}$ with $b$ rational. (ii) The set of even integers. (iii) The set of positive integers.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

An integral domain is a commutative ring with an identity element

1 \neq 0

such that the ring has no zero divisors. A field is an integral domain in which every non-zero element has a multiplicative inverse.

(i) The set of numbers of the form

b \sqrt{2}

with

b

rational

Integral Domain Test

Closure under Addition: Given $a \sqrt{2} + b \sqrt{2} = (a + b) \sqrt{2}$ , where $a + b$ is rational, the set is closed under addition.
Closure under Multiplication: Given $a \sqrt{2} \times b \sqrt{2} = a b \times 2$ , where $a b$ is rational, the set is closed under multiplication.
Commutative: Ordinary addition and multiplication are commutative.
Identity: The additive identity is $0$ and the multiplicative identity is $1$ , which is not in the form $b \sqrt{2}$ unless $b = 0$ . Therefore, this set does not contain a multiplicative identity, and it is not an integral domain.

Field Test

Since it’s not an integral domain, it can’t be a field.

(ii) The set of even integers

Integral Domain Test

Closure under Addition: The sum of two even integers is even, so the set is closed under addition.
Closure under Multiplication: The product of two even integers is even, so the set is closed under multiplication.
Commutative: Ordinary addition and multiplication are commutative.
Identity: The additive identity is $0$ and the multiplicative identity is $1$ , which is not even. Therefore, this set does not contain a multiplicative identity, and it is not an integral domain.

Field Test

Since it’s not an integral domain, it can’t be a field.

(iii) The set of positive integers

Integral Domain Test

Closure under Addition: The sum of two positive integers is positive, so the set is closed under addition.
Closure under Multiplication: The product of two positive integers is positive, so the set is closed under multiplication.
Commutative: Ordinary addition and multiplication are commutative.
Identity: The additive identity is $0$ which is not positive, and the multiplicative identity is $1$ which is positive. Therefore, this set contains a multiplicative identity but not an additive identity, and it is not an integral domain.

Field Test

Since it’s not an integral domain, it can’t be a field.

Summary

None of the given sets form an integral domain with respect to ordinary addition and multiplication, and therefore, none of them are fields.

Verified Answer

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2015

1(a) (i) How many generators are there of the cyclic group \(G\) of order 8? Explain.

Expert Answer

1(a) (ii) Taking a group \(\{e, a, b, c\}\) of order 4, where \(e\) is the identity, construct composition tables showing that one is cyclic while the other is not.

Expert Answer

2(a) If \(R\) is a ring with unit element 1 and \(\phi\) is a homomorphism of \(R\) onto \(R'\), prove that \(\phi(1)\) is the unit element of \(R'\).

Expert Answer

4(a) Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields: (i) The set of numbers of the form \(b \sqrt{2}\) with \(b\) rational. (ii) The set of even integers. (iii) The set of positive integers.

Expert Answer

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UPSC Previous Years Maths Optional Papers with Solution | Paper-02

UPSC Previous Years Maths Optional Papers with Solution | Paper-01

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