UPSC Optional (Maths) Paper-02 Algebra Solution

2011

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UPSC Algebra

1(a) Show that the set \[ G=\left\{f_1, f_2, f_3, f_4, f_5, f_6\right\} \] of six transformations on the set of Complex numbers defined by \[ \begin{aligned} &f_1(z)=z, f_2(z)=1-z \\ &f_3(z)=\frac{z}{(z-1)}, f_4(z)=\frac{1}{z} \\ &f_5(z)=\frac{1}{(1-z)} \text { and } f_6(z)=\frac{(z-1)}{z} \end{aligned} \] is a non-abelian group of order 6 w.r.t. composition of mappings.

Expert Answer

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

To show that the set

G = {f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}}

forms a group under composition of mappings, we need to verify the following group axioms:

Closure

We need to show that the composition of any two functions in

G

is also in

G

For example, let’s consider

f_{3} \circ f_{4}

. The composition

f_{3} (f_{4} (z))

is given by:

f_{3} (f_{4} (z)) = f_{3} (\frac{1}{z}) = \frac{\frac{1}{z}}{\frac{1}{z} - 1} = \frac{1}{1 - z}

After calculating, we find that

f_{3} \circ f_{4} = f_{5}

, which is in

G

Similarly, we can show that the composition of any two functions in

G

will result in another function in

G

. Therefore,

G

is closed under composition.

Associativity

Composition of functions is always associative. That is, for any

f, g, h \in G

(f \circ g) \circ h = f \circ (g \circ h)

Identity Element

The function

f_{1} (z) = z

serves as the identity element for

G

under composition. For any

f \in G

f_{1} \circ f = f

and

f \circ f_{1} = f

Inverse Element

Each function in

G

must have an inverse that is also in

G

$f_{1}$ is its own inverse.
$f_{2}$ is its own inverse.
$f_{3}$ and $f_{6}$ are inverses of each other.
$f_{4}$ and $f_{5}$ are inverses of each other.

Non-Abelian

To show that

G

is non-abelian, we need to find

f, g \in G

such that

f \circ g \neq g \circ f

Let’s consider

f_{3}

and

f_{4}

$f_{3} \circ f_{4} = f_{5}$
$f_{4} \circ f_{3} = f_{6}$

Since

f_{5} \neq f_{6}

G

is non-abelian.

Conclusion

Since

G

satisfies all the group axioms and is non-abelian, it is a non-abelian group of order 6 with respect to composition of mappings.

Verified Answer

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1(e) (i) Prove that a group of Prime order is abelian.

Expert Answer

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

To prove that a group

G

of prime order

p

is abelian, we need to show that for all

a, b \in G

a \cdot b = b \cdot a

Step 1: Group Order is Prime

Let

p

be a prime number, and let

G

be a group of order

p

Step 2: Elements in the Group

Since

G

has order

p

, it has

p

elements. One of these elements is the identity element

e

Step 3: Properties of Subgroups

By Lagrange’s theorem, the order of any subgroup of

G

must divide the order of

G

. Since

p

is prime, the only divisors of

p

are 1 and

p

itself. Therefore, any subgroup of

G

must have order 1 or

p

Step 4: Subgroups are Trivial or the Whole Group

A subgroup of order 1 is trivial and contains only the identity element

e

. A subgroup of order

p

must be

G

itself.

Step 5: Every Element Generates the Group

Take any element

a \in G

where

a \neq e

. The subgroup generated by

a

, denoted

⟨ a ⟩

, must have order greater than 1 (since

a \neq e

). By Step 4,

⟨ a ⟩

must therefore be the whole group

G

Step 6: Commutativity

Take any two elements

a, b \in G

. Both

a

and

b

generate

G

(by Step 5). Therefore,

b

can be written as

a^{n}

for some integer

n

, and

a

can be written as

b^{m}

for some integer

m

Now,

a \cdot b = a \cdot a^{n} = a^{n + 1}

and

b \cdot a = a^{n} \cdot a = a^{n + 1}

Since

a \cdot b = b \cdot a

G

is abelian.

Conclusion

We have shown that any group of prime order

p

must be abelian.

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1(e) (ii) How many generators are there of the cyclic group $ (G, .) $ of order 8?

Expert Answer

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

In a cyclic group

G

of order

n

, the number of generators is equal to the number of integers

k

that are relatively prime to

n

. These integers

k

are in the range

1 \leq k < n

Step 1: Identify the Order of the Group

The order of the group

G

is given as 8.

Step 2: Find Integers Relatively Prime to 8

We need to find the integers

k

that are relatively prime to 8 and lie in the range

1 \leq k < 8

The integers in this range are

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

Out of these, the integers that are relatively prime to 8 are

{1, 3, 5, 7}

Step 3: Count the Number of Generators

The number of generators of

G

is equal to the number of integers that are relatively prime to 8, which is 4.

Conclusion

There are 4 generators of the cyclic group

G

of order 8.

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2(a) Give an example of a group $ G $ in which every proper subgroup is cyclic but the group itself is not cyclic.

Expert Answer

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

A classic example of a group

G

where every proper subgroup is cyclic but

G

itself is not cyclic is the Klein four-group

V_{4}

Definition of

V_{4}

The Klein four-group

V_{4}

is defined as the set

{e, a, b, c}

with the following multiplication table:

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

Proper Subgroups are Cyclic

The proper subgroups of

V_{4}

are:

${e, a}$ generated by $a$
${e, b}$ generated by $b$
${e, c}$ generated by $c$

Each of these subgroups is cyclic, as they are generated by a single element.

V_{4}

is Not Cyclic

The group

V_{4}

itself is not cyclic because there is no single element that generates the entire group. Each of

a, b, c

only generates a subgroup of order 2, and

e

generates only the trivial subgroup

{e}

Conclusion

The Klein four-group

V_{4}

is an example of a group where every proper subgroup is cyclic, but the group itself is not cyclic.

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3(a) Let $ F $ be the set of all real-valued continuous functions defined on the closed interval $[0,1]$. Prove that $ (F,+,.) $ is a Commutative Ring with unity with respect to addition and multiplication of functions defined pointwise.

Expert Answer

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

To prove that

(F, +, \cdot)

is a commutative ring with unity, we need to verify the following properties:

Closure under Addition and Multiplication

For any

f, g \in F

f + g

and

f \cdot g

are also real-valued continuous functions defined on

[0, 1]

. Therefore,

F

is closed under addition and multiplication.

Associativity of Addition and Multiplication

For any

f, g, h \in F

(f + g) + h = f + (g + h) and (f \cdot g) \cdot h = f \cdot (g \cdot h)

Commutativity of Addition and Multiplication

For any

f, g \in F

f + g = g + f and f \cdot g = g \cdot f

Existence of Additive and Multiplicative Identity (Unity)

The zero function

f (x) = 0

for all

x \in [0, 1]

serves as the additive identity, and the function

f (x) = 1

for all

x \in [0, 1]

serves as the multiplicative identity (unity).

For any

f \in F

f + 0 = f and f \cdot 1 = f

Existence of Additive Inverse

For any

f \in F

, the function

- f

defined by

- f (x) = - f (x)

for all

x \in [0, 1]

is the additive inverse of

f

f + (- f) = 0

Distributive Law

For any

f, g, h \in F

f \cdot (g + h) = (f \cdot g) + (f \cdot h)

Conclusion

Since all these properties are satisfied,

(F, +, \cdot)

is a commutative ring with unity with respect to addition and multiplication of functions defined pointwise.

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4(a) Let $ a $ and $ b $ be the elements of a group, with $ a^2=e $, $ b^6=e $ and $ a b = b^4 a $. Find the order of $ ab $, and express its inverse in each of the.

Expert Answer

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

To find the order of

a b

and its inverse, we need to consider the given relations:

$a^{2} = e$
$b^{6} = e$
$a b = b^{4} a$

Step 1: Find the Order of

a b

To find the order of

a b

, we need to find the smallest positive integer

n

such that

(a b)^{n} = e

Let’s substitute the values into the formula and simplify:

(a b)^{2} = a b a b

Using the relation

a b = b^{4} a

, we get:

(a b)^{2} = b^{4} a \cdot b^{4} a = b^{4} \cdot b^{4} \cdot a \cdot a = b^{8} \cdot a^{2} = b^{2} \cdot e = b^{2}

Similarly, we can find

(a b)^{3}

(a b)^{3} = (a b)^{2} \cdot a b = b^{2} \cdot b^{4} a = b^{6} \cdot a = e \cdot a = a

Finally, let’s find

(a b)^{6}

(a b)^{6} = (a b)^{3} \cdot (a b)^{3} = a \cdot a = a^{2} = e

So, the smallest positive integer

n

such that

(a b)^{n} = e

n = 6

. Therefore, the order of

a b

is 6.

Step 2: Find the Inverse of

a b

The inverse of an element

x

in a group is an element

x^{- 1}

such that

x \cdot x^{- 1} = e

To find

(a b)^{- 1}

, we need to find an element that, when multiplied by

a b

, gives

e

Since

(a b)^{6} = e

, we know that

(a b)^{- 1} = (a b)^{5}

Let’s substitute the values into the formula and simplify:

(a b)^{5} = (a b)^{2} \cdot (a b)^{2} \cdot a b = b^{2} \cdot b^{2} \cdot b^{4} a = b^{8} \cdot a = b^{2} \cdot a

After calculating, we find that

(a b)^{- 1} = b^{2} a

Conclusion

The order of

a b

is 6, and its inverse is

b^{2} a

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2011

Expert Answer

1(e) (i) Prove that a group of Prime order is abelian.

Expert Answer

1(e) (ii) How many generators are there of the cyclic group \( (G, .) \) of order 8?

Expert Answer

2(a) Give an example of a group \( G \) in which every proper subgroup is cyclic but the group itself is not cyclic.

Expert Answer

3(a) Let \( F \) be the set of all real-valued continuous functions defined on the closed interval \([0,1]\). Prove that \( (F,+,.) \) is a Commutative Ring with unity with respect to addition and multiplication of functions defined pointwise.

Expert Answer

4(a) Let \( a \) and \( b \) be the elements of a group, with \( a^2=e \), \( b^6=e \) and \( a b = b^4 a \). Find the order of \( ab \), and express its inverse in each of the.

Expert Answer

Get Complete Solution Mathematics Optional Paper | UPSC

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

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

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