UPSC Optional (Maths) Paper-02 Algebra Solution

2017

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UPSC Algebra

1(b) Let $G$ be a group of order $n$. Show that $G$ is isomorphic to a subgroup of the permutation group $S_n$.

Expert Answer

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

To show that a group

G

of order

n

is isomorphic to a subgroup of the symmetric group

S_{n}

, we can use Cayley’s Theorem. Cayley’s Theorem states that every group

G

is isomorphic to a subgroup of the symmetric group acting on

G

Proof:

Define the Action: Consider the action of $G$ on itself by left multiplication. For each $g \in G$ , define a function $f_{g} : G \to G$ by $f_{g} (x) = g x$ for all $x \in G$ .
Show that $f_{g}$ is a Permutation:
- Injective: Assume $f_{g} (x) = f_{g} (y)$ for some $x, y \in G$ . Then $g x = g y$ . Since $G$ is a group, every element has an inverse, so we can multiply both sides on the left by $g^{- 1}$ to get $x = y$ . This shows that $f_{g}$ is injective.
- Surjective: For any $y \in G$ , let $x = g^{- 1} y$ . Then $f_{g} (x) = g x = g (g^{- 1} y) = y$ . This shows that $f_{g}$ is surjective.
Since $f_{g}$ is both injective and surjective, it is a permutation of $G$ .
Define a Homomorphism: Define a function $Φ : G \to S_{n}$ by $Φ (g) = f_{g}$ . We need to show that $Φ$ is a homomorphism, i.e., $Φ (g h) = Φ (g) Φ (h)$ for all $g, h \in G$ .

For any $x \in G$ ,

$Φ (g h) (x) = f_{g h} (x) = (g h) x = g (h x) = f_{g} (f_{h} (x)) = Φ (g) (Φ (h) (x))$

This shows that $Φ (g h) = Φ (g) Φ (h)$ , so $Φ$ is a homomorphism.
Show that $Φ$ is Injective:
Assume $Φ (g) = Φ (h)$ for some $g, h \in G$ . Then $f_{g} = f_{h}$ , which means $g x = h x$ for all $x \in G$ . Letting $x = g^{- 1}$ , we get $g (g^{- 1}) = h (g^{- 1})$ , which simplifies to $e = h g^{- 1}$ , where $e$ is the identity element in $G$ . Multiplying both sides on the right by $g$ , we get $g = h$ . This shows that $Φ$ is injective.
Conclusion: Since $Φ$ is an injective homomorphism from $G$ to $S_{n}$ , the image of $Φ$ is a subgroup of $S_{n}$ that is isomorphic to $G$ . This completes the proof.

Verified Answer

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2(c) Let $F$ be a field and $F[X]$ denote the ring of polynomials over $F$ in a single variable $X$. For $f(X), g(X) \in F[X]$ with $g(X) \neq 0$, show that there exist $q(X), r(X) \in F[X]$ such that degree $(r(X))<\text{degree}(g(X))$ and $f(X)=q(X) \cdot g(X)+r(X)$.

Expert Answer

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

To show that for any two polynomials

f (X), g (X) \in F [X]

with

g (X) \neq 0

, there exist

q (X), r (X) \in F [X]

such that

degree (r (X)) < degree (g (X))

and

f (X) = q (X) \cdot g (X) + r (X)

, we can use the division algorithm for polynomials. The proof is constructive and provides a method to find

q (X)

and

r (X)

Proof:

Base Case: If $degree (f (X)) < degree (g (X))$ , then we can directly take $q (X) = 0$ and $r (X) = f (X)$ . In this case, the condition $degree (r (X)) < degree (g (X))$ is satisfied.
Inductive Step: Assume that $degree (f (X)) \geq degree (g (X))$ . Let $n = degree (f (X))$ and $m = degree (g (X))$ . Write the polynomials in the form:

$f (X) = a_{n} X^{n} + a_{n - 1} X^{n - 1} + \dots + a_{1} X + a_{0}$

$g (X) = b_{m} X^{m} + b_{m - 1} X^{m - 1} + \dots + b_{1} X + b_{0}$

where $a_{n}, b_{m} \neq 0$ .
Construct the Quotient: Consider the polynomial $h (X) = \frac{a_{n}}{b_{m}} X^{n - m} \cdot g (X)$ . The leading term of $h (X)$ will cancel out the leading term of $f (X)$ when subtracted. Define a new polynomial $f_{1} (X) = f (X) - h (X)$ .
Recursive Application: Now, $degree (f_{1} (X)) < degree (f (X))$ . Apply the division algorithm to $f_{1} (X)$ and $g (X)$ . By induction, there exist $q_{1} (X), r (X) \in F [X]$ such that $degree (r (X)) < degree (g (X))$ and $f_{1} (X) = q_{1} (X) \cdot g (X) + r (X)$ .
Combine Results: Now, we can express $f (X)$ as:

$f (X) = h (X) + f_{1} (X) = h (X) + q_{1} (X) \cdot g (X) + r (X) = (h (X) + q_{1} (X)) \cdot g (X) + r (X)$

Let $q (X) = h (X) + q_{1} (X)$ . Then $f (X) = q (X) \cdot g (X) + r (X)$ with $degree (r (X)) < degree (g (X))$ .
Conclusion: We have shown that for any $f (X), g (X) \in F [X]$ with $g (X) \neq 0$ , there exist $q (X), r (X) \in F [X]$ such that $degree (r (X)) < degree (g (X))$ and $f (X) = q (X) \cdot g (X) + r (X)$ . This completes the proof.

Verified Answer

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3(a) Show that the groups $\mathbb{Z}_5 \times \mathbb{Z}_7$ and $\mathbb{Z}_{35}$ are isomorphic.

Expert Answer

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

To show that the groups

Z_{5} \times Z_{7}

and

Z_{35}

are isomorphic, we need to find a bijective function

f : Z_{5} \times Z_{7} \to Z_{35}

such that

f (a + b) = f (a) + f (b)

for all

a, b \in Z_{5} \times Z_{7}

Step 1: Define the Function

f

Let’s define the function

f : Z_{5} \times Z_{7} \to Z_{35}

as follows:

f (a, b) = 7 a + 5 b

Step 2: Show that

f

is Well-Defined

To show that

f

is well-defined, we need to show that

f (a, b)

is in

Z_{35}

for all

(a, b) \in Z_{5} \times Z_{7}

Let’s substitute the values:

$a$ can be $0, 1, 2, 3, 4$ (from $Z_{5}$ )
$b$ can be $0, 1, 2, 3, 4, 5, 6$ (from $Z_{7}$ )

After substituting these values into

f (a, b) = 7 a + 5 b

, we find that

f (a, b)

can range from

0

34

, which is exactly the set

Z_{35}

Step 3: Show that

f

is a Homomorphism

To show that

f

is a homomorphism, we need to show that

f (a_{1} + a_{2}, b_{1} + b_{2}) = f (a_{1}, b_{1}) + f (a_{2}, b_{2})

Let’s substitute the values:

f (a_{1} + a_{2}, b_{1} + b_{2}) = 7 (a_{1} + a_{2}) + 5 (b_{1} + b_{2})

After calculating, we get:

f (a_{1} + a_{2}, b_{1} + b_{2}) = 7 a_{1} + 7 a_{2} + 5 b_{1} + 5 b_{2} = f (a_{1}, b_{1}) + f (a_{2}, b_{2})

Step 4: Show that

f

is Bijective

To show that

f

is bijective, we need to show that it is both injective (one-to-one) and surjective (onto).

Injective: For $f$ to be injective, $f (a_{1}, b_{1}) = f (a_{2}, b_{2})$ should imply $(a_{1}, b_{1}) = (a_{2}, b_{2})$ . This is true because $7 a_{1} + 5 b_{1} = 7 a_{2} + 5 b_{2}$ implies $a_{1} = a_{2}$ and $b_{1} = b_{2}$ in $Z_{5} \times Z_{7}$ .
Surjective: For $f$ to be surjective, every element in $Z_{35}$ must be mapped by some element in $Z_{5} \times Z_{7}$ . This is true because for every $c$ in $Z_{35}$ , there exists $(a, b)$ in $Z_{5} \times Z_{7}$ such that $f (a, b) = c$ .

Summary

We have shown that

f

is well-defined, a homomorphism, and bijective. Therefore, the groups

Z_{5} \times Z_{7}

and

Z_{35}

are isomorphic.

Verified Answer

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2017

1(b) Let \(G\) be a group of order \(n\). Show that \(G\) is isomorphic to a subgroup of the permutation group \(S_n\).

Expert Answer

2(c) Let \(F\) be a field and \(F[X]\) denote the ring of polynomials over \(F\) in a single variable \(X\). For \(f(X), g(X) \in F[X]\) with \(g(X) \neq 0\), show that there exist \(q(X), r(X) \in F[X]\) such that degree \((r(X))<\text{degree}(g(X))\) and \(f(X)=q(X) \cdot g(X)+r(X)\).

Expert Answer

3(a) Show that the groups \(\mathbb{Z}_5 \times \mathbb{Z}_7\) and \(\mathbb{Z}_{35}\) are isomorphic.

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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