Wednesday, October 1, 2025

PGTRB PRACTICE QUESTIONS ON ALGEBRA 10 questions

Algebra Concepts MCQ Quiz

Algebra Concepts MCQ Quiz

Test your knowledge of groups, rings, fields, and vector spaces

1. The set \( \mathbb{R} \) of all real numbers forms a group with respect to the binary operation \( * \) defined by \( a * b = a + b + 1 \). What is the identity element of this group?
a) 0
b) 1
c) -1
d) 2

Answer: c) -1

For \( e \) to be the identity element, we need \( a * e = a \) for all \( a \in \mathbb{R} \).

\( a * e = a + e + 1 = a \) ⇒ \( e + 1 = 0 \) ⇒ \( e = -1 \).

Verification: \( a * (-1) = a + (-1) + 1 = a \), and \( (-1) * a = (-1) + a + 1 = a \).

2. Which of the following sets forms a group under multiplication?
a) \( \{1, \omega, \omega^2\} \) where \( \omega \) is a complex cube root of unity
b) \( \{1, -1, 2, -2\} \)
c) \( \mathbb{Z} \) (the set of all integers)
d) \( \mathbb{N} \) (the set of natural numbers)

Answer: a) \( \{1, \omega, \omega^2\} \) where \( \omega \) is a complex cube root of unity

• Option a: Forms a cyclic group of order 3 under multiplication. It satisfies closure (\( \omega \cdot \omega^2 = \omega^3 = 1 \)), associativity, has identity (1), and every element has an inverse (\( \omega^{-1} = \omega^2 \)).

• Option b: Not closed under multiplication (\( 2 \cdot 2 = 4 \) which is not in the set).

• Option c: Most integers don't have multiplicative inverses in \( \mathbb{Z} \).

• Option d: Natural numbers don't have multiplicative inverses or additive identity for multiplication.

3. Let \( G = \{a \in \mathbb{R} \mid a \neq 1\} \) with binary operation \( a * b = a + b - ab \). What is the inverse of an element \( a \) in this group?
a) \( \frac{1}{a} \)
b) \( \frac{a}{a-1} \)
c) \( \frac{a}{1-a} \)
d) \( 1-a \)

Answer: b) \( \frac{a}{a-1} \)

Let \( a^{-1} \) be the inverse of \( a \). Then \( a * a^{-1} = e \), where \( e \) is the identity.

First, find the identity: \( a * e = a + e - ae = a \) ⇒ \( e - ae = 0 \) ⇒ \( e(1-a) = 0 \) ⇒ \( e = 0 \) (since \( a \neq 1 \)).

Now, \( a * a^{-1} = a + a^{-1} - a \cdot a^{-1} = 0 \)

⇒ \( a + a^{-1}(1 - a) = 0 \)

⇒ \( a^{-1}(1 - a) = -a \)

⇒ \( a^{-1} = \frac{-a}{1-a} = \frac{a}{a-1} \)

4. Which of the following is a normal subgroup of \( S_3 \) (the symmetric group of degree 3)?
a) \( \{e, (1 2)\} \)
b) \( \{e, (1 2 3), (1 3 2)\} \)
c) \( \{e, (1 3)\} \)
d) \( \{e, (2 3)\} \)

Answer: b) \( \{e, (1 2 3), (1 3 2)\} \)

\( S_3 = \{e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)\} \) has order 6.

• Option b is \( A_3 \), the alternating group of degree 3, which is the unique subgroup of order 3 in \( S_3 \).

• Subgroups of order 2 in \( S_3 \) (options a, c, d) are not normal. For example, for \( H = \{e, (1 2)\} \):

Take \( g = (1 3) \), then \( gHg^{-1} = (1 3)\{e, (1 2)\}(1 3) = \{e, (2 3)\} \neq H \).

Thus \( H \) is not normal in \( S_3 \).

5. The set of all \( 2 \times 2 \) matrices of the form \( \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \) where \( a, b \in \mathbb{R} \) and \( a^2 + b^2 \neq 0 \), forms a group under matrix multiplication. What type of group is this?
a) Finite non-abelian group
b) Finite abelian group
c) Infinite non-abelian group
d) Infinite abelian group

Answer: d) Infinite abelian group

• The set is infinite because there are infinitely many real numbers \( a \) and \( b \).

• The group is abelian because:

Let \( A = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \), \( B = \begin{pmatrix} c & d \\ -d & c \end{pmatrix} \)

\( AB = \begin{pmatrix} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{pmatrix} \)

\( BA = \begin{pmatrix} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{pmatrix} \)

So \( AB = BA \), hence the group is abelian.

This group is isomorphic to the multiplicative group of nonzero complex numbers.

6. Which of the following is an integral domain but not a field?
a) \( \mathbb{Z}_7 \)
b) \( \mathbb{Q} \)
c) \( \mathbb{Z} \)
d) \( \mathbb{R} \)

Answer: c) \( \mathbb{Z} \)

An integral domain is a commutative ring with unity and no zero divisors. A field is a commutative ring with unity where every nonzero element has a multiplicative inverse.

• \( \mathbb{Z}_7 \): Is a field (since 7 is prime)

• \( \mathbb{Q} \): Is a field

• \( \mathbb{Z} \): Is an integral domain but not a field (only ±1 have multiplicative inverses in \( \mathbb{Z} \))

• \( \mathbb{R} \): Is a field

7. Let \( V \) be a vector space of all \( 2 \times 2 \) matrices over \( \mathbb{R} \). What is the dimension of \( V \)?
a) 2
b) 3
c) 4
d) 8

Answer: c) 4

A basis for the vector space of all \( 2 \times 2 \) matrices over \( \mathbb{R} \) is:

\( \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\} \)

This set has 4 elements and any \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) can be written as a linear combination:

\( a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \)

Thus, the dimension is 4.

8. Which of the following polynomials is irreducible over \( \mathbb{Q} \)?
a) \( x^2 - 2 \)
b) \( x^2 + x + 1 \)
c) \( x^2 - 4 \)
d) Both a and b

Answer: d) Both a and b

A polynomial is irreducible over a field if it cannot be factored into polynomials of lower degree with coefficients in that field.

• \( x^2 - 2 \): Irreducible over \( \mathbb{Q} \) (no rational roots)

• \( x^2 + x + 1 \): Irreducible over \( \mathbb{Q} \) (discriminant = 1 - 4 = -3 < 0, no rational roots)

• \( x^2 - 4 \): Reducible over \( \mathbb{Q} \) since \( x^2 - 4 = (x-2)(x+2) \)

Thus, both a and b are irreducible over \( \mathbb{Q} \).

9. What is the splitting field of the polynomial \( x^4 - 5x^2 + 6 \) over \( \mathbb{Q} \)?
a) \( \mathbb{Q}(\sqrt{2}) \)
b) \( \mathbb{Q}(\sqrt{3}) \)
c) \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \)
d) \( \mathbb{Q}(\sqrt{6}) \)

Answer: c) \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \)

Factor the polynomial: \( x^4 - 5x^2 + 6 = (x^2 - 2)(x^2 - 3) \)

The roots are \( \pm\sqrt{2}, \pm\sqrt{3} \).

The smallest field containing \( \mathbb{Q} \) and all these roots is \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \).

This is the splitting field because the polynomial factors completely into linear factors over this field:

\( x^4 - 5x^2 + 6 = (x - \sqrt{2})(x + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3}) \)

10. Let \( T: \mathbb{R}^3 \to \mathbb{R}^2 \) be a linear transformation. If the dimension of the kernel of \( T \) is 1, what is the dimension of the range of \( T \)?
a) 1
b) 2
c) 3
d) Cannot be determined

Answer: b) 2

By the Rank-Nullity Theorem:

\( \text{dim}(V) = \text{rank}(T) + \text{nullity}(T) \)

Here, \( \text{dim}(V) = 3 \) (since \( V = \mathbb{R}^3 \))

\( \text{nullity}(T) = \text{dim}(\text{ker}(T)) = 1 \)

So, \( 3 = \text{rank}(T) + 1 \) ⇒ \( \text{rank}(T) = 2 \)

The rank of \( T \) is the dimension of its range, so the dimension of the range is 2.

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