Algebra MCQ Quiz - Objective Question with Answer for Algebra - Download Free PDF

Concept:

Number of Divisors of a Number:

• To find the total number of positive divisors of a number, we first perform its prime factorization.
• If a number is expressed as a product of prime powers: a = p^c₁ × q^c₂ × r^c₃..., then the number of positive divisors is:
• Number of divisors = (c₁ + 1)(c₂ + 1)(c₃ + 1)...
• This includes both 1 and the number itself as divisors.

Calculation:

Given number = 1080

⇒ 1080 ÷ 2 = 540

⇒ 540 ÷ 2 = 270

⇒ 270 ÷ 2 = 135

⇒ 135 ÷ 3 = 45

⇒ 45 ÷ 3 = 15

⇒ 15 ÷ 3 = 5

⇒ 5 ÷ 5 = 1

⇒ Prime factorization of 1080 = 2³ × 3³ × 5

Now, using the formula:

⇒ Number of divisors = (3 + 1)(3 + 1)(1 + 1)

⇒ Number of divisors = 4 × 4 × 2

∴ Hence, the number of positive divisors of 1080 is 32.

Explanation:

H be an abelian non-trivial proper subgroup of G with the property that H ∩ gHg−1 = {e} for all g / ∉ H

K = $\{g\in G:gh=hg\text{ for all }h\in H\}$,

H is normal in K , as K centralizes H

Option (1): K is a proper subgroup of H

Since  $H\subseteq K$  by definition, K cannot be a proper subgroup of H

Instead, K either equals H or is strictly larger.

Option(1) is not correct 

Option (2): H is a proper subgroup of K

K may equal H because the centralizer of H may consist solely of H itself, depending on the group structure

Option (3): K = H

Since  $H\cap gH{g}^{-1}=\{e\}forallg\notin H,$ 

no element outside H can commute with all elements of H

Therefore, the centralizer of H in G is exactly H , implying K = H

Option (4): There exists no abelian subgroup  $L\subseteq G$  such that K is a proper subgroup of L

Since K = H , and H is abelian and isolated (no elements outside H commute with all elements of H ),

there cannot exist a larger abelian subgroup L containing K as a proper subgroup.

Hence Option(3) and Option(4) are correct 

Explanation:

Option (1) is incorrect because normality does not imply that the subgroup is cyclic

Option (2)  is incorrect because G/N is not necessarily abelian for all normal subgroups N

Option (3)  is correct because the definition of a normal subgroup is  $gN{g}^{-1}=N$  for all  $g\in G$

Option (4)  is incorrect because a normal subgroup does not have to be trivial or the whole group;

normal subgroups can have other non-trivial forms

Hence Option(3) is the Correct Answer.

Explanation:  

Given |G| =  57 = 3 × 19

All Subgroup of Group is Cyclic

The Subgroup are of odd order so each element will have its inverse different from itself 

So, If G is abelian then there exists no proper subgroup H of G such that product of all elements of H is identity is false 

Hence Option(4) is the correct answer.

Explanation: 

1. G : Group of all non-zero real numbers under multiplication.

2. $H=\{x\in G\mid x=1$ or x is irrational}

3. $K=\{x\in G\mid x\ge 1\}$

Analysis For H :

1. Identity: The identity element of G is 1 , which belongs to H .

2. Closure: If  $x,y\in H$, we need  $xy\in H$ . However:

If x = 1 and y is irrational,  $xy=y\in H$

If both x and y are irrational, their product x y may be rational,

Hence, H is not closed under multiplication.
   
⇒ H is not a subgroup of G 

For K :

1. Identity: The identity 1 ∈  G satisfies x ≥ 1 , so 1 ∈ K

2. Closure: If x, y ∈ K , then x ≥ 1 and y ≥ 1 . Multiplying x and y , we get x y ≥ 1 , so x y ∈ K

3. Inverses: For x ≥ 1 , its multiplicative inverse x^{-1} ≤  1 . Thus,  $x^{-1}\notin K$  unless x = 1

Hence, K is not closed under inverses

⇒ K is not a subgroup of G

Neither H nor K is a subgroup of G .  

Hence Option(3) is the correct answer.

Concept:

If xⁿ = 1 then o(x) divides n

Explanation:

ℤ/105ℤ ≅ ℤ₁₀₅

105 = 3 × 5 × 7

So $U_{{\mathbb{Z}}_{105}}$ ≅ U(3) × U(5) × U(7) ≅ ℤ₂ × ℤ₄ × ℤ₆

Given x2 = 1 so o(x) divides 2 Hence o(x) = 1 or 2

Element of ℤ2 of order 1 and 2 is 2

Element of ℤ4 of order 1 and 2 is 2

Element of ℤ6 of order 1 and 2 is 2

Hence total such elements = 2 × 2 × 2 = 8 

Option (4) is correct

Concept:

Concept:

If G be a group such that o(G) = 2p where p is odd prime then G ≅ ${\mathbb{Z}}_{2p}$ or G ≅ ${\mathbb{D}}_p$

(ii) If o(G) = n where G is abelian group then class equation of G is n = 1 + 1 + ... + 1 (n times)

Explanation:

Given o(G) = 10 = 2.5

So here p = 5 which is odd prime.

Hence G ≅ ${\mathbb{Z}}_{10}$ or G ≅ ${\mathbb{D}}_5$

If G ≅ ${\mathbb{Z}}_{10}$ then it is cyclic so abelian. Therefore class equation of a group of order 10 is 10 = 1 + 1 + … + 1 (10-times)

If G ≅ ${\mathbb{D}}_5$ then there will be 5 rotation and 5 reflection.

So in this case class equation of a group of order 10 is 

10 = 1 + 2 + 2 + 5

Hence option (1) is correct

Explanation:

Let the operation, Δ i.e., A, B ∈ P(X) ⇒ A Δ B = (A ∪ B) \ (A ∩ B) for this,

(i) Closer: Let A, B ∈ P(x) then A Δ B = (A ∪ B) \ (A ∩ B) ∈ P(X)
So, P(x) is closed under Δ. 

(ii) Associativity: let A, B, C ∈ P(x), then (A Δ B) ΔC = ([(A ∪ B) \ (A ∩ B))] ∪ C) \[([A ∪ B) \ ((A ∩ B))] ∩ C)

A Δ (B Δ C) = (A ∪[(B ∪ C) | (B∩C)]) \ (A∩[(B∪C) | (B∩C)])

form, figures you can see,

(A Δ B) ΔC = A Δ (B Δ C)

(iii) Identity:

AΔϕ = (A ∪ ϕ) \ (A ∩ ϕ) = A \ ϕ = A

So, ϕ ∈ P(x) such that A Δ ϕ = A

(iv) Inverse:

A Δ A = (A ∪ A) \ (A ∩ A) = A \ A = ϕ

So, for A ∈ P(x),  A-1 = A.

∴ P(x) is group under Δ.

Now for * operation, A * B = A ∩ B, A, B ∈ P(x)

let x = {1, 2, 3} then P(x) = {ϕ, x, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}

Here, if we take, e = x

(∵ x ∩ A = A, A ∈ P(x))

But for e = x, inverse  of any A, A ∈ P(x)

∵ A ∩ B ≠ x (for any A, B ∈ P(x)A, B ≠ x)

So, P(x) is not a group under (*).

option (3) is true.

Explanation:

For this type of problems, just try to discard given options by taking suitable counter examples..

For option (1). Let n = 16 then

(n - 1)! + 3 = 15! + 3 = even + odd = odd

⇒ No any even integer divide it ie to (n - 1)! + 3

⇒ option (1) is false. (Because z not true for all)

For option (2). taken n = 17, then

∵ (n - 1)! = 16! , Here n = 17 is prime so it will never divide 16!

⇒ option (2) false.

For option (4). take n = 16, then n² = 16² + n! + 1 = 16 ! + 1 Here 16² is even while 16! + 1 is odd integer So 16² + 16! + 1 

⇒ option (4) is false.

For option (3) is true [consider any n ≥ 16 even]

Note: Proof for this option is too long, so just pay to understand with example.

$=\frac{p_1^{r_1}{p}_2^{r_2}\cdots p_n^{r_n}}{p_1^{r_1-1}\cdot (p_1-1)\cdot p_2^{r_2-1}(p_2-1)\dots p_n^{r_{n-1}}\left(p_{n-1}\right)}$

$=\frac{p_1{p}_2\cdots p_n}{(p_1-1)(p_2-1)\cdots (p_n-1)}$ = integer (given) = p/n¹/n

∵ (p₁ - 1) × p₁ ⇒ ∃ some other prime p₂ S.t (p₁ - 1)|p₂

But ∵  p₂ is also a prime, so not divisible by any of integer except 1 .

(there of one prime factor. is 2, then n tar as at most two distinct prime faster else one.

thus, option (3) is true

Concept:  

Maximal Ideal: A maximal ideal I in a ring R is an ideal such that the quotient ring R/I is a field. 
 

Prime Ideal: A prime ideal P in a ring R is an ideal such that if the product of two elements is in P ,

then at least one of the elements must be in P .

Explanation:

$R=\{\sum _{n\in \mathbb{Z}}{a}_n{X}^n\mid a_n\in \mathbb{Z},\text{ and }{a}_n\ne 0\text{ for finitely many }n\in \mathbb{Z}\}$

The addition and multiplication in the ring are defined as

$\left(\sum _{n\in \mathbb{Z}}{a}_n{X}^n\right)+\left(\sum _{n\in \mathbb{Z}}{b}_n{X}^n\right)=\sum _{n\in \mathbb{Z}}(a_n+b_n)X^n$

$\left(\sum _{n\in \mathbb{Z}}{a}_n{X}^n\right)\left(\sum _{n\in \mathbb{Z}}{b}_m{X}^m\right)=\sum _{k\in \mathbb{Z}}\left(\sum _{n+m=k}{a}_n{b}_m\right)X^k$

Option 1:

The addition in this ring is clearly commutative since the sum of polynomials in any ring is commutative.

Now consider the multiplication. In standard polynomial rings, multiplication is commutative as long as the

coefficients come from a commutative ring (in this case, integers $\mathbb{Z}$ ).

Since $\mathbb{Z}$ is commutative under multiplication, and the exponents $X^n$ follow commutative multiplication rules

(i.e., $X^n{X}^m=X^{n+m}$), the ring R is also commutative under multiplication.

Therefore, the statement R is not commutative is false.

Option 2:

 The ideal $X-1$ would be maximal if R/$X-1$  is a field.

However, this is not necessarily true in the ring R as described, since R/$X-1$ is unlikely

to be a field (it may reduce to a simpler ring, but not a field).
 

Option 3:

(X - 1, 2) is a standard type of ideal in certain polynomial rings, particularly over integers. For a prime ideal,

the condition that multiplication of elements should remain within the ideal must hold. 

Option 4:

$(X,5)$ as a Maximal Ideal: If $(X,5)$ is maximal, the quotient R/$(X,5)$ should be a field.

While in certain rings this could lead to a field, this needs further validation.

Therefore, option 3) is correct.

Explanation:

S = $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ So S² = $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$$\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ = $\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$ = - I, S3 = -S, S⁴ = I

Also, 𝑇1 = $\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]$

So, 𝑇1 = i𝑇1, ${𝑇}_1^2=-I$, ${𝑇}_1^3=-iI$, ${𝑇}_1^4=I$

Let Q₄ = {S, S², S³, I, i, -i, iS} then Q₄ is a quaternion group

Given 

𝐺1, 𝐺2, 𝐺3 be three subgroups of 𝐺𝐿2 (ℂ) given by

𝐺1 = < 𝑆, 𝑇1 >, where 𝑇1 = $\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]$,

𝐺2 = < 𝑆, 𝑇2 >, where 𝑇2 = $\left[\begin{array}{cc}i& 0\\ 0& -i\end{array}\right]$,

𝐺3 = < 𝑆, 𝑇3 >, where 𝑇3 = $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.

Then 𝐺1 = $\{S^i{T}_1^j|i,j\in \mathbb{Z}\}$ is non-abelian group of order 8

Similarly, 𝐺2 = $\{S^i{T}_2^j|i,j\in \mathbb{Z}\}$ is non-abelian group of order 8

𝐺3 = $\{S_1,T_3|S^4=I,T_3^2=I,T_{3}S=T_3{S}^{-1}\}$ ≈  D₄

So, |𝑍(𝐺3)| = |𝑍(D4)| = 2

Hence 𝑍(𝐺3) = {$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$} false because then |𝑍(𝐺3)| = 1

𝑍(𝐺1) is isomorphic to 𝑍(𝐺2) also 𝐺1 is not somorphic to 𝐺3

Hence (1), (2), (3) are false

(4) is correct

Concept:

A mapping ϕ: $\mathbb{N}$ →  $\mathbb{N}$ defined by ϕ(n) = {x ∈ $\mathbb{N}$ | 1 ≤ x

ϕ (pⁿ) = pⁿ - p^n-1

ϕ(mn) = ϕ(m)ϕ(n) if gcd(m, n) = 1 
Explanation:

ϕ(n) table:

From the table of ϕ(n) we can see that if we take n as a prime number greater than 3, then ϕ(n) > ϕ(n+1) and if we take n + 1 as a prime number greater than 3, then ϕ(n) < ϕ(n+1)  

∴ options (1) and (2) are correct.

ϕ(n) table:

So if we take N = 6, then ∀ n > 6, we have ϕ(N) < ϕ(n)

So option (3) is correct 

So the option that is not true is (4)

Concept:

If R is a commutative ring then,

ab = ba ∀ a,b ∈ R. 

M, which is an ideal of R, will be called the maximal ideal of R,

1) if M ⊂ R, M ≠ R (there is at least one element in R that does not belong to M)

2) There should be no ideal 'N', such that M ⊂ N ⊂ R. (there is no ideal between M and R). 

Analysis:

R/M is a field [∵ every finite integral domain is a field]

∴ R/M is a ring with unity

∴ 1 + M ≠ M

i.e., 1 ∉ M

Now, one belongs to R, but it does not belong to R.

∴ M ≠ R.

Let I be an ideal of R

Such that M ⊆  I ⊆  R

Let, M ≠ I

∃ a ∈ I, such that a ∉ M

∴ a + M ∉ M

Now, R/M is a field.

∴ Every, non-zero of R/M is revertible

∴ a + M is invertible

∴ ∃ b + M ∈ R/M such that

(a + M) (b + M) = 1 + M

ab + M = 1 + M

ab – 1 ∈ M ⊆  I      ---(1)

a ∈ I, b ∈ R

∴ ab ∈ I          ---(2)  (∵ I is an ideal)

From (1) and (2), we can write

ab – (ab – 1) ∈ I

∴ 1 ∈ I

Now, as unity belongs to ideal, so ideal becomes ring

∴ I = R

∴ M is a maximal ideal of R

If R is a commutative ring with unity then every maximal ideal is a prime ideal.

Concept:

If $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$ are cyclic groups, the number of group homomorphisms from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$ is given by

$\text{Number of homomorphisms}=gcd(n,m),$ where $gcd(n,m)$ is the greatest common divisor of  n and m.

Explanation:

The number of group homomorphisms from $\mathbb{Z}/150\mathbb{Z}$ to $\mathbb{Z}/90\mathbb{Z}$ is given by the greatest common divisor (gcd)

of the orders of the two groups. That is  $\text{Number of homomorphisms}=gcd(150,90)$
  
Prime factorization of 150: $150=2\times 3\times 5^2$

Prime factorization of 90: $90=2\times 3^2\times 5$

Now, the gcd of 150 and 90 is the product of the lowest powers of the common prime factors:
   
$gcd(150,90)=2\times 3\times 5=30$

The number of group homomorphisms from $\mathbb{Z}/150\mathbb{Z}$ to $\mathbb{Z}/90\mathbb{Z}$ is 30.

Thus, option 1) is correct.