Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

The group of all invertible 4 × 4 matrices with entries in the field of 3 elements

Concept:

The group of all invertible n × n matrices with entries in the field of p elements then cardinality of p-SSG is p^∑(n-1)

Calculation:

The group of all invertible 4 × 4 matrices with entries in the field of 3 elements

then the order of 3-SSG is \(\rm 3^{\frac{4(4-1)}{2}}=3^6=729\)

Hence the option (4) is correct.

Concept -

Algebraic Closure Theorem - every field has an algebraic closure, and when considering the complex numbers as the field K, this theorem implies that C is an algebraic extension of K.

Explanation:

The Algebraic Closure Theorem ensures that every field has an algebraic closure, and when considering the complex numbers as the field K, this theorem implies that C is an algebraic extension of K.

Since K is a proper subfield of ℂ and is not contained in ℝ, K must contain some complex numbers.

So, By the Algebraic Closure Theorem, ℂ is an algebraic extension of K.

Hence, Option 3 is Correct.

Concept:

Explanation:

(4): Let R is commutative

N = {x ∈ R|xn = 0 for some n ∈ ℕ}.

Now, Let x, y ∈ N then xn = 0 and yn = 0 for some n ∈ ℕ

Now, (x - y)ⁿ = xn + (-1)ⁿyn + \(^nC_1x^{n-1}y+...\) = 0

So, x - y ∈ N

Also, (xy)ⁿ = xⁿyⁿ = 0, (yx)n = ynxn = 0

Hence xy ∈ N

Therefore N is an ideal  

(4) is correct

(1): Let A = \(\begin{bmatrix}0&1\\0&0\end{bmatrix}\) and B = \(\begin{bmatrix}0&0\\1&0\end{bmatrix}\) then A, B ∈ N

But A + B = \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\) ∉ N as (A + B)ⁿ ≠ 0 for any integer n

Closer property not hold

(1) is false

Explanation -

We know that some results about the subgroup \(H = \{\frac{m}{2^n} + \mathbb{Z} | m \in \mathbb{Z}, \ \ n = 0, 1, 2.. \}\)

(i) H is proper normal subgroup of \(\frac{\mathbb{Q}}{\mathbb{Z}}\)

(ii) H is infinitely generated subgroup of \(\frac{\mathbb{Q}}{\mathbb{Z}}\)

(iii) H is non cyclic subgroup of \(\frac{\mathbb{Q}}{\mathbb{Z}}\)

Hence option (3) is true.

Concept :

The order of an element in a direct product of a finite number of finite groups is the lcm of the orders of the components of the element.

|(\(g_1 \),\(g_2\),....,\(g_n\))| = lcm(|\(g_1\)|,|\(g_2\)|,....,|\(g_n\)|) 

Suppose \(\phi\) is an isomorphism from a group G onto a group G'.Then G is Abelian iff G' is Abelian.

Calculation :

G = z3⊕z3⊕z3 

|G|=27

The possible order of elements in z3 is 1 and 3, so the possible order of elements in G is 1 and 3.

All the non-identity elements are of order 3.

Thus number of elements of order 3 in G is 26.

|H| = 27

Every non-identity element in H is also of order 3 since,

\(\left[\begin{array}{lll} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{lll} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right] =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)

Thus, number of elements of order 3 in H are 26.

We know that direct sum \(G_1\)⊕....⊕\(G_n\) is abelian iff \(G_1 ,....,G_n \) are abelian.

Thus, G is abelian.

But H is not abelian. (Take a=1, c=1 in one element and a=1,c=0 in another element and check they do not commute.)

So G and H are not isomorphic.

Thus options (1), (2) and (4) are correct.

Explanation:
The external direct product of two cyclic groups is not always cyclic.

For instance, the direct product of Zn and Zm (representing the cyclic groups of orders n and m respectively) is cyclic if and only if n and m are coprime.

If n m are not coprime (i.e., they share a common divisor greater than 1), the resulting group is not cyclic, although it is still abelian, because all cyclic groups are abelian. For example, Z₂ × Z₂ is isomorphic to the Klein four-group, which is not cyclic, but is abelian

Hence, option 3 is correct.

Concept:

(i) General Linear Group \(GL(n, \mathbb{Z}/p^k\mathbb{Z}) \):

The group of n ×  n invertible matrices with entries from \(\mathbb{Z}/p^k\mathbb{Z} \) has order given by:

\( |GL(n, \mathbb{Z}/p^k\mathbb{Z})| = (p^n - p^0)(p^n - p^1) \ldots (p^n - p^{n-1}) × (p^k)^n \)  

(ii) Euler’s Totient Function  ϕ : For p  a prime and k  a positive integer, \(ϕ(p^k) = p^k - p^{k-1} = p^{k-1}(p-1)\)

Explanation:

Calculate the order of  G :

Given G is the group of 3 × 3 invertible matrices with entries from \(\mathbb{Z}/27\mathbb{Z} \), we need to calculate\( |GL(3, \mathbb{Z}/27\mathbb{Z})| \)

For n = 3 and  p = 3,  k = 3:

\( |GL(3, \mathbb{Z}/27\mathbb{Z})| = (3^3 - 3^0)(3^3 - 3^1)(3^3 - 3^2) × (3^3)^3 \)  

\( 3^3 - 3^0 = 27 - 1 = 26 \space ,3^3 - 3^1 = 27 - 3 = 24 \space, 3^3 - 3^2 = 27 - 9 = 18 \)

Now, we multiply these terms along with (3³)³:

\( |GL(3, \mathbb{Z}/27\mathbb{Z})| = 26 × 24 × 18 × (27)^3 \)  

\(26 = 2 × 13 \space ,24 = 2^3 × 3 \space ,18 = 2 × 3^2 \space, 27 = 3^3 \implies (27)^3 = 3^{9} \)

So, the order of G is:

Option 1)

a is divisible by 3⁷ :

Since  3¹² is a factor of |G| , a  is divisible by 3⁷ .

This statement is true.

Option 2)

a  is divisible by  2⁴:

Since  2⁵ is a factor of  |G| , a  is divisible by 2⁴.

This statement is true.

Option 3)

 a  is divisible by 6⁴

Since  6⁴ = 2⁴ × 3⁴  and 2⁵ × 3¹² are factors of |G| ,  a is divisible by 6⁴. This statement is true.

option 4)  a is not divisible by 54:

Since 54 = 2 × 3³

So a is divisible by 54.

Hence this statement is not true.

Concept -

Cauchy Theorem - Let G be a finite group and let p be a prime number that divide the order of G then G contains a element of order p.

Explanation -

By Cauchy Theorem, any group of order 6 has an element of order 2 as well as order 3.

So, subgroup generated by these elements are cyclic of order 2 and 3, say H and K ,respectively.

Subgroup of order 3, i.e. K has index 2 and hence normal.

So, product of H and K is defined and is equal to G as it contains both H and K. So, either H is normal or not.

If H is normal, we get cyclic group of order 6, and H is not normal we get semi direct product of H and K which is isomorphic to S₃.

So, only two groups are there upto isomorphism.

Concept:

Explanation:

Given 𝜎 = 𝜎1 𝜎2 where 𝜎1 is a 4-cycle and 𝜎2 is a 3-cycle in 𝑆8.

𝜎1 and 𝜎2 are disjoint cycles then there exists a 𝜎3 of 1- cycle

So 𝜎 = 𝜎1 𝜎2 𝜎3  

conjugate to 𝜎 means their cycle decomposition will be the same with 𝜎.

Hence the number of elements in 𝑆8 that are conjugate to 𝜎 

= \(8!\over 4^1.3^1.1^1.1!.1!.1!\) = 3360

Answer is 3360

Explanation -

we know some results -

Let (G, * ) and (G', o) are two group  and \(f : G \to G'\) be a homomorphism then

(i)  The relation of isomorphism is an equivalence relation.

(ii) Image of abelian group under homomorphism is abelian.

(iii) If order of a is finite then this implies order of f(a) is also finite. where a is in G.

(iv) If order of f(a) is finite then order of a is need not be finite. where a is in G.

(v) Image of cyclic group under homomorphism is cyclic.

Hence option (4) is true.