Homomorphisms MCQ Quiz in தமிழ் - Objective Question with Answer for Homomorphisms - இலவச PDF ஐப் பதிவிறக்கவும்

Solution- 

A Topological space is called as locally compact if  every point has a compact neighbourhood. Therefore, A topological space is locally compact if every point x of X has neighbourhood.

Therefore, Correct Option is Option 2).

Explanation -

In an abelian group (also known as a commutative group), the operation of the group commutes. This means that for any two elements a and b in the group, the equation ab = ba holds true. Because of the commutative property, the inner automorphism corresponding to any element of an abelian group is the identity function.

let (G, *) be an abelian group, and let's denote by \(ϕ_a(x)\) the inner automorphism of G defined by some element a, where 'x' is an element of G. Then\(ϕ_a(x) = a x a⁻¹\)

However, since the group operation is commutative, we can rearrange the terms to obtain: \(ϕ_a(x) = a a⁻¹ x = e x = x\)  where 'e' is the identity element of the group.

Since \(ϕ_a(x) = x\) for any element x in the group,  the inner automorphism of an element of the Abelian group is the identity map for any element a in an abelian group G.

Hence option (i) is true.

Explanation:

Recall: (1) Homomorhis image of cyclic is cyclic.

Now |G| = 8 and G is cyclic.

If ϕ : G → S5 be onto group homomorhism.

Then ϕ (G) = S5

as G is cyclic ⇒ S5 is cyclic which is not True.

∴ \(\nexists\) any onto homomorhism from G to S5.

Also, ∵ o(G) ≠ o(S5) ⇒ \(\nexists\) one-one homomorhism

from G → S5. opt (2), (3) - True.

Now Let ϕ : G → S5 be a homomorhism.

as G is a cyclic group of order 8.

⇒ G ≈ ℤ8

So, for easyness, we take G = ℤ8

and ℤ8 = <1> ⇒ x = 1m for x ∈ ℤ8

also ϕ(ℤ8) ⊆ S5

ϕ(1m) = ϕ (1 + 1 + ....+1)(m times) = (ϕ(1))m

and o(ϕ(1))|1| ⇒ (ϕ(1))|8 and also o(ϕ(1))|5! = 120

⇒ 0(ϕ(1)) = 1, 2, 4, 8

Case I: ϕ(1) = I, identity map 

Case II: ϕ(1) = (12) ⇒ ϕ(m) = (ϕ(1))m = (12)m

then ϕ(ℤ8) = {I, (12)}

= (12)m+n = \(\left\{ \begin{matrix} I; & m, n \ both \ are \ odd \ or \ even \\\ (12); & m \ is \ odd \ \& \ n \ is \ even \ or \ vice-versa \end{matrix} \right.\)

and ϕ(m) ϕ(n) = \(\left\{ \begin{matrix} I; & m, n = odd, \ m, n = even \\\ (12); & m = odd, n = even \ or \ m = even, \ n = odd \end{matrix} \right.\)

Similarly, ϕ(1) = (ab), all 2-cycles in S5

Then ϕ(m) = (ab)m is a group homomorphism.

and Number of element of type \(ab = \frac{5!}{2.1^3. 3!} = 10\)

Case III: ϕ(1) = (abcd), (abcd) ∈ S5

Then ϕ(m) = \(\left\{ \begin{matrix} I; & m = 4k \\\ (abcd); & m = 4k + 1 \\\ (acb) ; & m = 4k + 2 \\\ (adcb); & m = 4k + 3 \end{matrix} \right.\)

and one can easily check ϕ(m) is homomorphism.

and such type of homomorphisms = No. of elements  of type (abcd) = \(\frac{5!}{1^2 4^1} = 30\)

Case IV: ϕ(1) = (ab)(cd), (ab)(cd) ∈ S5

Then ϕ(m) = \(\left\{ \begin{matrix} I; & m, n \ both \ are \ odd \ or \ even \\\ (ab)(cd); & m \ is \ odd \ \& \ n \ is \ even \ or \ vice-versa \end{matrix} \right.\)

and such type homomorphisms = \(\frac{5!}{2^2 . 2!} = 15\)

Here o(ϕ(1)) ≠ 8 ∵ ϕ(1) ∈ S5

and S5 have no element of order 8.

∴ Possible group homomorphisms

= 1 + 10 + 30 + 15

= 56

⇒ ∃ more than 20 different group homomorphism from G → S5

opt (1) - False

opt (4) - True

So, opt (2), (3) and (4) - correct

Explanation:

Statement 1: 

The Total Number of automorphism in \( (\mathbb{Z} , +) \) = 2

So, Total Number of distinct group Homomorphism from \( (\mathbb{Z} , +) \) onto \( (\mathbb{Z} , +) \) = 2 

⇒ (1) is false  

Statement 2: 

Since Every Group of Finite order may or May not be Cyclic 

Example : Klein Four Group (K₄)

⇒ (2) is false  

Hence Option(3) is the correct answer.

Concept:

Hom(ℤ_m, ℤn) = ℤd, where d = gcd(m, n) and d is the number of homomorphism from ℤm to ℤn.

Explanation:

(1): ℤ2, ℤ3

gcd(2, 3) = 1

So, Hom(ℤ2, ℤ3) = ℤ

(1) is TRUE

(2): ℤ2, ℤ4

gcd(2, 4) = 2

So, The number of homomorphism from ℤ2 to ℤ4 is 2

(2) is NOT TRUE

Explanation:

(1): Let f:ℝ → S¹, be a map defined by 

f(a) = e^2πia

Let a, b ∈ ℝ then 

f(a + b) = e^{2πi(a + b)} = e2πia . e2πib = f(a) . f(b)

So f is a homomorphism.

Let a ∈ ker(f) 

⇒ f(a) = 1 ⇒  e^2πia = 1 ⇒  a ∈ ℤ 

Hence ker(f) = ℤ 

Hence by fundamental theorem of homomorphism 

ℝ/ℤ ≅ S¹

(2): e^2πia ∈ P if and only if there exists n ∈ ℤ such that 

(e2πia)ⁿ = e2πia = 1

⇒ a ∈ ℚ 

Hence we have a surjective homomorphism f from ℚ to P given by f(a) = e2πia whose kernel is ℤ.

Hence by fundamental theorem of homomorphism 

ℚ/ℤ ≅ group of roots of unity P.

(3) is true. 

Concept:

Euler’s Totient function ϕ(n) = \(n\times \Pi_{\text{p/n, p is prime}}(1-\frac1p)\)

Explanation:

Let us consider the group elements {e, g, g², g³, …, gⁿ⁻¹}. 

Now, gⁱ is a generator if and only if gcd(i, n) = 1.

So, the total number of generators of the cyclic group ℤ_n is equal to the number of integers coprime with n (and less than n), which is Euler’s Totient function ϕ(n).

Therefore, the number of automorphisms of ℤ₁₀₀ 

= ϕ(100) = 100 × \((1-\frac12)(1-\frac15)\) = 100 ×. \(\frac12\times\frac45\) = 40.

They are a₁(x) = x, a₃(x) = x³, a₇(x) = x⁷, a₉(x) = x⁹, ...

Option (4) is true.

Concept use:

The No of Homomorphism from  ℤ/4ℤ to S4  is d(n) 

where d(n) is the Number of Elements of S_n which Divides the n .

 φ(n) = number of integers k such that 1 ≤ k ≤ n and gcd(k, n) = 1

Calculations:

Since, The Number of Elements in S₄

1. Elements of Order 1:

• Identity Permutation: There is only 1 identity permutation.

2. Elements of Order 2:

• 2-cycles: There are 6 2-cycles: (1 2), (1 3), (1 4), (2 3), (2 4), (3 4).
• Product of two disjoint 2-cycles: There are 3 such permutations: (1 2)(3 4), (1 3)(2 4), (1 4)(2 3).

So, there are a total of 9 elements of order 2.

3. Elements of Order 3:

• 3-cycles: There are 8 3-cycles: (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3).

4. Elements of Order 4:

• 4-cycles: There are 6 4-cycles: (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2).

Therefore, the total number of elements in S₄ whose order divides 4 is 1 + 9 + 6 = 16.

Hence, there are 16 homomorphisms from Z₄ to S₄.
Note we don't have element of 6, 8 & 12.

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.

Explanation - 

In a ring Z/nZ, each element can be represented by a unique integer in the set {0, 1, 2, …, n-1}.

Thus,  ϕ(n) = n * ϕ(1) for n in Z/6Z  which must satisfy ϕ(1) * 0 = ϕ(0) (which is always 0, since homomorphisms must map 0 to 0),

and ϕ(1) * 15 must be equal to ϕ(6) in S (which is also equal to 0 in S, since 6 is 0 in Z/6Z),

The value of ϕ(1) must be an element of S that, when multiplied by 15, yields 0 in S.

Only the elements {0, 5,10} of S satisfy this condition, but as homomorphisms need to map 1 of ring R to 1 of ring S and in S and equals 1,

only {1, 10} will make it a valid ring homomorphism because of the multiplicative identity.

Therefore, there are exactly 2 distinct ring homomorphisms from R to S.

Hence option (3) is true.