Test: Group Theory - 7 - Mathematics MCQ

Yes, the number of left cosets and the number of right cosets of a subgroup H in a group G are always equal.

Answer: Yes.

Solution: To understand why, let's define the left cosets and right cosets:

1. Left Cosets: The left cosets of H in G are of the form gH = {gh | h ∈ H} for each g ∈ G.

2. Right Cosets: The right cosets of H in G are of the form Hg = {hh' | h' ∈ H} for each g ∈ G.

Both the left cosets and right cosets form a partition of the group G. Each coset corresponds to a distinct equivalence class of elements of G, where two elements g₁ and g₂ are equivalent if and only if g₁H = g₂H (for left cosets) or Hg₁ = Hg₂ (for right cosets).

If F is a homomorphism of a group G into a group G' with kernel K, then the following properties hold:

1. Kernel Definition:
   The kernel of the homomorphism F, denoted as ker(F), is the set of all elements in G that are mapped to the identity element in G'.
   In mathematical terms:
   ker(F) = {g in G | F(g) = e'},
   where e' is the identity element of G'.

2. Normal Subgroup:
   The kernel K = ker(F) is a normal subgroup of G. This means that for all elements g, h in G:
   gKg⁻¹ = K.
   In other words, conjugating any element of the kernel by any element of G results in an element still within the kernel.

3. First Isomorphism Theorem:
   The image of F, denoted by F(G), is isomorphic to the quotient group G / ker(F).
   That is, G / ker(F) ≅ F(G).
   This theorem provides a key relationship between the structure of G, its kernel, and its image under the homomorphism.

4. Injectivity:
   The homomorphism F is injective (one-to-one) if and only if ker(F) = {e}, where e is the identity element of G.
   This means that if the kernel is trivial (contains only the identity element), the homomorphism is injective.

5. Homomorphism Property:
   For all elements a, b in G:
   F(ab) = F(a)F(b).
   This means the homomorphism preserves the group operation.

Conclusion:

• K is always a normal subgroup of G.

Hence option C is correct

In the given problem, G is a finite group, and H is a normal subgroup of G.

The order of a quotient group G/H is given by:
o(G/H) = o(G) / o(H)

where o(G) is the order (number of elements) of the group G, and o(H) is the order of the normal subgroup H.

Since G/H is the quotient group formed by the cosets of H in G, its order is the number of such cosets. Therefore, the order of G/H is the ratio of the order of G to the order of H.

Thus, o(G/H) = o(G) / o(H).

Let G be a group of order 15. By the Sylow theorems, the number of Sylow p-subgroups is denoted by n_p, where p is a prime divisor of the order of the group.

We are given that |G| = 15 = 3 × 5, so we have two prime divisors: 3 and 5.

1. For Sylow 3-subgroups: The number of Sylow 3-subgroups, n_3, satisfies two conditions from the Sylow theorems:

   • n₃ ≡ 1 (mod 3), i.e.,n₃ must be congruent to 1 modulo 3.
   • n₃ divides 5 (because the index of a Sylow 3-subgroup is 5, and the order of G is 15).

   The divisors of 5 are 1 and 5. We need to check which of these satisfy the condition n₃ ≡ 1 (mod 3):

   • For n₃ = 1, we have 1 ≡ 1 (mod 3), which satisfies the condition.
   • For n₃ = 5, we have 5 ≡ 2 (mod 3), which does not satisfy the condition.

   Hence, n₃ = 1.

So, the number of Sylow 3-subgroups of G is 1.

To determine whether the permutation $f = (1\ 2\ 3)(1\ 2)$ is odd or even, we need to analyze the permutation in terms of its cycle structure and the number of transpositions it can be decomposed into.

Step 1: Analyze the permutations

• The permutation $(1\ 2\ 3)$ is a 3-cycle. A 3-cycle can be written as two transpositions, so it is an odd permutation.

  $$(1\ 2\ 3) = (1\ 3)(1\ 2)$$

  Hence, the permutation $(1\ 2\ 3)$ is odd.

• The permutation $(1\ 2)$ is a simple transposition, which is an odd permutation.

Step 2: Combine the permutations

The permutation $f = (1\ 2\ 3)(1\ 2)$ is the composition of two odd permutations:

• The composition of two odd permutations results in an even permutation.

Conclusion:

The permutation $f$ is an even permutation.

The correct answer is: b) even permutation.

Indeed, for integers m,n (not both zero),
⟨m,n⟩ = gcd(m,n)Z
In this case, that means that
⟨2,7⟩ = gcd(2,7)Z
= 1Z
= Z

Let G = A₄, H = {(1 2)(3 4),(1 3)(2 4),(1 4)(2 3), e},

K = {e, (1 2) (3 4)} 

Then H is normal in G and K is normal in H. But K is not normal in G

Here is the extracted text without using LaTeX or KaTeX:

Step 1: Understanding the group Z and the subgroup H

• Z is the set of all integers under addition.
• H = {3x | x in Z} is the set of all integer multiples of 3. This is a subgroup of Z because:
  1. The identity element (0) is in H.
  2. H is closed under addition (if 3a is in H and 3b is in H, then 3a + 3b = 3(a + b) is in H).
  3. Every element in H has an additive inverse in H.

Thus, H is a normal subgroup of Z, since Z is an abelian group, and all subgroups of abelian groups are normal.

Step 2: Quotient group Z/H
The quotient group Z/H consists of cosets of the form x + H, where x is in Z. Two elements z1, z2 in Z belong to the same coset if their difference is an element of H, i.e., z1 - z2 is in H.

For example:

• 0 + H = H = {3x | x in Z} is the coset containing all multiples of 3.
• 1 + H contains all integers of the form 1 + 3x, where x is in Z.
• 2 + H contains all integers of the form 2 + 3x, where x is in Z.

Step 3: Structure of Z/H
Since every integer x in Z is congruent to one of 0, 1, 2 modulo 3, the cosets of H in Z are exactly H, 1 + H, and 2 + H.

Therefore, the quotient group Z/H has three distinct cosets:

• H = {3x | x in Z}
• 1 + H = {1 + 3x | x in Z}
• 2 + H = {2 + 3x | x in Z}

we have o(HK) = 
Implies 
Implies o(H ∩ K) = 3

By the Lagrange’s theorem, the order of subgroup divides the order of group. We claim that all elements of G cannot be of order 2. Suppose it so.

Let a,b ∈ G be two different element of order 2. Let H = < a >, K = < b > be the cyclic groups generated by a and b, then
o(H) = 2, o(K) = 2
Since all the elements of group G are of order 2, it must be abelian.
∴ HK= KH ⇒ HK is subgroup of G And so

By Lagrange’s theorem, o(HK) would divide o(G).Which is not true hence our assumption is not correct.
Again, since G is finite, o(a) I o(G) for all a ∈ G
⇒ there exist at least one element a ∈ G such that  o(a) = 5 or 10.
If o(a) = 5, then H = < a > is subgroup of order 5.

If o(a) = 10, then H = < a2 > is a subgroup of order 5.

The order of the element (2̅, 2̅) in Z₄ × Z₆ is determined by finding the least common multiple (LCM) of the orders of 2̅ in Z₄ and Z₆:

1. Order in Z₄:
   The order of 2̅ in Z₄ is the smallest n such that n · 2 ≡ 0 (mod 4).
   2 · 2 = 4 ≡ 0 (mod 4) ⇒ Order = 2.

2. Order in Z₆:
   The order of 2̅ in Z₆ is the smallest n such that n · 2 ≡ 0 (mod 6).
   3 · 2 = 6 ≡ 0 (mod 6) ⇒ Order = 3.

3. LCM of 2 and 3:
   LCM(2, 3) = 6.

Thus, the order of (2̅, 2̅) is 6, corresponding to option B.

A simple group is a nontrivial group whose only normal subgroups are the trivial group and the group itself. For a homomorphism from a simple group G to another group H, the kernel of the homomorphism must be a normal subgroup of G. Since G is simple, the kernel can only be either the trivial subgroup or the entire group G.

• If the kernel is the entire group G, the homomorphism is trivial, mapping every element of G to the identity element in H.
• If the kernel is the trivial subgroup, the homomorphism is injective (one-to-one) because the trivial kernel implies no two distinct elements in G are mapped to the same element in H.

Thus, any homomorphism from a simple group must either be trivial or injective. Therefore, the correct answer is: C.

Option A:
9 does not divide 12.
If 9  divided 12.. In the case of rings, even if n divides m, the operations would still be modulo m, which is different from modulo n. So even then, unless there's a specific embedding, it's not clear.
For example, if you take m = 6 and n = 3. Then Z/3Z is not a subring of Z/6Z because the operations are different. For instance, in Z/3Z, 1 + 1 = 2, but in Z/6Z, it's also 2. Wait, but multiplication: 2*2=4 in Z/6Z, but in Z/3Z, 2*2=1. So multiplication isn't preserved. So even if n divides m, it's not a subring. Therefore, maybe the key reason is that Z/9Z is not a subset because the elements are different equivalence classes.

Option B:
The gcd being 3 relates to the Chinese Remainder Theorem or ring isomorphisms, but it does not directly address why Z/9Z cannot be a subring of Z/12Z.

Option C:
This is irrelevant. The structure of Z/nZZ/nZ as a ring does not depend on nn being a power of another number.

Option D :
Z/9Z is not a subset of Z/12Z.
Well, Z/9Z has elements {0,1,2,...,8} modulo 9, and Z/12Z has {0,1,2,...,11} modulo 12.
Are the elements of Z/9Z actually elements of Z/12Z? Let's see.
No,  Because in Z/9Z, the elements are equivalence classes modulo 9, while in Z/12Z, they are modulo 12. So even though the numbers 0-8 are present in both, their equivalence classes are different.

In a cyclic group of order n, an element a^k is a generator if and only if gcd(k, n) = 1. Given that both a (corresponding to k = 1) and a² (corresponding to k = 2) are generators, we must have:

1. gcd(1, n) = 1 (always true).
2. gcd(2, n) = 1.

For gcd(2, n) = 1, n must be odd (if n were even, gcd(2, n) = 2 ≠ 1, so a² would not generate the entire group).

Answer:
A: n must be odd

Explanation:

• If n is even, a² has order (n / gcd(2, n)) = n / 2 < n, so a² cannot generate the group.
• If n is odd, gcd(2, n) = 1, ensuring a² is a generator.
• Primality of n is not required (e.g., n = 9, which is odd and composite, satisfies the condition).

Step-by-Step Explanation:

1. Understanding the Problem:
   We need to find the number of elements g in a cyclic group G of order 6 such that G = ⟨g⟩. These elements are called generators of the group.

2. Key Property of Cyclic Groups:
   In a cyclic group of order n, the number of generators is given by Euler's totient function φ(n), which counts integers between 1 and n that are coprime to n.

3. Calculating φ(6):

   • The divisors of 6 are 1, 2, 3, 6.
   • Numbers coprime to 6 (i.e., with gcd = 1): 1 and 5.
   • Thus, φ(6) = 2.

4. Generators of G:
   If G = {e, a, a², a³, a⁴, a⁵}, the generators are elements a^k where gcd(k, 6) = 1:

   • gcd(1, 6) = 1 → a¹ = a.
   • gcd(5, 6) = 1 → a⁵.

   Hence, there are 2 generators.

Answer:
D: 2

To determine why the structure {P(S), ∪} is not a group, we verify the group axioms:

1. Closure:
   For any A, B ⊆ S, A ∪ B is also a subset of S. Thus, P(S) is closed under ∪.
   → A is incorrect.

2. Associativity:
   The union operation ∪ is associative: (A ∪ B) ∪ C = A ∪ (B ∪ C) for all A, B, C ⊆ S.
   → B is incorrect.

3. Inverses:
   For {P(S), ∪} to be a group, every subset A ⊆ S must have an inverse B ⊆ S such that:
   A ∪ B = ∅.

   However, if A ≠ ∅, no such B exists because A ∪ B must contain all elements of A. The only subset with an inverse is ∅ (itself).
   Inverses fail for non-empty subsets.
   C is correct.

4. Identity Element:
   The empty set ∅ acts as the identity element because A ∪ ∅ = A for all A ⊆ S.
   → D is incorrect.

we have o(AB) = 
Implies 
= 30/10 = 3

A homomorphism is determined by φ(1) = a ∈ Z₁₂, where 8a ≡ 0 mod 12. Solutions: a = 0, 3, 6, 9.

Thus, 4 homomorphisms.