If G is a group, Z its center and if G/Z is cyclic then Ga)must be abelianb)must be non abelianc)must be normal subgroupd)must be subgroupCorrect answer is option 'A'. Can you explain this answer?

Question

Answer

Explanation:

To understand why option 'A' is the correct answer, let's break down the given information.

G is a group:
A group is a set of elements with a binary operation (usually denoted as *) that satisfies the following properties:
1. Closure: For any two elements a and b in the group G, the result of their operation a*b is also in G.
2. Associativity: For any three elements a, b, and c in G, the operation is associative, meaning (a*b)*c = a*(b*c).
3. Identity: There exists an identity element e in G such that for any element a in G, a*e = e*a = a.
4. Inverse: For every element a in G, there exists an inverse element a^(-1) in G such that a*a^(-1) = a^(-1)*a = e.

Z is the center of G:
The center of a group G, denoted as Z(G), is the set of all elements in G that commute with every element in G. In other words, for any element z in Z(G) and any element g in G, we have zg = gz.

G/Z is cyclic:
The quotient group G/Z is cyclic, which means that there exists an element in G/Z, denoted as gZ, such that every element in G/Z can be written as (gZ)^n for some integer n. In other words, the elements of G/Z can be generated by repeatedly multiplying gZ by itself.

Implications:
Given this information, we can draw the following implications:

1. G/Z being cyclic implies that every element in G/Z can be written as (gZ)^n:
This means that for any element x in G/Z, there exists an integer n such that x = (gZ)^n. Since Z is the center of G, every element in Z commutes with every element in G. Therefore, every element in G/Z commutes with every element in G.

2. Every element in G/Z commutes with every element in G:
Since every element in G/Z can be written as (gZ)^n for some integer n, and every element in G/Z commutes with every element in G, it follows that every element in G commutes with every other element in G.

Conclusion:

Based on the implications above, we can conclude that if G/Z is cyclic, then every element in G commutes with every other element in G. This implies that G is an abelian group, meaning it is commutative. Therefore, option 'A' is the correct answer.

Answer

We have given that G/Z is a cyclic group, so let Zg is a generator of the cyclic group G/Z, where g ∈ G.
W e now show that G is an abelian group i.e.,

Since a ∈ G, so Za ∈ G/Z. But G/Z is a cyclic group which is gen era ted by Zg. Thus there exists an integer m such that

Za = (Zg)^m = Zg^m [∵ Z is a normal subgroup of G]

Again

a ∈ Za and Za = Zg^m ⇒ a ∈ Z g^m.

Now

a ∈ Z g m ⇒

z₁ ∈ Z such that a ∈ z₁ g^m

Similarly, for b ∈ G, b = z₂gⁿ, where z₂ ∈ z and n is any integer.

Now

Again

Answer

The answer is A

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