Suppose N is a normal subgroup of a group G. Which one of the following is true?a)If G is an infinite group then G/N is an infinite groupb)If G is a nonabelian group then G/N is a nonabelian groupc)If G is a cyclic group then G/N is an abelian groupd)If G is an abelian group then G/N is a cyclic groupCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Explanation:

To prove that option 'C' is true, we need to show that if G is a cyclic group, then G/N is an abelian group.

Definition:

A group G is said to be cyclic if there exists an element g in G such that every element of G can be written as a power of g.

Proof:

Let G be a cyclic group and let g be a generator of G. This means that every element of G can be written as g^n, where n is an integer.

Now consider the quotient group G/N, where N is a normal subgroup of G. The elements of G/N are the cosets of N in G, denoted by gN for all g in G.

Claim 1:

For any two elements g_1N and g_2N in G/N, their product is commutative.

Proof of Claim 1:
Let g_1N and g_2N be two elements of G/N. This means that g_1N = g_1N and g_2N = g_2N for some g_1, g_2 in G.

Then, (g_1N)(g_2N) = (g_1g_2)N = (g_2g_1)N = (g_2N)(g_1N)

Therefore, the product of any two elements in G/N is commutative.

Claim 2:

The identity element of G/N is N.

Proof of Claim 2:
The identity element of G/N is the coset eN, where e is the identity element of G.

Since N is a normal subgroup, we have eN = N.

Therefore, the identity element of G/N is N.

Claim 3:

For any element gN in G/N, its inverse is (g^-1)N.

Proof of Claim 3:
Let gN be an element of G/N. This means that gN = gN for some g in G.

The inverse of gN is the coset (g^-1)N, where g^-1 is the inverse of g in G.

Since N is a normal subgroup, we have (g^-1)N = (g^-1)N.

Therefore, the inverse of gN is (g^-1)N.

Conclusion:

From the above claims, we can conclude that G/N is an abelian group when G is a cyclic group.

Therefore, option 'C' is true.

Answer

If any group is a cyclic group then it is Abelian group also. Here G, is a CYCLIC group.

We know that Quotient group(G/N) of CYCLIC group is a CYCLIC & ABELIAN group is ABELIAN.

Now, Quotient group (G/N) is CYCLIC group {since G is cyclic} implies it must be an ABELIAN group.

Therefore, Quotient group (G/N) is an ABELIAN group. So, option 'C' is correct.

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