Group of Order 49

Explanation:

To prove that option 'A' is the correct answer, we need to show that G is abelian.

Definition: A group G is said to be abelian if the group operation is commutative, i.e., for any elements a, b in G, ab = ba.

Proof:

We know that the order of a group is equal to the number of elements in the group.

Given that the order of group G is 49, we can use Lagrange's theorem to deduce certain properties of G.

Lagrange's Theorem: If G is a finite group and H is a subgroup of G, then the order of H divides the order of G.

Using Lagrange's theorem, we can conclude that the orders of the subgroups of G can only be 1, 7, or 49.

Now, let's consider the center of the group, denoted as Z(G), which is the set of elements that commute with all other elements in G.

Claim: The order of the center of G, denoted as |Z(G)|, is greater than 1.

Proof of Claim:

Suppose |Z(G)| = 1, i.e., the center of G contains only the identity element.

Let's consider an element g in G, where g ≠ e (identity element).

Since G is a group, every element must have an inverse. Therefore, the inverse of g is also an element of G.

Now, consider the element g and its inverse g^(-1).

Since g and g^(-1) are distinct elements, they must belong to different cosets of the center. This is because if g and g^(-1) belonged to the same coset, g * g^(-1) would commute with all elements, which contradicts the fact that g and g^(-1) are distinct.

Therefore, there are at least 2 distinct cosets of the center of G, each containing 24 elements.

But this implies that the order of G is at least 48, which is a contradiction since the order of G is given as 49.

Hence, the claim is proved, and |Z(G)| > 1.

Conclusion:

Since the center of G has order greater than 1, it means that there exist elements in G that commute with all other elements.

Therefore, G is abelian.

Hence, option 'A' is the correct answer.