Solution:

To prove that the center of G is isomorphic to Z(77), we need to understand the properties of the center of a group.

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

Properties of the Center:
1. Z(G) is a subgroup of G.
2. If G is abelian, then Z(G) = G.
3. If G/Z(G) is cyclic, then G is abelian.

Order of the Group:
Since the order of G is 77, it can be decomposed into prime factors as 77 = 7 * 11.

Using Property 2:
If G is abelian, then Z(G) = G. However, we cannot conclude that G is abelian based on its order alone.

Using Property 3:
If G/Z(G) is cyclic, then G is abelian. We need to determine the order of G/Z(G) to apply this property.

Order of G/Z(G):
By the First Isomorphism Theorem, the order of G/Z(G) is equal to the order of G divided by the order of Z(G).

Since the order of G is 77 and Z(G) is a subgroup of G, the order of Z(G) must divide 77.

Possible Orders of Z(G):
The possible orders of Z(G) are the divisors of 77, which are 1, 7, 11, and 77.

Case 1: Z(G) has order 1:
If Z(G) has order 1, then G/Z(G) has order 77/1 = 77. This means that G/Z(G) is cyclic.

Case 2: Z(G) has order 7:
If Z(G) has order 7, then G/Z(G) has order 77/7 = 11. This means that G/Z(G) is cyclic.

Case 3: Z(G) has order 11:
If Z(G) has order 11, then G/Z(G) has order 77/11 = 7. This means that G/Z(G) is cyclic.

Case 4: Z(G) has order 77:
If Z(G) has order 77, then G/Z(G) has order 77/77 = 1. This means that G/Z(G) is trivial.

Conclusion:
Based on the possible orders of Z(G) and the corresponding orders of G/Z(G), we can conclude that:

If Z(G) has order 1, then G/Z(G) is cyclic, which implies that G is abelian.
If Z(G) has order 7, then G/Z(G) is cyclic, which implies that G is abelian.
If Z(G) has order 11, then G/Z(G) is cyclic, which implies that G is abelian.
If Z(G) has order 77, then G/Z(G) is trivial, which does not imply that G is ab