Introduction:
An abelian group is a group where the group operation is commutative. In other words, for any elements a and b in the group G, the operation a*b is equal to b*a. In this case, we have an abelian group G of order 25, and we need to determine the number of normal subgroups it has, excluding the identity subgroup and G itself.

Solution:

Definition of a Normal Subgroup:
A subgroup H of a group G is said to be a normal subgroup if it satisfies the condition that for every element g in G, the conjugate of H by g is a subset of H. Mathematically, this can be written as gHg^(-1) ⊆ H for all g in G.

Identity Subgroup and G:
The identity subgroup is always a normal subgroup of any group, including an abelian group. Similarly, the group G itself is also a normal subgroup of G.

Proof that there is only one normal subgroup:
To prove that there is only one normal subgroup of G other than the identity subgroup and G, we need to consider the possible orders of subgroups that can exist in an abelian group of order 25.

Prime Factorization of 25:
The order of an abelian group is the number of elements it contains. In this case, the order of G is 25. The prime factorization of 25 is 5 * 5.

Number of Subgroups of Order 5:
Using Lagrange's theorem, the order of any subgroup must divide the order of the group. Therefore, the possible orders of subgroups in G are 1, 5, and 25. Since G is an abelian group, all subgroups are normal.

Identity Subgroup and G:
The identity subgroup has an order of 1, and G itself has an order of 25. These are the two normal subgroups that were excluded in the question.

Subgroups of Order 25:
Since G is an abelian group, it is isomorphic to the additive group of integers modulo 25. Therefore, the only subgroup of order 25 is the group itself, which is already considered.

Subgroups of Order 5:
The only remaining possibility is to have a normal subgroup of order 5. Let's assume that such a subgroup H exists.

Conjugates of H:
Since H is a normal subgroup, for every element g in G, the conjugate of H by g is a subset of H, i.e., gHg^(-1) ⊆ H. In this case, the conjugate of H by g is actually H itself, as G is abelian. Therefore, gHg^(-1) = H for all g in G.

Intersection of Conjugates:
The intersection of any two conjugates of H must also be a subgroup of G. In this case, the intersection of H and gHg^(-1) for any g in G is the identity subgroup, as gHg^(-1) = H. Therefore, the intersection of any two conjugates of H is the identity subgroup.

Cayley's Theorem:
Cayley's theorem