Statement: If N is a normal subgroup of G and H is any subgroup of G, then NH is a subgroup of G.

Explanation:
To prove that NH is a subgroup of G, we need to show that NH satisfies the three criteria for being a subgroup:

1. Closure: For any two elements x, y in NH, their product xy must also be in NH.
2. Identity: NH must contain the identity element of G.
3. Inverses: For any element x in NH, its inverse x^(-1) must also be in NH.

Proof:

1. Closure:
Let x, y be arbitrary elements in NH. Since H is a subgroup of G, we have x ∈ NH and y ∈ NH. This means that there exist elements n₁, n₂ in N and h₁, h₂ in H such that x = n₁h₁ and y = n₂h₂.
Now, consider the product xy:
xy = (n₁h₁)(n₂h₂) = n₁(h₁n₂)h₂
Since N is a normal subgroup of G, we know that h₁n₂ is also in N. Therefore, xy can be written as xy = n₃h₂, where n₃ = n₁(h₁n₂) is an element in N.
Since n₃ is in N and h₂ is in H, we have shown that xy can be expressed as an element of NH. Hence, NH is closed under multiplication.

2. Identity:
Since N is a subgroup of G, it must contain the identity element of G denoted as e. Therefore, e is in N. Additionally, since H is also a subgroup of G, it must contain the identity element of G. Let's denote it as h. Hence, the product eh is an element of NH. Since eh = h = he, we have shown that NH contains the identity element of G.

3. Inverses:
Let x be an arbitrary element in NH. This means that there exist elements n in N and h in H such that x = nh.
Since N is a normal subgroup of G, we know that n^(-1) is also in N. Therefore, (nh)^(-1) = h^(-1)n^(-1) is an element in NH. Hence, NH is closed under taking inverses.

Since NH satisfies all three criteria for being a subgroup, we can conclude that NH is a subgroup of G. Therefore, the correct answer is option 'A'.