Proof:

Let N be a normal subgroup of a group G. Suppose N is cyclic, which means that N is generated by a single element. Let's denote this generator as n.

Claim: Every subgroup of N is normal in G.

Proof of Claim:

Let H be a subgroup of N. We want to show that H is normal in G, i.e., for every g in G, gHg^(-1) = H.

Since N is cyclic, every element in N can be written as n^k for some integer k. Similarly, every element in H can be written as n^m for some integer m.

Consider an arbitrary element g in G. We need to show that gHg^(-1) = H. To prove this, we will show that gHg^(-1) is a subgroup of H and H is a subgroup of gHg^(-1).

gHg^(-1) is a subgroup of H:

Let h_1, h_2 be arbitrary elements in H. We want to show that g(h_1g^(-1))(h_2g^(-1))^(-1) is also in H.

Since H is a subgroup of N, h_1 and h_2 can be written as h_1 = n^k and h_2 = n^m for some integers k and m.

Now, g(h_1g^(-1))(h_2g^(-1))^(-1) = g(n^k g^(-1))(n^m g^(-1))^(-1)
= g(n^k g^(-1))(g^(-1))^(-1)(n^(-m))^(-1)
= g(n^k g^(-1))(g n^(-m))^(-1)
= g(n^k g^(-1))(g^(-1) n^(-m))

Using the properties of the group operation, we can rewrite this expression as:
g(n^k g^(-1))(g^(-1) n^(-m)) = (g n^k g^(-1))(g n^(-m))

Since N is a normal subgroup of G, we know that gn^k g^(-1) and gn^(-m) are both in N. Therefore, their product (g n^k g^(-1))(g n^(-m)) is also in N.

Since H is a subgroup of N, this means that (g n^k g^(-1))(g n^(-m)) is also in H. Thus, g(h_1g^(-1))(h_2g^(-1))^(-1) is in H for arbitrary h_1, h_2 in H, which implies that gHg^(-1) is a subgroup of H.

H is a subgroup of gHg^(-1):

Let h be an arbitrary element in H. We want to show that g(hg^(-1))g^(-1) is also in H.

Since H is a subgroup of N, h can be written as h = n^k for some integer k.

Now, g(hg^(-1))g^(-1) = g(n^k g^(-1))g^(-