To prove that the cyclic subgroup H generated by a is a normal subgroup of N(a), we need to show that for any element g in G, the conjugate of H by g, denoted by gHg^-1, is also a subgroup of N(a) that is equal to H.

First, note that the elements of N(a) commute with a. Therefore, if h = a^n is an element of H, then we have:

ghg^-1 = ga^ng^-1 = (gag^-1)^n

Since g and a commute with each other, gag^-1 is also an element of N(a), and hence (gag^-1)^n is an element of H. This shows that gHg^-1 is a subset of H.

Next, let h = a^n be an arbitrary element of H and let g be an arbitrary element of N(a). Then we have:

ghg^-1 = ga^ng^-1 = aga^(n-1)g^-1

Since a is in N(a), it commutes with g and therefore ag = ga. Hence, we have:

ghg^-1 = ga^(n-1)g^-1a = (gag^-1)^{n-1}ga = (gag^-1)^{n-1}h

Since gag^-1 is an element of N(a), we have (gag^-1)^{n-1} is an element of N(a) that commutes with a, and hence is also an element of H. This shows that ghg^-1 is an element of gHg^-1, and hence H is a subset of gHg^-1.

Therefore, we have shown that gHg^-1 is a subset of H and H is a subset of gHg^-1, which implies that gHg^-1 = H. Hence, H is a normal subgroup of N(a).