Originally Posted by
kathrynmath Let N be a subgroup of the center of G. Show that if G/N is a cyclic group, then G must be abelian.
suppose G/N is cyclic. this means G/N is generated by a single coset xN, for some x in G.
so every element of G is of the form x^kn, for some integer k, and some n in N, and since N is contained in the center of G, n is in the center of G.
so let g,h be in G. we can write g = x^kn, h = x^mn', for integers k,m and n,n' in Z(G).
thus gh = (x^kn)(x^m)n' = (x^k)(nx^m)n'
= (x^k)(x^m)nn' (since n(x^m) = (x^m)n)
= x^(k+m)(nn') = (x^m)(x^k)(nn')
= (x^m)(x^k)(n'n) (since elements of N commute with all of G, including members of N)
= (x^m)(x^kn')n = (x^m)(n'x^k)n (since n' commutes with all of G, and thus x^k in particular)
= (x^m)n'(x^k)n = hg, so G is abelian.
I feel like there must be an easier way to do this.