# Thread: Let N be subgroup of the center of G. Show if G/N is a cyclic group, then G abelian

1. ## Let N be subgroup of the center of G. Show if G/N is a cyclic group, then G abelian

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.

2. 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.

Not really, imo. Of course, you can always shorten down some lines and explanations, but the above is the standard proof I know.

Tonio

3. Ok thanks