Question

Let G be a group with identity e. Let Z(G) =ag=ga∀g∈G (the "center"ofG).

a) Show that Z(G)normal subgroup to G.
b) Show that if G is non-abelian then G/Z(G) is non-cyclic
c) Show that if G is a finite non-abelian group of order where p is a prime and Z(G){e} then |Z(G)|=p.

Answer 1

Z(G) is shown to be a normal subgroup of G by demonstrating that conjugation by any group element does not change its members. If G is non-abelian, G/Z(G) cannot be cyclic as the commutativity of its cosets would contradict the non-abelian character of G. For a finite non-abelian group of order p^n where p is prime, if Z(G) is more than just the identity, its order is proven to be p.

Step-by-step explanation:

The student's question involves the concepts of group theory within abstract algebra, specifically regarding the center of a group (Z(G)) and properties of normal subgroups and cyclic groups.

1. To show that Z(G) is a normal subgroup of G, we demonstrate that for every element a in Z(G) and every group element g in G, the product gag-1 is also in Z(G). Since by definition of the center, ag = ga for all g in G, the conjugation doesn't change a, thus a remains in Z(G), proving it's a normal subgroup.
2. To show that if G is non-abelian then G/Z(G) is non-cyclic, we assume the contrary that G/Z(G) is cyclic and produce a contradiction. If G/Z(G) were cyclic, it would be generated by some coset gZ(G). For any elements x, y in G, their cosets would commute, xZ(G)yZ(G) = yZ(G)xZ(G), making G abelian, which contradicts the non-abelian nature of G.
3. Lastly, if G is a finite non-abelian group of order pn where p is a prime and Z(G) = {e}, then |Z(G)| = p. This can be shown using the class equation and properties of p-groups, leading to the conclusion that Z(G), being nontrivial, must have order p.