Question

Hg/h is cyclic then g is abelian where h is subgroup of centre

Answer 1

Answer:

36 Prove that if G/Z(G) is cyclic then G is abelian. If G/Z(G) is cyclic with generator xZ(G). Every element in G/Z(G) can be written as xkz for some k ∈ Z and z ∈ Z(G). Now, let g, h ∈ G, then g = xaz and h = xbw for z, w ∈ Z(G)...........................

Answer 2

Answer:

Hs cannot be a normal subgroup of A(S). ... aHa-1 = H for every element a ∈ A(S). Define g : S → S as follows. g(x) =.. ... Then if G/Z is cyclic, show that G is Abelian.....................