Proof that N(G) = Z(G) for every G of cubefree order.

If G is abelian, then N(G) = Z(G) trivially, since Z(G) <= N(G).
It also follows for centerless groups, as Baer (1956) shows
Z(G) = 1 implies N(G) = 1.

Hence we only have to consider the case when G is nonabelian
with nontrivial center.


Notation: let Int(X,Y) denote the intersection of 
sets X and Y.

Lemma 1: If G is cubefree, then Int(Z(G),G') = 1.

Proof: As G is nonabelian, Z(G) < G and the derived subgroup G' > 1.
As G is cubefree, every Sylow subgroup P has a maximum order p^2,
and hence every P is abelian. It is known that if P is abelian, 
then p does not divide |Int(Z(G),G')|. As every P is abelian, 
no p dividing |G| divides |Int(Z(G),G')| and consequently Int(Z(G),G') = 1. 


Lemma 2: If Int(Z(G),G') = 1, then Z(G/Z(G)) = 1.

Proof: Let gZ(G) and hZ(G) be two cosets of G/Z(G) not equal 
to Z(G). Commutativity implies [gZ(G),hZ(G)] = Z(G). Let z be an 
element of Z(G). Rewriting gZ(G) as gz and hZ(G) as hz, [gZ(G),hZ(G)] 
can be rewritten as
	
	gzhz(z^-1)(g^-1)(z^-1)(h^-1) = gh(g^-1)(h^-1) = [g,h].
	
Therefore if [gZ(G),hZ(G)] = Z(G), then [g,h] is an element of
Z(G). But since Int(Z(G),G') = 1, the only commutator in Z(G) is
the identity [e,e]. Hence [g,h] = [e,e], and gZ(G)hZ(G) = hZ(G)gZ(G) 
implies gZ(G) = hZ(G), and consequently Z(G/Z(G)) = 1.


Theorem: N(G) = Z(G) if G is cubefree.

Proof: As Z(G/Z(G)) = 1, the second center Z_2 in the upper central
series (UCS) of G is equal to Z_1: 
	
	Z(G/Z(G)) = Z_2/Z_1 = 1, hence Z_2 = Z_1 = Z(G).
	
Z(G) is therefore the terminal member of the UCS of G. Defining an
upper norm series (UNS) of G analogously, where N_(i+1)/N_i = N(G/N_i), a
result of Baer (1956) states that if Z_k is the terminating member
of a UCS, then Z_k = N_k, N_k the terminating member of the correpsonding 
UNS. As Z_1 = Z(G) is the terminating member of the UCS of G, so then

	Z(G) = Z_1 = N_1 = N(G)
	
and the theorem is proved.