A375199 Number of groups G of order n such that |N(G)| <> |Z(G)|, where N(G) is the intersection of the normalizers of all subgroups of G and Z(G) is the center of G.

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 39, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 10, 3, 0, 0, 0, 0, 0, 0, 2

Offset: 1

Author

Miles Englezou, Aug 11 2024

Keywords

• nonn

Comments

The intersection of the normalizers of all subgroups of G is also called the Baer norm.

N(G) = Z(G) for every group of cubefree order. (See the Miles Englezou link for a proof.)

Examples

			a(3) = 0 since Z(C3) = N(C3) = C3, and C3 is the only group of order 3.
a(8) = 1 since |Z(Q8)| = 2 and |N(Q8)| = 8, and for other groups G of order 8 we get |N(G)| = |Z(G)|.
		

References

• R. Baer, Norm and hypernorm, Publ. Math. Debrecen, 4 (1956), 347-350.

Links

• Wikipedia, Norm (group)
• Miles Englezou, Proof that N(G) = Z(G) for groups of cubefree order

Crossrefs

Cf. A051532, A046099, A004709.

Programs

GAP

U:=[];; LoadPackage("sonata");;
for n in [1..64] do
    T:=[];;
    for i in [1..NrSmallGroups(n)] do
        S:=[];;
        G:=SmallGroup(n,i);;
        for k in [1..Length(Subgroups(G))] do
            S:=Concatenation(S,[Normaliser(G,Subgroups(G)[k])]);
        od;
        if Size(Intersection(S))<>Order(Centre(G)) then
            T:=Concatenation(T,[i]);
        fi;
    od;
    U:=Concatenation(U,[Size(T)]);
od;	
Print(U);

Formula

|N(G)| >= |Z(G)|. If n is a term of A051532 then a(n) = 0, since G = Z(G) = N(G).

By Baer (1956), Z(G) = 1 implies N(G) = 1. Hence no centerless group G satisfies |N(G)| <> |Z(G)|.

a(n) > 0 only when n is divisible by a cube (i.e., when n is a term of A046099). Equivalently, a(n) = 0 when n is a term of A004709.