Intersection of all normalizers

Question

This is probably standard for group-theorists: Let $G$ be a finite group. Is it true that the intersection of all normalizers of subgroups equals the center? If so, where do I find a proof? What about the same question for infinite groups?

The original question can be reformulated as follows: Let $G$ be a finite group and fix $g\in G$. Assume that for every $x\in G$ there exists a natural number $k(x)$ such that $$ gxg^{-1}=x^{k(x)}, $$ does it follow that $k(x)=1$ for all $x$? One gets the reformulation by applying the original statement to cyclic groups. It suffices to consider cyclic groups, as an element that normalizes all cyclic groups, normalizes every subgroup.

Answer 1

No. There are non-abelian groups $G$ for which all subgroups are normal, such as the quaternion group of order 8. So the intersection of all normalizers is just $G$.

Answer 2

The intersection of all normalizers of subgroups of a group $G$ is called the norm of $G$. By a result of Schenkman [E. Schenkman, On the norm of a group, Illinois J. Math., 7 (1960) 150-152] the norm of a group always lies in $Z_2(G)$ of the upper central series. So the question has positive answer if for example the center $Z(G)=Z_2(G)$.

The Schenkman's result has been improved by Cooper [C.D.H. Cooper, Power automorphisms of a group, Math. Z., 107 (1968) 335-356.] as follows: an automorphism which leaves every subgroup of a group $G$ invariant induces the trivial automorphism in the central factor group $G/Z(G)$.

Answer 3

As was pointed out by Rickard and Weidner, the existence of Hamiltonian groups like $Q_8$ shows that the answer to the original question is "no". But all Hamiltonian groups have even order, so this leaves open the question for odd order groups. The answer is "no" for these groups too.

Let $p$ be an odd prime, and let $G$ be the unique nonabelian group of order $p^3$ and exponent $p^2$. Then $Z(G)$ has order $p$, but $G$ has an elementary abelian subgroup $E$ of order $p^2$, consisting of all elements $x \in G$ with $x^p = 1$. I claim that $E$ normalizes every subgroup of $G$. To see this, let $H \subseteq G$. If $|H| \geq p^2$ then $H$ is normal in the whole group $G$, and otherwise $H \subseteq E$, and since $E$ is abelian, $E$ normalizes $H$.

Answer 4

A Hamiltonian group is a non abelian group, in which every subgroup is normal. For example the Quaternion group $$\{1,i,j,k,-1,-i,-j,-k\}$$ is Hamiltonian. This gives a counterexample to your first question.

Answer 5

See R. Baer. He Proved: N(G)=E if and only if Z(G)=E.

Answer 6

There is a number of papers dealing with groups with many normal subgroups. For example, let $G$ be a nonabelian group all of whose maximal abelian subgroups are normal. It is easy to see that then $G$ is nilpotent, so to move further, one may assume that $G$ is a $p$-group. If $p>2$, then $G$ is regular. Indeed, if $x,y\in G$, then $X=\langle x\rangle^G$ and $Y=\langle y\rangle^G$ are abelian normal subgroups of $G$. Therefore, $\text{cl}(XY)\le2$ (Fitting) so $XY$ is regular, and we conclude that $G$ is also regular in view of $x,y$ are chosen arbitrary. However, I think that the class of $G$ is unbounded. Another interesting class consists of nonabelian $p$-groups all of whose minimal nonabelian subgroups are normal. Such $p$-groups were classified by two teams of Chinese $p$-group theorists (see, in volume 5 of Berkovich-Janko series \S 228).