Take the 2-minute tour ×  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.  share|improve this question  4
Answers 4  up vote 10 down vote accepted  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 $G$.  share|improve this answer This should
generalize easily to infinite groups, ri