02-02. Centralizers and normalizers, stabilizers and kernels
Definition. centlizer
- $G$,
- group
- $A \neq \emptyset$,
- subset
$C_{G}(A)$ is calledthe centralizer of $A$ in $G$.
■
Proposition.
- $G$,
- $A \neq \emptyset \subseteq G$,
- subset
proof.
$\Box$
Proposition.
- $G$,
- $A \neq \emptyset \subseteq G$,
- subset
$C_{A}(G)$ is a subgroup of $G$.
proof.
$C_{A}(G) \neq \emptyset$ since $1 \in C_{G}(A)$. Suppose that $a, b \in C_{G}(A)$.
\[\begin{eqnarray} \forall c \in A, \ ab c (ab)^{-1} & = & a b c b^{-1} a^{-1} \nonumber \\ & = & a c a^{-1} \nonumber \\ & = & c \nonumber \end{eqnarray}\]$C_{G}(A)$ is closed under the operation. Suppose that $a \in C_{G}(A)$.
\[\begin{eqnarray} \forall c \in A, \ a^{-1} c (a^{-1})^{-1} & = & a^{-1} c a \nonumber \\ & = & a^{-1} a c \quad (\because a \in C_{G}(A)) \nonumber \\ & = & c . \nonumber \end{eqnarray}\]$\Box$
Definition. center
- $G$,
- group
$Z(G)$ is called the center of $G$.
■
Remark
\(C_{G}(G) = Z(G) .\)
■
Definition. normalizer
- $G$,
- group
■
Remark
\(C_{G}(A) \subseteq N_{G}(A) .\)
$gag^{-1} \in A$ is weaker condition than $gag^{-1} = a$. $C_{G}(A)$ requires that $gag^{-1}$ exactly is equal to $a$.
\[\forall b \in B, \ bBb^{-1} \subseteq B \Rightarrow B \subseteq N_{G}(B) .\]If $A \trianglelefteq G$, $N_{G}(A) = G$.
■