Centralizers and Normalizers

Centralizers

Let G be a group. $A \subseteq G$ is a subset. Let $C_G(A) = \{ g \in G : gag^{-1} = a\} (a \in A). C_G(A)$ is called the centralizer of A and is the set of elements of G that commutes with A: $gag^{-1} = a \rightarrow ga = ag$.

Proposition: The centralizer of A in G is a subgroup of G $(C_G(A) \leq G)$

Proof: if $gh \in C_G(A)$ then $gh^{-1} \in C_G(A)$. Let $g \in C_G(A)$ then for all $a \in A, gag^{-1} = a$ so $a = g^{-1}ag$ which implies $g^{-1} \in C_G(A)$. Let $h, g \in C_G(A)$ then for $a \in A, gh^{-1}a(gh^{-1})^{-1} = gh^{-1}ahg^{-1} = gag^{-1} = a$ so $gh^{-1} \in C_G(A)$ which means $C_G(A)$ is a subgroup.

A quick note on notation: if $A = \{a\}$ the centralizer of A is denoted $C_G(\{a\})$.

Normalizers

Let G be a group, $A \subseteq G$ is a subset. For $g \in G$ let $gAg^{-1} = \{gag^{-1} : a \in A\}$. Then $N_G(A) = \{g \in G: gAg = A\}$ is the normalizer of G.

Remark: $N_G(A) = \{g \in G: \forall a \in A, gag^{-1} \in A, g^{-1}ag \in A \}$

Propositon: The normalizer of A in G is a subgroup of G $(N_G(A) \leq G)$

Proof: if $g \in N_G(A)$ then $g^{-1} \in N_G(A)$. Let $g, h \in N_G(A)$, then $gh^{-1}A(gh^{-1})^{-1} = gh^{-1}Ahg^{-1} = gAg^{-1} = A$

Remarks

1) $C_G(G)$ is a specific kind of centralizer, called the center of G and it is the set of elements of G that commute, usually denoted Z(G) 2) if G is abelian $C_G(A) = N_G(A) = Z(G) = G$ 3) $C_G(A) \leq N_G(A)$. 4) $C_G(A) = \cup_{a \in A} C_G(a)$. 5) $\{a^n : n \in Z \} \leq C_G(a)$.

Example: computations in $D_8$

Let $A = \{1, r^2, r^3\}$ We can easily see that $A \subseteq C_{D_8}(A)$. If $s \in C_{D_8}(A)$ then $srs^{-1} = r^{-1}ss^{-1} = r^{-1} = r^3 \neq r$ hence $s \not\in C_{D_8}(A)$. Since every element of $D_8$ is a power of r or s that means $A = C_{D_8}(A)$.

We know $A = C_{D_8}(A) \in N_{D_8}(A)$ by definition, and $srs^{-1} = r^{-1}ss^{-1} = r^{-1} = r^3 \in A$ so $s \in N_{D_8}(A)$. Since every element of $D_8$ is a power of r or s that means $D_8 = N_{D_8}(A)$.

We know $Z(D_8) \in C_{D_8}(A) = A$. The identity is always in Z(G), hence $1 \in Z(D_8)$. Since $srs^{-1} = r^{-1}ss^{-1} = r^{-1} = r^3$ then $r, r^3 \not\in Z(D_8)$. Finally we can see that $sr^2s^{-1} = r^{-2}ss^{-1} = r^{-2} = r^2$ so $Z(D_8) = \{1, r^2\}$.