Basic Theorems Regarding Centralizers of a Subset of a Group

# Basic Theorems Regarding Centralizers of a Subset of a Group

Recall from The Centralizer of a Subset A of a Group G, CG(A) page that if G is a group and A is a nonempty subset of G then the centralizer of A in G is defined to be:

(1)\begin{align} \quad C_G(A) = \{ g \in G : gag^{-1} = a \quad \forall a \in A \} \end{align}

We proved that $C_G(A)$ is always a subgroup of $G$, that is:

(2)\begin{align} \quad C_G(A) \leq G \end{align}

We will now prove some basic results regarding centralizers.

Proposition 1: Let $G$ be a group. Then $C_G(Z(G)) = G$. |

**Proof:**Note that if $h \in Z(G)$ then $gh = hg$ for all $g \in G$, i.e., $ghg^{-1} = h$ for all $g \in G$. Since this holds true for all $h \in Z(G)$ and for all $g \in G$ we see that $C_G(Z(G)) = G$. $\blacksquare$

Proposition 2: Let $G$ be a group and let $A$ and $B$ both be nonempty subsets of $G$. If $A \subseteq B$ then $C_G(B) \leq C_G(A)$. |

**Proof:**Suppose that $A \subseteq B$. Let $g \in C_G(B)$. Then $gbg^{-1} = b$ for all $b \in B$. Since $A \subseteq B$ we have more specifically that $gag^{-1} = a$ for all $a \in A$. Thus $g \in C_G(A)$ and so $C_G(B) \leq C_G(A)$. $\blacksquare$

Proposition 3: Let $G$ be a group and let $H$ be a subgroup of $G$. Then $H \leq C_G(H)$ if and only if $H$ is abelian. |

**Proof:**$\Rightarrow$ Suppose that $H \leq C_G(H)$. Let $h_1, h_2 \in H$. Since $h_1 \in H \leq C_G(H)$ we have that $h_1h'h_1^{-1} = h'$ for all $h' \in H$. In particular we have that $h_1h_2h_1^{-1} = h_2$, i.e., $h_1h_2 = h_2h_1$. So $H$ is abelian.

- $\Leftarrow$ Conversely, suppose that $H$ is abelian. Let $h \in H$. Then $hh'h^{-1} = h'$ for all $h' \in H$. That is, $h \in C_G(H) = \{ g \in G : gh'g^{-1} = h' \quad \forall h' \in H \}$, so $H \leq C_G(H)$.

Proposition 4: Let $G$ be a group and let $A$ be a nonempty set. Then $Z(G) \leq C_G(A)$. |

**Proof:**Note that $A \subseteq G$. So by Proposition 2 we have that:

\begin{align} \quad Z(G) = C_G(G) \leq C_G(A) \quad \blacksquare \end{align}