Basic Theorems Regarding Normalizers of a Subset of a Group

# Basic Theorems Regarding Normalizers of a Subset of a Group

Recall from The Normalizer of a Subset A of a Group G, NG(A) page that if $G$ is a group and $A$ is a nonempty subset of $G$ then the normalizer of $A$ in $G$ is defined to be:

(1)\begin{align} \quad N_G(A) = \{ g \in G : gAg^{-1} = A \} \end{align}

We proved that $N_G(A)$ is always a subgroup of $G$ and that $C_G(A)$ is always a subgroup of $N_G(A)$ so that:

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

We will now prove some other results regarding normalizers. Compare the below results with those regarding centralizers on the Basic Theorems Regarding Centralizers of a Subset of a Group page.

Proposition 1: Let $G$ be a group. If $a \in G$ then $N_G(a) = C_G(a)$, i.e., the normalizer of a singleton set in $G$ is the centralizer of that singleton in $G$. |

**Proof:**This follows immediately from the definitions of $N_G(a)$ and $C_G(a)$. $\blacksquare$

Proposition 2: Let $G$ be a group. Then $N_G(Z(G)) = G$. |

**Proof:**From the Basic Theorems Regarding Centralizers of a Subset of a Group page we know that $C_G(Z(G)) = G$. Since $G = C_G(Z(G)) \leq N_G(Z(G)) \leq G$ we conclude that $N_G(Z(G)) = G$. $\blacksquare$

Proposition 3: Let $G$ be a group and let $H$ be a subgroup of $G$. Then $H \leq N_G(H)$. |

*Note that the normalizer can be constructed of ANY nonempty subset $A$ of $G$. In the special case when $A = H$ is also a subgroup of $G$, proposition 3 tells us that $H$ will be contained in $N_G(H)$.*

**Proof:**By definition we have that:

\begin{align} \quad N_G(H) = \{ g \in G : gHg^{-1} = H \} \end{align}

- If $H$ is a subgroup of $G$ then $H$ is closed under its operation. In particular, for all $h \in H$ we have that $hHh^{-1} \subseteq H$. In fact, for all $h \in H$, if $\varphi_{h} : H \to H$ is defined for all $h_1 \in H$ by $\varphi_h(h_1) = hh_1h^{-1}$ then $\varphi_h$ is bijection with inverse $\varphi_{h^{-1}}$ since for all $h_1 \in H$ we have that:

\begin{align} (\varphi_h \circ \varphi_{h^{-1}})(h_1) &= \varphi_h(\varphi_{h^{-1}}(h_1)) \\ &= \varphi_h(h^{-1}h_1h) \\ &= h(h^{-1}h_1h)h^{-1} \\ &= h_1 \\ &= \mathrm{id}_H(h_1) \end{align}

(5)
\begin{align} \quad (\varphi_{h^{-1}} \circ \varphi_h)(h_1) &= \varphi_{h^{-1}}(\varphi_h(h_1)) \\ &= \varphi_{h^{-1}}(hh_1h^{-1}) \\ &= h^{-1}(hh_1h^{-1})h \\ &= h_1 \\ &= \mathrm{id}_H(h_1) \end{align}

- So for each $h \in H$ we actually have that $hHh^{-1} = \varphi_h(H) = H$. Thus if $h \in H$ we have that $h \in N_G(H)$, i.e., $H \leq N_G(H)$. $\blacksquare$

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

**Proof:**Let $A$ be a nonempty subset of $G$. Let $g \in Z(G)$. Then $ghg^{-1} = h$ for all $h \in G$. In particular we have that $gag^{-1} = a$ for all $a \in A \subseteq G$.

- Let $\varphi_g : A \to A$ be defined for all $a \in A$ by $\varphi_g(a) = gag^{-1}$. Then $\varphi_g = \mathrm{id}_A$ since $\varphi_g(a) = gag^{-1} = a$ for all $a \in A$, and so $\varphi_g$ is bijective.

- Thus $gAg^{-1} = \varphi_g(A) = A$. So $g \in N_G(A)$ which shows that $Z(G) \leq N_G(A)$. $\blacksquare$