The Normalizer of a Subset A of A Group G, NG(A)

# The Normalizer of a Subset A of A Group G, NG(A)

 Definition: Let $G$ be a group and let $A$ be a nonempty subset of $G$. The Normalizer of $A$ in $G$ denoted $N_G(A)$ is defined to be the set $N_G(A) = \{ g \in G : gAg^{-1} = A \}$.

As once might assume - $N_G(A)$ will also be a subgroup of $G$.

 Proposition 1: Let $G$ be a group and let $A$ be a nonempty subset of $G$. Then $N_G(A)$ is a subgroup of $A$.

Note that in general, $N_G(A)$ MIGHT NOT BE A NORMAL subgroup of $G$!

(1)
\begin{align} \quad g_1g_2A(g_1g_2)^{-1} = g_1(g_2Ag_2^{-1})g_1^{-1} = g_1Ag_1^{-1} = A \end{align}
• So $g_1g_2 \in N_G(A)$.
• Furthermore, it is clear that $1 \in N_G(A)$ since $1A1^{-1} = A$.
• Lastly, let $g \in N_G(A)$. Then $gAg^{-1} = A$, which is equivalent to $g^{-1}Ag = A$. So $g^{-1} \in N_G(A)$. Hence $N_G(A)$ is a subgroup of $A$. $\blacksquare$
 Proposition 2: Let $G$ be a group and let $A$ be a nonempty subset of $G$. Then $C_G(A)$ is a subgroup of $N_G(A)$.
• Proof: Let $g \in C_G(A)$. Then $gag^{-1} = a$ for all $a \in A$. So $gAg^{-1} = A$, which shows that $g \in N_G(A)$. Hence $C_G(A)$ is a subgroup of $N_G(A)$. $\blacksquare$