Problem

Let \(H\) be a subgroup of order \(2\) in \(G\). Show that \(N_G(H) = C_G(H)\). Deduce that if \(N_G(H) = G\) then \(H \le Z(G)\).

Solution

By definition of normalizer, \(N_G(H) = \{g \in G \mid gHg^{-1} = H\}\).

By definition of centralizer, \(C_G(H) = \{g \in G \mid ghg^{-1} = h \text{ for all } h \in H\}\).

Since \(|H| = 2\), \(H = \{1, a\}\) where \(a \ne 1\). For any \(g \in G\), $$ gHg^{-1} = H \iff \{g1g^{-1}, gag^{-1}\} = \{1, a\} \iff gag^{-1} = a $$

and $$ ghg^{-1} = h \text{ for all } h \in H \iff g1g^{-1} = 1\ \land\ gag^{-1} = a \iff gag^{-1} = a. $$

Therefore, $$ N_G(H) = \{g \in G \mid gag^{-1} = a\} = C_G(H) $$

$$\tag*{$\blacksquare$}$$

Suppose \(N_G(H) = G\). Then for all \(g \in G\), \(gag^{-1} = a\), or \(ga = ag\). Therefore, by definition of center,

$$ h \in H = \{1,a\} \implies gh = hg\text{ for all }g \in G \implies h \in Z(G). $$

Hence \(H \le Z(G)\).

$$\tag*{$\blacksquare$}$$