Exercise 1707Algebra › GroupsL1Added 3 Nov 2026

The normalizer

#normalizer#subgroup#normal

Statement

Let $H$ be a subgroup of $G$ . The normalizer of $H$ is $N_G(H)=\{g\in G:gHg^{-1}=H\}$ . Show that $N_G(H)$ is a subgroup of $G$ containing $H$ , and $H\triangleleft N_G(H)$ .

Hint

Show $N_G(H)$ is closed under product and inverse.

Show solutionHide solution

Solution

Contains $H$ . For $h\in H$ , $hHh^{-1}=H$ , so $h\in N_G(H)$ .

Subgroup. If $g_1,g_2\in N_G(H)$ : $(g_1g_2)H(g_1g_2)^{-1}=g_1(g_2Hg_2^{-1})g_1^{-1}=g_1Hg_1^{-1}=H$ , so $g_1g_2\in N_G(H)$ . And $g^{-1}Hg=H\iff gHg^{-1}=H$ , so $g^{-1}\in N_G(H)$ .

Normal. By definition, $H\triangleleft N_G(H)$ .

Conclusion. $N_G(H)$ is a subgroup of $G$ containing $H$ , and $H$ is normal in $N_G(H)$ .

The normalizer

Statement

Solution

More problems in Groups