Question

If H is a subgroup of a groupG, then the normalizer of H is the set$\text{N}\left(\text{H}\right)\text{=}\left\{\text{x}\in {\text{G|x}}^{-\mathbf{1}}\text{Hx=H}\right\}$(as in exercise notation27). Prove that N(H)is a subgroup of G that contain H.

Short answer

Answer

It is proved that N(H) is a subgroup of G that contains H.

Step-by-step solution

Step by Step Solution Step 1: Required Theorem

Theorem 7.11:

A non-empty subset H of group G is a subgroup of G provided that,

1. If $\text{a,b}\in \text{H}$, the $\text{ab}\in \text{H}$;and
2. If $\text{a}\in \text{H}$then${\text{a}}^{-\mathbf{1}}\in \text{H}$

Step 2: Prove that N(H) is a subgroup of G that contains H

$$\begin{gathered} ItisknownthatN\left(H\right)isanon-emptysubsetofG. \\ Let,a,b\in N\left(H\right)then{b}^{-1}{a}^{-1}Hab=b^{-1}Hb=H \\ Therefore,N\left(H\right)isclosedunderoperation,andbH{b}^{-1}=b{b}^{-1}Hb{b}^{-1}=H \end{gathered}$$

So, N(H) is also closed under inverse.

Thus, by using Theorem 7.11, we can say that N(H) is a subgroup of G.

Also, for all $x,y\in H$.

$$\begin{gathered} Weknowthat{x}^{-1}Hx\subseteq Handx=x^{-1}xx \\ Therefore,H\subseteq x^{-1}Hx \\ Thus,byu\sin gTheorem7.11,wecannowsaythatH\subseteq N\left(H\right) \end{gathered}$$

Hence, it is proved that N(H) is a subgroup of G that contains H.