Question

Complete the proof of Theorem 9.24.

Short answer

The theorem 9.24 is proved.

Step-by-step solution

Statement

If A is a subgroup of a group G, then N(A) is a subgroup of G and A is a normal subgroup of N(A).

Here, $\mathrm{N}\left(\mathrm{A}\right)=\left\{\mathrm{g}\in \mathrm{G}|{\mathrm{g}}^{-1}\mathrm{Ag}=\mathrm{A}\right\}$is the normalizer of set A.

Prove that N(A) is a subgroup of G

Let $\mathrm{g},\mathrm{h}\in \mathrm{A}$.

Then, find $\left(\mathrm{gh}\right)\mathrm{A}$as:

$\begin{array}{rcl}\left(\mathrm{gh}\right)\mathrm{A}& =& \mathrm{g}\left(\mathrm{Ah}\right)\\ & =& \left(\mathrm{Ag}\right)\mathrm{h}\\ & =& \mathrm{A}\left(\mathrm{gh}\right)\end{array}$

So, $\mathrm{gh}\in \mathrm{N}\left(\mathrm{A}\right)$.

Also, $\mathrm{gA}=\mathrm{Ag}$.

This implies, ${\mathrm{Ag}}^{-1}={\mathrm{g}}^{-1}\mathrm{A}$.

Thus, ${\mathrm{g}}^{-1}\in \mathrm{N}\left(\mathrm{A}\right)$.

Therefore, N(A) is a subgroup.

Prove that A is the normal subgroup of N(A)

From the definition, $\mathrm{N}\left(\mathrm{A}\right)=\left\{\mathrm{g}\in \mathrm{G}|{\mathrm{g}}^{-1}\mathrm{Ag}=\mathrm{A}\right\}$.

It is clear that, A is the normal subgroup of N(A).

Therefore, N(A) is a subgroup of G and A is a normal subgroup of N(A).