Exercise 4.4.10 (Action of $G/A$ on A)

Proof. Let $a\in A$, and suppose that $\bar{g}=\bar{h}$, where $g,h\in G$. Then $\mathit{𝑔𝐴}=\mathit{h𝐴}$, thus $h^{-1}\mathit{𝑔𝐴}=A$, so $h^{-1}g\in A$.

Since $A$ is abelian,

Therefore

This shows that the map

is well defined: $\mathit{𝑔𝑎}{g}^{-1}$ does not depend of the choice of the representative $g$ in the class $\bar{g}$ (and $\mathit{𝑔𝑎}{g}^{-1}\in A$ for all $g\in G$ since $A⊴G$).

We verify that $f$ defines an action $\bar{g}\cdot a=f(\bar{g},a)$: for all $a$ in $A$ and for all $g,h$ in $G$,

(i)

$\bar{1}\cdot a=1a{1}^{-1}=a,$

(ii)

moreover $$\begin{array}{llll}\hfill \bar{g}\cdot (\bar{h}\cdot a)& =\bar{g}\cdot \mathit{h𝑎}{h}^{-1}& \hfill & \\ \hfill & =\mathit{𝑔h𝑎}{h}^{-1}{(\mathit{𝑔h})}^{-1}& \hfill & \\ \hfill & =\overline{\mathit{𝑔h}}\cdot a.& \hfill & \end{array}$$

So $\bar{G}$ acts by conjugation on $A$ by $\bar{g}\cdot a=\mathit{𝑔𝑎}{g}^{-1}$.

Consider the following counterexample when $A$ is not abelian.

Take $G=S_4$, $A=A_4$ and $g=(12),h=(13),a=(12)(34)\in A$.

Then $H⊴G$. Since $g$ and $h$ are odd permutation, $g{A}_4=h{A}_4=S_4-A_4$, so $\bar{g}=\bar{h}$, but

$$\begin{array}{llll}\hfill \mathit{𝑔𝑎}{g}^{-1}& =(12)(12)(34){(12)}^{-1}=(21)(34),& \hfill & \\ \hfill \mathit{h𝑎}{h}^{-1}& =(13)(12)(34){(13)}^{-1}=(32)(14),& \hfill & \end{array}$$

thus $\mathit{𝑔𝑎}{g}^{-1}\ne \mathit{h𝑎}{h}^{-1}$: the action is not well defined.