Let $G$ be a group and $A$ be a normal abelian subgroup, where $\overline{G}=G/A$. Show that $\overline{g}a=gag^{-1}$ is a left action of $\overline{G}$ on $A$.
Let $\overline{g_1},\overline{g_2}\in\overline{G}$ be arbitrary. The following computations show that $\overline{g}a=gag^{-1}$ is a left action:
\begin{align} \overline{1}a&=1a1^{-1} \\ &=a \end{align}
\begin{align} \overline{g_1}(\overline{g_2}a)&=\overline{g_1}(g_2ag_2^{-1}) \tag{$\clubsuit$}\\ &=g_1g_2ag_2^{-1}g_1^{-1} \\ &=(g_1g_2)a(g_1g_2)^{-1} \\ &=\overline{g_1g_2}a \end{align}
I encounter difficulty in showing the action is well-defined; that is, the action of a coset on an element of $A$ is uniquely determined regardless of what representative we choose.
To prove the action is well-defined, we let $g_1,g_1^*\in\overline{g_1}$ and $g_2,g_2^*\in\overline{g_2}$ and show $\overline{g_1g_2}a=\overline{g_1^*g_2^*}a$. We have that $g_1^*=g_1a_1$ and $g_2^*=g_2a_2$ for some $a_1,a_2\in A$.
The following computation is where I encounter difficulty:
\begin{align} \overline{g_1^*}(\overline{g_2^*}a)&=\overline{g_1a_1}(\overline{g_2a_2}a) \\ &=g_1a_1(g_2a_2aa_2^{-1}g_2^{-1})a_1^{-1}g_1^{-1} \\ &=g_1a_1(g_2(a_2a_2^{-1})ag_2^{-1})a_1^{-1}g_1^{-1} \tag{$\ast$}\\ &=g_1a_1(g_2ag_2^{-1})a_1^{-1}g_1^{-1} \tag{$\star$}\\ &=\overline{g_1^*}(\overline{g_2}a) \\ &\neq\overline{g_1g_2}a \end{align}
We used the fact $A$ is abelian to make the simplification in $(\ast)$, but we cannot do the same to simplify $(\star)$ because every element of $G$ does not necessarily commute with every element of $A$.
Another thought is that $g_2ag_2^{-1}\in A$ since $A$ is a normal subgroup of $G$. We could continue our computation as follows:
\begin{align} g_1a_1(g_2ag_2^{-1})a_1^{-1}g_1^{-1}&=g_1(a_1a_1^{-1})(g_2ag_2^{-1})g_1^{-1}\\ &=g_1g_2ag_2^{-1}g_1^{-1} \\ &=(g_1g_2)a(g_1g_2)^{-1} \\ &=\overline{g_1g_2}a \end{align}
We conclude $\overline{g_1^*g_2^*}a=\overline{g_1^*}(\overline{g_2^*}a)=\overline{g_1g_2}a$ since we already proved the action for arbitrary cosets in $(\clubsuit)$.