I'm learning about group actions from Dummit and Foote, and I have a question about the following remark that the authors make in section 4.1:

> 
> In particular, an action of [a group] $G$ on [a nonempty set] $A$ may also be viewed as a faithful action of the quotient group $G/\ker\varphi$ on $A$.
> 
> 
> 

Here, $\varphi: G \rightarrow S\_A$ is the permutation representation associated to the action of $G$ on $A$, i.e. $\varphi(g) = \sigma\_g$ for all $g \in G$, where for each $g \in G$, the map $\sigma\_g: A \rightarrow A$ is defined by $\sigma\_g(a) = g \cdot a$ for all $a \in A$.

**My question**: What exactly is the action of $G/\ker \varphi$ on $A$?

My guess: It's the map $(\overline{g}, a) \mapsto g \cdot a$ for all $\overline{g} \in G/\ker \varphi$ and all $a \in A$....Is this correct?

**Follow-up question**: Assuming my guess for the above question is correct (or at least makes sense), I made the following "conjecture" which I haven't seen explicitly stated in the text yet, and was wondering if the conjecture and argument is correct:

*Claim*: Let $G$ be a group acting on a nonempty set $A$ (denote the action by $g \cdot a$ for all $g \in G$ and $a \in A$), and let $H$ be a normal subgroup of $G$. Then the map from $G/H \times A \rightarrow A$ given by

$$(\overline{g}, a) \mapsto g \cdot a$$
for all $\overline{g} \in G/H$ and $a \in A$ is a group action of $G/H$ on $A$.

*Proof attempt*: Let $\overline{g\_1}, \overline{g\_2} \in G/H$ and let $a \in A$.   

0. First, we want to show that this map is well-defined: Does $\overline{g\_1} = \overline{g\_2}$ imply that $g\_1 \cdot a = g\_2 \cdot a$? Well, $\overline{g\_1} = \overline{g\_2} \iff g\_1 g\_2^{-1} \in H$. Indeed, noting that $1\_G = 1\_H \in H$, we have $g\_1 g\_2^{-1} = g\_1 1\_H g\_2^{-1} \in H$ because $H$ is a normal subgroup of $G$.
1. Verifying the first axiom of a group action:
\begin{align\*}
 \overline{g\_1} \cdot (\overline{g\_2} \cdot a) &= g\_1 \cdot (g\_2 \cdot a) && (\text{Defn of action of } G/H \text{ on } A) \\[3pt]
 &= (g\_1 g\_2) \cdot a && (\text{Axiom 1 applied to action of $G$ on $A$}) \\[3pt]
 &= \overline{g\_1 g\_2} \cdot a. && (\text{Defn of action of } G/H \text{ on } A)
\end{align\*}
2. Verifying the second axiom:
\begin{align\*}
 1\_{G/H} \cdot a &= \overline{1\_G} \cdot a \\[3pt]
 &= 1\_G \cdot a && (\text{Defn of action of $G/H$ on $A$}) \\[3pt]
 &= a. && (\text{Axiom 2 applied to action of $G$ on $A$})
\end{align\*}

Is my claim and proof correct? If so, would this be considered called the "canonical" action of $G/H$ on $A$?

**Update 10/12/21:** Thanks to @ancientmathematician for providing a counter-example. Here's my modified conjecture (this time defined in terms of left cosets):

As before, let $G$ be a group acting on a nonempty set $A$, and denote the group action by $g \cdot a$ for all $g \in G, a \in A$. Let $H$ be a normal subgroup of $G$ and let $\overline{g} := gH$ for all $g \in G$. Consider the map
$$(\overline{g},a) \mapsto g \cdot a.$$

*New Claim*: The above map is well-defined if and only if the kernel of the action of $G$ on $A$ is $H$.   

*Proof attempt*:   

$\implies$ direction: Suppose $H$ is the kernel of the action of $G$ on $A$. Let $g\_1, g\_2 \in G$, let $a \in A$, and suppose $\overline{g\_1} = \overline{g\_2}$. Then $g\_1^{-1} g\_2 \in H$, and thus $g\_1^{-1} g\_2$ is in the kernel of the action. Therefore,
\begin{align\*}
 (g\_1^{-1}g\_2) \cdot a &= a && (\text{By definition of being in the kernel}) \\[3pt]
 g\_1 \cdot ((g\_1^{-1}g\_2) \cdot a) &= g\_1 \cdot a \\[3pt]
 (g\_1 g\_1^{-1} g\_2) \cdot a &= g\_1 \cdot a && (\text{Axiom 1 of a group action}) \\[3pt]
 g\_2 \cdot a &= g\_1 \cdot a.
\end{align\*}

Since $a \in A$ was arbitrary, this shows that $\overline{g\_1} = \overline{g\_2} \implies g\_1 \cdot a = g\_2 \cdot a$ for *all* $a \in A$.   

$\impliedby$ direction: Let $g\_1, g\_2 \in G$ and suppose $g\_1 \cdot a = g\_2 \cdot a$ for all $a \in A$. Then
\begin{align\*}
 g\_1^{-1} \cdot (g\_1 \cdot a) &= g\_1^{-1} \cdot (g\_2 \cdot a) \\[3pt]
 (g\_1^{-1} g\_1) \cdot a &= (g\_1^{-1} g\_2) \cdot a && (\text{Axiom 1 of a group action}) \\[3pt]
 1\_G \cdot a &= (g\_1^{-1} g\_2) \cdot a \\[3pt]
 a &= (g\_1^{-1} g\_2) \cdot a && (\text{Axiom 2 of a group action})
\end{align\*}

So we have $a = (g\_1^{-1} g\_2) \cdot a$ for *all* $a \in A$, and thus $g\_1^{-1} g\_2$ is in the kernel of the action of $G$ on $A$; but this is just $H$, so $g\_1^{-1} g\_2 \in H$ which then implies $\overline{g\_1} = \overline{g\_2}. \quad \square$

It seems that my proofs of (1) and (2) applied to this new situation would still be valid, so I take it that this is a well-defined group action of $G/H$ on $A$ if and only if the kernel of the action of $G$ on $A$ is $H$. Do I have this correct?