If $(g,a) \mapsto g \cdot a$ is an action of $G$ on $A$ and $H \vartriangleleft G$, is $(\overline{g},a) \mapsto g \cdot a$ an action of $G/H$ on $A$?

Question

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?

Answer 1

Your map is not well defined.

Let $G=S_3$ acting on $A=\{1,2,3\}$ and let $H=A_3$.

Then $(12)H=(13)H$ but $(12)\cdot1\ne (13)\cdot1$.

Answer 2

• Here is what you can do if the action is transitive.

The condition you are looking for is that $H$ is contained in the stabilizer of an element of $A$, say $H = G_a$ for some $a \in A$. Note that $G_{ga} = g G_a g^{-1}$, so that if $H$ is normal, this is independent of the choice of element $a$. Then it's true, otherwise it's false.

For example, $G$ acts on itself by left-multiplication, and this action is transitive. If $G$ is finite, then the action of $G/H$ would also be transitive since $\overline g e = g$, but how can the image of the map $(\overline g, e) \mapsto ge$, which is $G$, be larger than its domain $G/H$?

It's well-known that $A$ is $G$-isomorphic (this notion is defined analogous to multiplicativity of vector space homomorphisms) to the action of $G$ on $G/G_a$ via $ga \mapsto g G_a$. If $H \le G_a$, then the map will still be well-defined, and all the action properties just carry over. Otherwise it's not well-defined.

• If the action is not transitive:

The set $A$ may be partitioned into the orbits, and your quotient action has to be defined on every orbit. So the condition is that for every orbit $H$ is contained in the stabilizer of one element.