$S_3$ acting on a subgroup -- confirming my computation is correct

Question

Concrete examples often force me to revisit the fundamentals. In this instance, I want to make sure that my computation of $S_3$ acting on a subgroup is correct.

For a group action by a group $G$ on a set $A$, the element $g \cdot a$ must be in $A$. Here I want to use a subgroup of $S_3$ as my $A$. In particular, let $A = \{1, (1 \, 2 \, 3), (1 \, 3 \, 2)\}$ and perform the action $(1 \, 2) \cdot (1 \, 2 \, 3)$. The group element acting in this case says "1 goes to 2 and 2 goes to 1." Hence, in the element being acted upon, I replace 1 with 2 and then 2 with 1 and get $(2 \, 1 \, 3)$, which is the same as $(1 \, 3 \, 2)$ and is in my $A$.

Just to be clear, this is not the same result as $(1 \, 2)(1 \, 2 \, 3) = (2 \, 3)$, which would be the product of the two elements of $G$ (and where the result is not in $A$). Phrased another way, $g \cdot a \neq ga$ where the two elements happen to be permutations and the operation is in the context of a group element acting upon an element.

Sorry if this question seems too trivial, but perhaps someone else shares the same uncertainty and could benefit from the concrete example.

EDIT: From the comments below I can see the need for some context. My motivation for this question arose from Exercise 10 in Section 4.1 of Dummit and Foote. The problem begins as follows: -- Let $H$ and $K$ be subgroups of the group $G$. For each $x \in G$ define the $H \, K double \, coset$ of $x$ in $G$ to be the set $H \, x \, K = \{hxk \, | \, h \in H, k \in K\}$.

To get a feel for how this might work, I decided to let $G = S_3$, $H = \{1, (1 \, 2)\}$, and $K = \{1, (1 \, 2 \, 3), (1 \, 3 \, 2)\}$. My first step was to compute the left cosets of $K$. One of those involves $x = (1 \, 2)$ for the coset $xK$.

This is where I wondered about how a group action of $S_3$ on my set $K$ would work and how such a computation should be done. (I know how to compute left cosets. My question pertains to the originally described group action.)

Does this help explain the situation?

Answer 1

The action of a group $G$ on a set $X$ is a function $G\times X\to X$; as any function, in order to describe such an action you must provide enough information so that the value of the function at each and every element of the domain is clear.

Here you are trying to have $G=S_3$ act on the set $A$ that consists of the elements $1$, $(123)$, and $(132)$. But even though these elements form a subgroup of $G$, you are not really thinking of it as a subgroup: you are really just thinking of it as a set on which $G$ will act. That's fine, we can certainly do that.

Now, you tell us that the element $(12)\in G$ will act by sending $(123)$ to $(132)$. That is all that you tell us about the action.

We are able to deduce a few things about the action from this. To avoid potential confusion with permutation multiplication, I will write the action with $\bullet$, and the elements of $A$ as literals, 1, (123), and (132).

1. Because it is an action, we know that $1\in G$ will act as the identity, so we know that $1\bullet$1$ = $1; $1\bullet$(123)$=$(123); and $1\bullet$(132)$=$(132).

2. Because it is an action, we know that applying $(12)$ twice should be the same as applying $1$; so that means that because $(12)\bullet$(123) $=$(132), then we must have $(12)\bullet$(132)$=$(123).

3. Because it is an action on $A$, we know that $(12)$ must induce a permutation on $A$; so we also know that $(12)\bullet$1$=$1, because that is all that is left.

Now, so far this is all that you've told us.

Unfortunately, this does not uniquely determine an action of $S_3$ on $A$. Right now, it is only an action of $\langle (12)\rangle$ on $A$. There are several actions of $S_3$ on $A$ which satisfy 1, 2, and 3 above, but they are all different. For example:

1. We could make $(23)$ and $(13)$ act the same way as $(12)$, and make $(123)$ and $(132)$ act the same way as $1$. It is not hard to verify that this gives you an action of $S_3$ on $A$, in which every element of $S_3$ fixes 1.

2. We could make $(13)$ act by exchanging 1 and (123), which forces $(23)$ to act by exchanging 1 and (132) (since $(23)=(12)(13)(12)$); this also determines the action of $(123)$, since $(13)(12)=(123)$ (composing right to left), so $(123)$ would have to send 1 to (123), (123) to (132), and (132) to $1$. Then $(132)$ would have to act by 1$\mapsto$(132)$\mapsto$(123) $\mapsto$ 1.

3. We could instead let $(13)$ act by exchanging 1 and (132); this forces $(23)$ to exchange 1 and (132); $(123)$ to act by 1$\mapsto$(132)$\mapsto$(123)$\mapsto$1; and $(132)$ to act by 1$\mapsto$(123)$\mapsto$(132)$\mapsto$1.

The information you have provided does not let us decide among the three possibilities, so we you have not identified a specific action of $S_3$ on $A$. So while what you have described could be part of an action of $S_3$ on $A$, it does not yet constitute sufficient information to determine whether you have actually provided an action, and if so which action it is.

(The above actions are all the possibilities: the action corresponds to an endomorphism of $S_3$ hat sends $(12)$ to a transposition, and the only possibilities once you fix the transposition you want is to let the $3$-cycles act trivially and all three transpositions to act the same way (action 1), or to identify the elements (123) and (132) with the points $1$ and $2$ in some order, and 1 with $3$, and let $S_3$ act on this set like it would on the set $\{1,2,3\}$. The two possible identifications correspond to actions 2 and 3 above.)

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In general, any group can act on any of its subgroups trivially. It may also act on any of its subgroups if you just don't care about the group structure of the subgroup: it is just acting on it as a set, so all you need is a homomorphism from $G$ to $S_H$, the set of permutations on $H$.

It's possible that no action but the trivial action exists. For example, $A_5$ cannot act nontrivially on its subgroup $\{1,(12)(34), (13)(24), (14)(23)\}$, because there is no nontrivial homomorphism $A_5 \to S_4$.