$S_3$ acting on a subgroup -- confirming my computation is correct
Solution 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).
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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). -
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). -
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:
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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. -
We could make $(13)$ act by exchanging
1and(123), which forces $(23)$ to act by exchanging1and(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 send1to(123),(123)to(132), and(132)to $1$. Then $(132)$ would have to act by1$\mapsto$(132)$\mapsto$(123)$\mapsto$1. -
We could instead let $(13)$ act by exchanging
1and(132); this forces $(23)$ to exchange1and(132); $(123)$ to act by1$\mapsto$(132)$\mapsto$(123)$\mapsto$1; and $(132)$ to act by1$\mapsto$(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.)
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 $A_5 \to S_4$.