Group action of Free Product Group

Question

Suppose I have two groups G and H, and K is their free product, K = $G*H$ Suppose G acts on a set X with action $\phi$ and H acts on X by action $\psi$, then what is the action of K on X.

I think the action of $K$ on X should be for any word $g_1h_1, \dots h_ng_n$ repeatedly apply the action of $g_i$ or $h_i$ for every word in the group $K$. I think this is a group action because it preserves the relations that generate the group K since it must preserve all relation for the group G and H. I am not sure is this line of reasoning is correct. From the universal property of the free product, I know a group action exists, I do not know if this is the correct action that is induced.

The reason I am unsure is the generalization of this problem. If G and H act on different sets $X_1$ and $X_2$ that are not disjoint, I should be able to define the same group action as before of K action on $X_1 \cup X_2$, but this feels strange, and I do not know if it is correct.

To sum up:

1. Is the group action of K on X defined correctly?
2. How does K act on $X_1 \cup X_2$ given G acts on $X_1$ and H on $X_2$?

Answer 1

1. Yes, this is the action that is defined for $G \ast H$.

Claim: A group action of a group $G$ on a set $X$ can be alternatively defined as a group homomorphism

$$ G \rightarrow S(X)$$

where $S(X)$ is the group of permutations (self-bijections) of $X$.

Proof: A group element is mapped to the permutation "act with g": $$g.(-) : X\rightarrow X, x \mapsto g.x$$ the axiom for a group action $$g.h.x = (gh).x$$ is now precisely stating that the map $g \mapsto g.(-)$ is a group homomorphism to the group of permutations. $\square$

Recall: The universal property of a free product of groups is the following:

A free product of two groups $G$ and $H$ is a group $G \ast H $, such that group homomorphisms out of it

$$ G \ast H \rightarrow G'$$

are precisely pairs of group homomorphisms out of $G$ and $H$ : $$ \varphi: G \rightarrow G' \qquad \psi: H \rightarrow G' $$

If you go through the construction of a free product of groups, which is words in $G$ and $H$ modulo the equivalence relation as specified by the individual group operations, you will find that the induced map out of the free product is given by: $$ [ g_1h_1 \dots g_n h_n ] \mapsto \varphi(g_1) \psi(h_1) \dots \varphi(g_n) \psi(h_n) $$ Which is to say "simply apply the group homomorphism to every letter of the word in $G'$. Moreover one has to check - I suspect this is the source of your doubts, so I suggest you do this - that this assignment is independent of the choice of representative and defines a group homomorphism. In the current situation $G'= S(X)$.

2. Canonically there is no such action, since the maps $$G \rightarrow S(X_1) \qquad H \rightarrow S(X_2)$$ target different permutation groups. But there is a way to fix this: we can identify permutations $S(X_1) \hookrightarrow S(X)$ by those permutations which fix elements of $X_2 \setminus X_1$. Analogously for $S(X_2)$. These two group homomorphisms give rise to an action of $G\ast H$ on $X$ which is given as in the previous example, with the special ingredient, that elements of $G$, $H$ act trivially on $X_2\setminus X_1$,$X_1\setminus X_2$ respectively, that is to say that:

$$g.x_2 = x_2 \quad \forall x_2 \in X_2\setminus X_1 $$ and $$h.x_1 = x_1 \quad \forall x_1 \in X_1 \setminus X_2 $$