Does the set A in a group action need to be related to the group G?

To verify if a map is a group action, first check that its domain is $G \times A$ and its codomain is $A$. Then, verify the two properties: $g_1 \cdot (g_2 \cdot a) = (g_1 g_2) \cdot a$ for all $g_1, g_2 \in G$ and $a \in A$, and $1 \cdot a = a$ for all $a \in A$.

Conditions: The map is defined from $G \times A$ to $A$.; $G$ is a group and $A$ is a set.

Yes, the $S_3$ permutation action on $A = \{x_1, x_2, x_3\}$ is faithful because its kernel is $\{1\}$, the singleton containing only the identity element.

Conditions: $G = S_3$, the symmetric group of degree 3.; $A = \{x_1, x_2, x_3\}$.; The action is defined by $\sigma x_i = x_{\sigma(i)}$.

For the trivial action $ga = a$, the kernel is the entire group $G$.

Conditions: The action is the trivial action $ga = a$ for all $g \in G$ and $a \in A$.

The kernel of the group action is the set of group elements $g$ such that $ga = a$ for all $a \in A$. It is denoted as $\{g \mid ga = a\}$.

Conditions: A group action of $G$ on $A$ is given.; The set consists of elements $g \in G$ satisfying the fixing condition for all $a \in A$.

Yes, the map $ga = a$ for all $g \in G$ and $a \in A$ is a group action, known as the trivial action. It satisfies both the compatibility and identity properties of a group action.

Conditions: $G$ is a group and $A$ is a set.; The map is defined by $ga = a$ for all $g \in G$ and $a \in A$.