I've got a question about this:
Show kernel of action is a subgroup of G if G acts on A
I put the solution below after some definitions which are used in the solution and also my question.
DEFINITIONS
A binary operation $\star$ on a set G is a function $\star : G$x$G \to G$. For any $a,b \in G$ we write $a \star b$ for $\star (a,b)$
Group denoted $(G,\star)$ is an ordered pair where $G$ is a set and $\star$ is a binary operation on $G$ satisfying axioms i,ii,iii: associativity, identity and inverse.
Group action is a map $G$x$A \to A$ denoted $g.a$ such that $g_{1}.(g_{2}.a)=(g_{1}g_{2}).a$ and $1.a=a$
`$.$` is not a binary operation and $ga$ is always a member of A
Kernel of action K = $\{ g \in G | ga=a \}$
SOLUTION TO PROBLEM
Show kernel of action is a subgroup of G if G acts on A
$1 \in K$ since $1.a=a$ for all $a \in A$ Suppose $k_{1},k_{2} \in K$ and Let $a \in A$ $\Rightarrow (k_{1}k_{2}).a=k_{1}.(k_{2}.a)=k_{1}.a=a$ $\Rightarrow k_{1},k_{2} \in K$ Now let $k \in K $ and $a \in A$ $\Rightarrow k^{-1}.a=k^{-1}.(k.a) = (k^{-1}.k).a = 1.a = a$ $\Rightarrow k^{-1} \in K$ $\Rightarrow K$ is a subgroup of $G$
QUESTION
This is where my confusion lies.
$k^{-1}.(k.a) = (k^{-1}.k).a$
I understand that this "associativity" applies to groups by axiom (i) but we already stated that `.` is not a binary operation. So how can we utilize this associativity alongside the `.` operator/operands, as in this line?
Here is a simpler approach.
An action of $G$ on $A$ induces a group homomorphism $G \to Sym(A)$ (and vice-versa).
The kernel of the action is exactly the kernel of this homomorphism and so is a (normal) subgroup of $G$.