I was studying about "group actions" in group theory. The definition of a group action is given as follows:
Let $G$ be a group and $S$ be a non-empty set . A (left) action of $G$ on $S$ is a function $\cdot :G\times S\longrightarrow S$(usually denoted by $\cdot(g,x)\longrightarrow g\cdot x$) such that
i) $(g_1g_2)\cdot x=g_1\cdot (g_2\cdot x)$ and
ii. $e\cdot x=x$, where $e$ is the identity of $G$
for all $x\in S,g_1,g_2\in G$. (Note: If no confusion arises then we write $gx$ for $g\cdot x$)
However, I don't get this definition at all. First of all, if we assume $G=\{g_1,g_2,g_3,\dots \}$ and $S=\{x_1,x_2,x_3,\dots\}$, then $G\times S=\{(g_1,x_1),(g_2,x_2),(g_3,x_3),\dots\}$. Now, if we replace $\cdot$ by $f$ (in order to make the notations simple), then we have $f:G\times S\longrightarrow S$. Also, it is given that this $f:G\times S\longrightarrow S$ notation is usually denoted by $f(g,x)\longrightarrow gfx(=gf(x))$; thus, $f(g,x)\longrightarrow gf(x)$. Now, the definition stated two conditions: $(g_1g_2)f(x)=g_1f(g_2f(x))$ and $ef(x)=x$.
Now, first off if we consider $f(g,x)$, then its obviously a "two-variable" function, so what's the significance of $gf(x)$ in "$f(g,x)\longrightarrow gf(x)$"? Also, how do they conclude $gf(x)\in S$ (as the range of $f$ is $S$)? Also, what is meant by the conditions?
The domain of the function $f$ is $G\times S$ so writing "$g_1g_2f(x)$" and "$ef(x)$" makes no sense at all i.e. the function must take two variables as an input but clearly it's not the case here. I am not quite getting it . . .
I tried searching if there were any posts made previously on this site that answers my questions about this particular topic but I couldn't find any . . .