I am reading Abstract Algebra. I cannot visualise the following example:
Let $n$ be a positive integer, and consider the set $S_n$ of all permutations from the set $n = {1, 2, \ldots , n}$ to itself. Let $n = 3$, and consider the group $S_3$ . Two of the permutations in this group are $\phi_1$ and $\phi_3$ , where $\phi_1$ sends $1$ to itself and transposes $2$ and $3$, and $\phi_3$ sends $3$ to itself and transposes $1$ and $2$.
Let's apply the group operation to this pair of permutations, looking at $\phi_1\circ \phi_3$ and $\phi_3\circ \phi_1$ . The effect that $\phi_1\circ \phi_3$ has on $1$ is $(\phi_1\circ \phi_3)(1) = \phi_1(\phi_3(1)) = \phi_1(2) = 3$, but the effect that $\phi_3\circ \phi_1$ has on $1$ is $(\phi_3\circ \phi_1)( 1 ) = \phi_3(\phi_1(1)) = \phi_3(1) = 2$. Hence $\phi_1\circ \phi_3 \neq \phi_3 \circ \phi_1$.
I did not understand what happened here.
My Attempt: $S_3$ should have $6$ elements. What are $\phi_1$ and $\phi_3$ here?
I don't see how/why $\phi_3(1)=2$ and $\phi_1(2)=3$. Please advise.