TITLE: Need help to visualise a set (reading Abelian groups)
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: Going through the reflections in the order pictured, you can see that you end with what I labeled as $r_1$  (most books use $\rho$ to denote rotations.  paint doesn't have greek. >_<)
Going through the reflections in the other order will give you what I labeled as $r_2$.
The point of the exercise is to show that the reflections aren't abelian because you can do things in a different order and get different results and aren't a subgroup because you can get something other than a reflection (the rotations by themselves are however an abelian subgroup).