Homomorphism from $\phi : S_4 \to S_3$ Can someone explain how the homomorphism $\phi : S_4 \to S_3$ works?
Artin defines the partitions $\pi_1 : \{1,2\} \cup \{3,4\} $, $\pi_2 : \{1,3\} \cup \{2,4\} $, $\pi_3 : \{1,4\} \cup \{2,3\} $. Then he defines the map $\phi$ from $S_4$ to the group of permutations of the set $\{ \pi_1, \pi_2, \pi_3 \}$.
What I do not understand is how exactly this map is defined. For example, according to the book the $4$-cycle $p=(1234)$ acts on subsets of order two like this:
$\{1,2\} \leadsto \{2,3\}$,    $\{1,3\} \leadsto \{2,4\}$,  $\{1,4\} \leadsto \{1,2\}$
$\{2,3\} \leadsto \{3,4\}$, $\{2,4\} \leadsto \{1,3\}$, $\{3,4\} \leadsto \{1,4\}$  
Then it says that $p$ corresponds to the transposition $(\pi_1 \pi_3)$ of the permutation group of partitions. So what exactly is happening here?
 A: Let's examine more closely how $p$ acts on the different $\pi_i$s:
$$p(\pi_1)=p \left( \{1,2\} \cup \{3,4\} \right)=\{p(1),p(2)\} \cup \{p(3),p(4)\}=\{2,3\} \cup \{4,1\}=\{1,4\} \cup \{2,3\}=\pi_3 $$
Similarly, one can show that
$$p(\pi_2)=\pi_2 $$
and $$p(\pi_3)=\pi_1.$$
Thus, when viewed as a permutation of $\{\pi_1,\pi_2,\pi_3\}$, $p$ has the cycle decomposition $( \pi_1 \; \pi_3)$.
A: Let $\{a, b\}\cup \{c, d\}$ be one of $\pi_1, \pi_2, \pi_3$, where $\{a, b, c, d\} = \{1, 2, 3, 4\}$. Then $\phi$ is defined by 
$$\phi(p) (\{a, b\}\cup \{c, d\}) = \{p(a), p(b)\} \cup \{p(c), p(d)\}.$$
For example, if $p = (1234)$, then 
$$\begin{split}
\phi(p) (\pi_1) &= \phi(p) (\{1, 2\} \cup \{3, 4\}) \\
&= \{p(1), p(2)\} \cup \{p(3), p(4)\} \\
&= \{2, 3\} \cup \{4, 1\} \\
&= \{1, 4\} \cup \{2, 3\} \\
&= \pi_3.
\end{split}$$
Similarly 
$$\begin{split}
\phi(p) (\pi_2) &= \phi(p) (\{1, 3\} \cup \{2, 4\}) \\
&= \{p(1), p(3)\} \cup \{p(2), p(4)\} \\
&= \{2, 4\} \cup \{3, 1\} \\
&= \{1, 3\} \cup \{2, 4\} \\
&= \pi_2.
\end{split}$$
and $\phi(p)(\pi_3) = \pi_1$. Thus $\phi(p)$ corresponds to $(\pi_1\pi_3)$.
A: Take some $p \in S_4$, as $p: \{1,2,3,4\} \to \{1,2,3,4\}$ is a bijection, it maps disjoint 2-element subsets to disjoint two element subsets. Now let $\pi_i = \{a,b\} \cup \{c,d\}$ be a partition of $\{1,2,3,4\}$, $p$ maps this partition to $\{p(a), p(b)\} \cup\{p(c), p(d)\}$, which is some partition $\pi_j$ - as $\pi_1$, $\pi_2$, $\pi_3$ are all partitions of $\{1,2,3,4\}$ into two-element subsets - now we define $\phi(p)(i) = j$. This define a permutation $\phi(p) \colon \{1,2,3\} \to \{1,2,3\}$.
To the example: If $p = (1234)$, we have 
\begin{align*}
  p(\{1,2\}) &= \{2,3\}\\
  p(\{3,4\}) &= \{1,4\} & \leadsto \phi(p)(1) = 3
\end{align*}
Along the same lines $\phi(p)(2) = 2$ and $\phi(p)(3) = 1$. Hence $\phi(p) = (13)$.
