In Algebra by Artin, there is an example (2.5.13), where he defines a homomorphism from $S_4$ to $S_3$.
there are three ways to partition the set of four indices {${1, 2, 3, 4}$} into pairs of subsets of order two, namely
$y_1: \{1, 2\} \cup \{3, 4\}$,
$y_2: \{1, 3\}\cup \{2, 4\}$,
$y_3: \{1, 4\} \cup\{2, 3\}$.
An element of the symmetric group $S_4$ permutes the four indices, and by doing so it also permutes these three partitions. This defines the map $f$ from $S_4$ to the group of permutations of the set $\{y_1, y_2, y_3\}$, which is the symmetric group $S_3$.
For example,the 4-cycle $p = (1\,2\,3\,4)$ acts on subsets of order two as follows:
{1, 2} $\rightarrow$ {2, 3}
{1, 3} $\rightarrow$ {2, 4}
{1, 4} $\rightarrow$ {1, 2}
{2, 3}$\rightarrow$ {3, 4}
{2, 4} $\rightarrow$ {1, 3}
{3, 4}$\rightarrow$ {1, 4}.
My question is how do you work with these operations? and how do you permute them?
For example, Artin writes that $(1,2)(3,4)$ is in the Kernel. So this means that $(1,2)(3,4)$ maps to the identity. Does that mean it maps to {1,2,3,4}? (weren't we mapping to $S_3$?) Would this map $y_1$ to $y_3$ ? How do I calculate this?
$(12)(34)$ is in the kernel of the defined mapping because it takes $y_i\to y_i\,,i=1,2,3$. For instance, since $y_1=\{1,2\}\cup\{3,4\}$, and $(12)(34)$ takes $\{1,2\}\to \{1,2\}$ and $\{3,4\}\to\{3,4\}$, it takes $y_1$ to $y_1$.
If we look at $y_2$, we see that $\{1,3\}\to\{2,4\}$ and $\{2,4\}\to\{1,3\}$. Thus $(12)(34)$ takes $y_2$ to $y_2$.
Similarly for $y_3$.
So, $(12)(34)\to e\in S_3$ under the mapping.
As far as working with them, consider $(1234)$. From your calculations, we get $y_1\to y_3$. And $y_2\to y_2$. And finally $y_3\to y_1$. Thus, under the mapping Artin has defined, we get $(1234)\in S_4\to (13)\in S_3$.