Compute character table of $S_4$ in Fulton-Harris This is an example in Fulton-Harris 2.3 at p.g. 18: computing the character table of $S_4$. I have computed $\chi_{trivial},\chi_{sgn},\chi_{std},\chi_{std\cdot sgn}$. The last character, denoted as $\chi_{W}$, can be computed by orthogonality relation. However, I have trouble understanding the explanation in Fulton-Harris:

The key is the $2$ in the last column for $\chi_{W}$: this says that the action of $(12)(34)$ on the two-dimensional vector space $W$ is an involution of trace $2$, and so must be the identity. Thus, $W$ is really a representation of the quotient group
$$S_4/\{1,(12)(34),(13)(24),(14)(23)\}\cong S_3$$
[One may see this isomorphism by letting $S_4$ act on the elements of the conjugacy class of $(12)(34)$.] $W$ must then be just the standard representation of $S_3$ pulled back to $S_4$ via this quotient. 

My questions are

*

*Why did we select the normal subgroup as $V=\{1,(12)(34),(13)(24),(14)(23)\}$? If there are other normal subgroups containing $(12)(34)$, can we find the representation of another quotient group?

*I don't see explicitly this isomorphism is the same as a group action of $S_4$ on the conjugacy class of $(12)(34)$. Would anyone like to explain a bit?

Appreciate your help!
 A: The observation is that, since $(12)(34)$ maps to the identity, then it's in the kernel of $S_4\to \operatorname{Aut}(W)$, so all three conjugates of $(12)(34)$ must be in the kernel as well (in addition to $1$ of course).
This set $V$ is a union of conjugacy classes, so it's a normal subgroup.  Whenever you have a normal subgroup in the kernel of a homomorphism, that homomorphism factors through the quotient:
$$S_4 \twoheadrightarrow S_4/V \to \operatorname{Aut}(W)$$
(The universal factoring is through the quotient by the kernel itself.)  What this means is that this $W$ representation may be regarded as coming from $W$ as a representation of the smaller group $S_4/V$.
The question now is which group is $S_4/V$? Let $C$ be the conjugacy class of $(12)(34)$, and consider the action of $S_4$ on $C$ by conjugation.  A group action, recall, is a homomorphism $S_4\to \operatorname{Bij}(C)$, and since $C$ has three elements, the group of bijections $\operatorname{Bij}(C)$ is isomorphic to $S_3$ by choosing some bijection between $C$ and $\{1,2,3\}$. Thus, this group action is "actually" a homomorphism $S_4\to S_3$.  What is its kernel?  You can check that $V$ is an abelian subgroup, so it acts on each element of $C$ trivially, hence $V$ is in the kernel.  You can also check that all six bijections of $C$ can be realized by conjugating by some element of $S_4$, so $S_4\to\operatorname{Bij}(C)$ is surjective.
Since $\lvert V\rvert =4$, $\lvert S_4\rvert=4!$, and $\lvert S_3\rvert=3!$, it follows that $K$ is the kernel, so by the first isomorphism theorem $S_4/V\cong S_3$.
One way I like to think about this is to imagine a tetrahedron with vertices labeled 1-4, and to think about the elements of $C$ as being pairs of opposite edges, so for example $(12)(34)$ is the pair of edges connecting vertices 1&2 and vertices 3&4.  Conjugations by $S_4$ correspond to rotating and reflecting this tetrahedron (a conjugation relabels the vertices arbitrarily: rotate and reflect to bring the tetrahedron back to "standard" position).  Then you can realize that the three sets of opposite edge pairs have the property that they meet at each vertex.  The data of the $S_4$ action on $C$ reduces to how the three edges at vertex 1 are permuted.
