About normality of $V = \{1,(12)(34),(13)(24),(14)(23)\}$ in $S_4$

Question

I am trying to follow the argument in this answer. Quoting:

Since all elements in $V$ are involutions, it's easy to check $V$ is a subgroup. Note that any conjugation will not change the order of an element. If $\exists s,v$ such that $svs^{-1}\notin V$, $svs^{-1}$ must be of the form $(i\ j)$. Then by symmetry, all elements of form $(i\ j)$ will be in the conjugacy class containing $V$. So the conjugacy class containing $V$ will consist of $6+3+1=10$ elements, which doesn't divide $|S_4|=24$. Contradiction.

What I don't see is why "by symmetry, all elements of form $(ij)$ will be in the conjugacy class of $V$". How to see this?

Answer 1

The "symmetry" is just an application of the following general fact:

two permutations in $S_n$ are conjugate if and only if they have the same cyclic structure.

The proof of the normality of $V$ in $S_4$ can be made shorter by observing that the non-trivial elements of the $V$ are precisely the permutations in $S_4$ whose cyclic structure is the product of two transpositions, so the claim immediately follows.

Answer 2

If one can obtain (12) by computing $sv s^{-1}$ one can also obtain (23), (34), ... because S4 and your V are symmetrical in that every possible permutation of a form is contained in them.

Elaboration: permutations of order 2 in $S_4$ must be products of disjoint 2-cycles and since we're in $S_4$ there are exactly two forms of them: $(ij)(kl)$ and $(ij)$. The former is exhausted in constructing V and so if a conjugate of $v \in V$ is not in $V$ it must be of the latter form. This actually cannot be done, and the proof relies on that for a reductio ad absurdum reasoning.

Answer 3

Suppose we have $s \in S_4, v\in V$ such that $svs^{-1} = (i\ j) \not\in V$. Then for any $a,b \in \{1,2,3,4\}$ we have:

$(a\ i)(b\ j)(svs^{-1})(a\ i)(b\ j) = (a\ i)(b\ j)(i\ j)(a\ i)(b\ j) = (a\ b)$ in the conjguacy class of $v$ as well, since this equals:

$tvt^{-1}$ with $t = (a\ i)(b\ j)s$.

(With the usual caveats if $\{a,b\} \cap \{i,j\} \neq \emptyset$, i'm sure you can figure out these cases on your own).