Need help proving $C_{S_4}(V_4)=V_4$. One obvious method to show $C_{S_4}(V_4)=V_4$ is to directly check that each element of $S_4$ that is not in $V_4$ fails to commute with some element of $V_4$. But is there a more efficient way to showing this true?
By $V_4$, I meant $\{(), (1 2)(3 4), (13)(24),(14)(23)\}$.
 A: If you mean the normal $V_4$ inside $S_4$, (the one generated by the three elements with cyclic structure $(**)(**)$), you can use the fact that the centralizer of a subgroup is a normal subgroup of the normalizer (in this case, the whole $S_4$). So the centralizer is either $V_4$ itself, $A_4$ or $S_4$. A quick check 
$$(123)(12)(34)= (134) \neq (243) =  (12)(34)(123)$$
tells you that not all elements of $A_4$ commute with $V_4$, hence $C_{S_4}(V_4) = V_4$
$\quad$
If you do not mean the normal $V_4$, there are three of these (with elements $1, (**), (\star\star), (**)(\star\star)$ that commute). 
Now, the index of the centralizer of an element is equal to the number of its conjugates. It is well known that in $S_4$ all transpositions form a single conjugacy class, and there are $6$ of them, so the centralizer of the element $(**)$ has index $6$ and, therefore, order $\frac{24}{6}=4$.
Since $V_4$ is obviously contained in its centralizer, and the centralizer of a subgroup is obviously contained in the centralizer of one element of such subgroup, this is enough to conclude $C_{S_4}(V_4) = V_4$.
A: (1) $V_4$ is a $2$-subgroup of $S_4$ hence it is contained in a Sylow-$2$ subgroup of $S_4$.
(2) Sylow-$2$ subgroup of $S_4$ is dihedral group $D_8$ (symmetries of square with vertices $1,\cdots,4$.)
(3) Since $D_8$ is non-abelian, the $V_4$ can commute inside a $D_8$ only with $V_4$.
(4) In particular, $V_4$ do not commute with any odd permutation in $S_4$ (since $(abcd)$ and $(a'b')$ can be realized as symmetries of square, hence as elements of some Sylow-$2$ subgroup.)
(5) $V_4\leq A_4$ and $V_4$ do not commutes with any $3$-cycle.
(6) Thus $C_{S_4}(V_4)=V_4$.
A: Remember that the action of $S_4$ on its elements is by relabeling. We will concentrate on elements of order $2$:


*

*Those of the form $(1,2)$:


The only way to relabel this element while preserving it is to swap $1$ and $2$, or to swap $3$ and $4$, this gives us the group $\{(), (1,2), (3,4), (1,2)(3,4)\}$ (a Vierergruppe). This gives us an easy algorithm to find a centrralizer of this cycle type. Note that it is also the centralizer of $(3,4)$.


*

*Those of the form $(1,2)(3,4)$:


The same relabelings as above apply, but there are additional ones, the cyclic relabeling  $(1,3,2,4)$ also leaves this permutation invariant giving the centralizer $\{(), (1,2), (1,3)(2,4), (1,4,2,3), (3,4), (1,2)(3,4), (1,3,2,4),
  (1,4)(2,3)\}$


*

*Conclusion:


If $V_4$ is generated by $(1,2)$ and $(3,4)$ then the intersection of the centralizers is one of two identical groups, namely the group $V_4$ itself.
If $V_4$ is generated by $(1,2)(3,4)$ and $(1,3)(2,4)$ then the intersection do not contain the simple transpositions anymore as well as the cyclic permutations, leaving only the the squares of cyclic permutations and the original generators.
