Centralizer of $(13)(24)$ in $A_4$ Compute the centralizer of $(13)(24)$ in $A_4$
My attempt:
I calculated order of the centralizer as
$$o(C_{A4}(13)(24))=\frac{o(A_4)}{o(cl(13)(24))}$$
where $cl$ reprsents the conjugacy class
I calculated $$o(cl(13)(24)=3$$
and therfore,
$$o(C_{A4}(13)(24))=4$$
\begin{align}C_{A_4}((13)(24)) &=\{g \in G : g(13)(24)g^{-1}=(13)(24) \}  \\
&= \{g \in G : (g(1) g(3))(g(2)g(4))=(13)(24) \}
\end{align}
The answer is given as $\{(1),(12)(34),(13)(24),(14)(23)\}$
But I  couldn't wrap my head around as to why only these elements are centralizer elements
 A: You’ve correctly shown that if $g \in A_4$ centralizes $(1\ 3)(2\ 4)$, then we have $(g(1)\ g(3))(g(2)\ g(4)) = (1\ 3)(2\ 4)$, which is a great start! You just need to follow the trail a little further.
There are no obvious restrictions on $g(1)$ alone, so let’s see if such a $g$ exists with $g(1) = 1$. Then $(g(1)\ g(3)) = (1\ g(3))$, and we must have $g(3) = 3$. We can try choosing $g(2) = 2$, at which point we must choose $g(4) = 4$ and $g(3) = 3$ reasoning as above, so that $g = 1 \in A_4$. Had we tried to choose $g(2) = 4$ instead (still supposing $g(1) = 1$), we would end up with $g = (2\ 4) \not \in A_4$.
A similar chain of reasoning implies that if we try (from scratch) $g(1) = 2$, then we must have $g = (1\ 2)(3\ 4)$. 
Overall, there are four possibilities for $g(1)$, and each leads to a unique $g \in A_4$ centralizing $(1\ 3)(2\ 4)$.
This is not the most high-brow or elegant approach to computing the centralizer (at least, I assume...), but it hopefully gives a sort of “hands-on” understanding.
