Simplicity of the $A_5$

Question

In the proof given in Abstract Algebra by Dummit and Foote, page 128, it states that $(1~2)(3~4)$ commutes with $(1~3)(2~4)$ but does not commute with any element of odd order in $A_5$. So it follows that the centralizer of $A_5$ is a group of order $4$. I understand the first sentence, but I don't understand why the centralizer cannot be of order $12$ without explicitly checking the conjugations.