Centraliser of (12)(34) in A5

Question

In the proof that A5 is simple, Dummit and Foote p.128 claim that (1) it is easy to see that (12)(34) commutes with (13)(24) but does not commute with any element of odd order in A5. And further (2) that it follows that |CA5((12)(34)|=4.
Part (1) has been addressed already on StackExchange, for (2) a clarification was asked, but that question was never answered. I repeat the point raised by Ecotistician: I don't see how (2) follows without checking which 2-cycles do commute with (12)(34).

Answer 1

The centraliser is of order $4$, namely $$ \{e, (12)(34), (13)(24), (14)(23)\}. $$ Another way to see it, is to compute the centraliser of $\tau=(12)(34)$ first in $S_5$, which is the subgroup generated by $(1324)$ and $(13)(24)$, and then intersect it with $A_5$ to obtain the desired centraliser.