I want to analyze the Sylow structure of $A_6$ (the alternating group of order 6),
which has order $|A_6|=\frac{|S_6|}{2}=\frac{6!}{2}=360=2^3 3^2 5$.
Starting with $p=5$.
We have that there are $\binom 65(5-1)!=144$ elements of order 5 in $A_6$, but given that 5 is prime, we have that the elements of order 5, are the 5-cycles, the group generated by one of this 5-cycles has $\varphi(5)=5-1=4$ generators, so we have that $|{\rm Syl}_5(A_6)|=\frac{144}{4}=36$.
If $P\in{\rm Syl}_5(A_6)$, we have that $36=|{\rm Syl}_5(A_6)|=|A_6:N_{A_6}(P)|=\frac{360}{|N_{A_6}(P)|}$, so in consequence $|N_{A_6}(P)|=10$ and I want to prove that $N_{A_6}(P)\cong D_{10}$.
By classification of groups of order 10, we have only two options for $N_{A_6}(P)$, which are $D_{10}$ (Symmetry group of a pentagon) or $C_{10}$ (Cyclic group of order 10). The difference between those is that the latter is abelian. I don't know if part of the argument I need has to do with the fact that $A_6$ is not simple.