Show that if there were one Sylow 3-subgroup, then A4 would be Abelian. Show that if there were one Sylow $3$-subgroup, then $A_4$ would be Abelian. 
I know the order of $A_4$ is $12$ and therefore $A_4$ can have $1$ or $4$ Sylow $3$-subgroup. However, I don't know how to prove the above statement. 

Turning the comment into an answer so it doesn't remain unanswered. Knowing that $A_4$ is generated by the $3$-cycles you can arrive at a contradiction. If $n_3 = 1$ it means every $3$ cycle is a power of $(1,2,3)$ for example, as every $3$-cycle has order exactly $3$. Combining these two, you got that $A_4$ is generated by $(1,2,3)$. This means it is abelian.
Actually you can show much easily that there can not be one Sylow 3-subgroup if you just realized that the $3$-cycles $(1,2,3)$ and $(1,2,4)$ are not a power of one another. This means that $n_3 = 4$. It is a very interesting and instructive exercise the classification of groups of order $12$ if you haven't thought of it. Keith Conrad has an amazing article summarizing it, GROUPS OF ORDER 12.
