$A_4$ is not the direct product of its sylow subgroups? This is an exercise in Hungerford. I've tried proving this as follows (forgetting for the moment that in $A_4$, $n_2=1$ and $n_3=4$):
By Sylow III, $n_2=1$ or $3$.
If $n_2=3$, the number of elements of order $2$ in $A_4$ is $n_2 (2-1)=3(2-1)=3.$ Then the number of elements of order not $3$ is $12-3=9$. Let $P$ be a Sylow $3$-subgroup. $|P|=3$ Number of elements we have to fill up this subgroup with exceeds the number of elements of this subgroup, hence the Sylow $3$-subgroup is not normal and $A_4$ not the direct product of its sylow subgroups.
Suppose next $n_2=1$. Again by Sylow III, $n_3=1$ or $4$. If $n_3=1$, then there are 6 elements in $A_4$, which is a contradiction. Hence $n_3 = 4$ then the Sylow $3$-subgroup is not normal and again $A_4$ is not the direct product of its Sylow subgroups.
Any corrections and/or alternative proofs are welcome.

There are many incorrect statements in your proof. 
Anyway, this is beside the point as you can prove this statement quickly without counting Sylow subgroups, or doing any other hard analysis:
Assume $A_4$ is a direct product of its Sylow subgroups. The Sylow 2-subgroups of $A_4$ have order 4, hence are abelian. Similarly, the Sylow 3-subgroups are abelian. If $A_4$ is a direct sum of Sylow subgroups, then $A_4$ must be abelian, a contradiction.
OR
If $A_4$ is a direct product of its Sylow subgroups, then each Sylow subgroup is normal (i.e. $n_p=1$ for $p=2,3$). However, $(123)$ and $(124)$ generate distinct Sylow 3-subgroups. Hence, $A_4$ is not a direct sum of its Sylow subgroups.
