Lets say a group $G$ consists of 3 Sylow groups, $H_1,H_2,H_3$. Each of order $p_1,p_2,p_3$, that are prime numbers and different. Since we only have one of each Sylow group for each p, the second sylow theorem gives that each group is normal. And we can show that each group commute:$H_iH_j=H_jH_i$. But then my text also states that this means that the group G is abelian, why is this?
Here is the backgroud for my question, but it is not necessary to read it.: http://s30.postimg.org/l6aevmj4v/syl.png (PS: They interchange 73 and 37 here, but it is not important.)
For any $x \in H_1, y\in H_2$, we have $$ xyx^{-1}y^{-1} \in H_1\cap H_2 $$ since both are normal. However, $|H_1\cap H_2|$ divides both $p_1$ and $p_2$, and so must be 1. Hence, $xy=yx$. Thus the map $$ H_1\times H_2\times H_3 \to G \text{ given by } (x_1,x_2,x_3) \mapsto x_1x_2x_3 $$ is a homomorphism. This gives the isomorphism you need.