I want to prove the following:
If every Sylow subgroup of $G$ is a normal subgroup, then $G$ is isomorphic to the product of its Sylow subgroups.
So far, I have gotten to realize that for every prime factor $p_i$ of $|G|$, there is exactly one Sylow subgroup and that these form a partition of $G$. However, I am having trouble finding the isomorphism between $G$ and the product which I will call $G_1 \times \ldots \times G_r$. My idea was just to use the mapping $(g_1, \ldots, g_r) \mapsto g_1\ldots g_r$, which is a bijection, but I don't see whether it is a homomorphism. In general, not all elements of $G_i$ commute with elements of $G_j$, so one cannot simply change the order of the elements in the product.