Abelian group as direct product of its p-Sylow subgroups. 
I have read the Sylow 3 theorems, but I don't think I fully understand what they mean. Could someone help clarify them for me. Especially how they might apply to this question. Thanks. 
 A: If the number $|G|$ has the arithmetic factorisation $$|G|=p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r},$$ 
then by Sylow-1 it has, for each $i$, a subgroup $H_i<G$ with $p_i^{n_i}$ elements. 
With this subgroups you can prove that 
$$G=H_1H_2\cdots H_r.$$
A: So we want to show that any ﬁnite abelian group is isomorphic to the direct product of its Sylow subgroups.
Let $G$ be a finite abelian group with $|G| = p_1^{i_1} ... p_m^{i_m}$ where $p_i$ with $1 \le i \le m$ are distinct primes. Note that by Sylow theorem, Sylow subgroups exist, and as all of their orders are in power of different primes, by Lagrange theorem none of those Sylow subgroups could be the subgroup of the others. So the order of any group could be written as the preceding statement.
We proceed by induction by assuming any abelian group with fewer distinct primes than $m$ is isomorphic to its Sylow subgroups.
Let $H$ be the subgroup of $G$ generated by Sylow $p_j$-subgroups for $1 \le j \le m-1$, i.e. $|H| = p_1^{j_1} ... p_{m-1}^{j_{m-1}}$ where $p_j$ with $1 \le j \le m-1$ are distinct primes.
Let $K$ be the Sylow $p_m$-subgroup of $G$. |H| does not divide $p_m^{i_m}$ because $p_m$ is a distinct prime. By Lagrange theorem, $H \cap K = 1$ because neither $H$ nor $K$ could be the subgroup of the other. As $G$ is an abelian group, both $H$ and $K$ are normal subgroups of $G$, so by Recognition theorem we get that:

$G \cong H \times K$.

Continue the arguments with $H$ replacing the role of $G$ above, we are done when we reach $G = p_1^{i_1}$ which is trivially true.
