On the automorphisms of a finite abelian group

Question

I read the following statement in an article:

Let $G$ be a finite abelian group. Then $G$ can be decomposed into a direct product of abelian $p$-groups i.e. $G\cong G_1 \times G_2 \times ... \times G_k$. From here it follows that $\operatorname{Aut}(G)\cong \operatorname{Aut}(G_1) \times \operatorname{Aut}(G_2)\times ... \operatorname{Aut}(G_k)$.

There are a couple of things which I do not understand. The fact that $G$ can be decomposed into a direct product of abelian $p$-groups seems all right to me from the classification theorem of finite abelian groups. But the next part regarding the automorphism group seems to assume that the orders of $G_1, G_2,... G_k$ are all relatively prime. I don't understand why this is the case.
EDIT: After the answers I received I think I managed to understand how this works. These $G_i$'s are just the Sylow subgroups as someone pointed out and their orders are obviously relatively prime. I believe that this is used in proving the classification theorem of finite abelian groups, but we may as well just use the part I quoted in order to get a nice form for $\operatorname{Aut}(G)$.

Answer 1

The orders of the $G_i$ are indeed relatively prime. This comes from the theorem, whose proof you may want to go over, and the definition of $p$-group.

One of many sources: http://ramanujan.math.trinity.edu/bmiceli/teach/algebra1S14/AbelianGroups.pdf

Answer 2

The order of $G_i$ are relatively prime, but not the orders of ${\rm Aut}(G_i)$.

For a given cyclic group $G\cong C_n$ we have $$ {\rm Aut}(C_n)\cong (\Bbb Z/n\Bbb Z)^{\times} $$ of order $\phi(n)$, see here:

Order of automorphism group of cyclic group

Hence for the group $C_{p^k}$ the automorphism group has order $\phi(p^k)=p^k-p^{k-1}$. These orders are not coprime in general for different primes $p$ and $q$.

In general, the automorphism group is given as follows:

The automorphism group of a direct product of abelian groups is isomorphic to a matrix group