## Every finite abelian group is the direct product of its Sylow subgroups

Prove that a finite abelian group is (isomorphic to) the direct product of its Sylow subgroups.

For a finite group $G$, we call the number of distinct primes dividing $|G|$ the breadth of $G$. We proceed by induction on the breadth of $A$.

First, note that a Sylow $p$-subgroup of an abelian group is normal, hence unique.

For the base case, if $A$ is a finite abelian group of breadth 1, we have $|A| = p^\alpha$. Clearly then $A$ is its own (unique) Sylow subgroup, and $A$ is trivially isomorphic to the direct product of itself.

For the inductive step, suppose that every finite abelian group of breadth $k$ is isomorphic to the direct product of its Sylow subgroups. Let $A$ be a finite abelian group of breadth $k+1$, and say $|A| = p_1^{\alpha_1} \cdots p_{k+1}^{\alpha_{k+1}}$. Denote by $P_i$ the (unique) Sylow $p_i$-subgroup of $A$. Note that $P_1\cdots P_k \cap P_{k+1} = 1$ by Lagrange. By the induction hypothesis, $P_1 \cdots P_k \cong P_1 \times \cdots \times P_k$. Thus $P_1 \cdots P_{k+1} = A$, and by the recognition theorem, $A \cong P_1 \cdots P_k \times P_{k+1}$ $\cong P_1 \times \cdots \times P_{k+1}$.

The conclusion holds for finite abelian groups of arbitrary breadth by induction.