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.

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