Let \(G\) be a finite abelian group that is not cyclic. Show that \(G\) contains a subgroup isomorphic to \(\mathbb{Z}_p \oplus \mathbb{Z}_p\) for some prime \(p\).
Solution
Write \(G\) as the direct sum of its Sylow subgroups, \[ G=\bigoplus_{p\mid |G|}G_p. \]
1. At least one Sylow subgroup \(G_p\) is noncyclic.
If every \(G_p\) were cyclic, then their orders would be pairwise coprime. The direct product of finite cyclic groups of pairwise coprime orders is cyclic, so \(G\) would be cyclic, contrary to hypothesis.
2. A noncyclic finite abelian \(p\)-group contains a subgroup isomorphic to\(\ZZ_p\oplus\ZZ_p\).
By the structure theorem for finite abelian \(p\)-groups, \[ G_p\cong \ZZ_{p^{a_1}}\oplus\cdots\oplus\ZZ_{p^{a_r}} \] with \(a_i\ge1\). Since \(G_p\) is not cyclic, one has \(r\ge2\).
In the first two summands, the elements \[ x=(p^{a_1-1},0,\ldots,0),\qquad y=(0,p^{a_2-1},0,\ldots,0) \] both have order \(p\). They lie in distinct direct summands, so \[ \langle x,y\rangle=\langle x\rangle\oplus\langle y\rangle \cong\ZZ_p\oplus\ZZ_p. \]
3. Hence \(G\) contains a subgroup isomorphic to \(\ZZ_p\oplus\ZZ_p\) for someprime \(p\).
Choose the prime supplied by 1 and apply 2 inside the subgroup \(G_p\le G\).