Show that if \(G_1\) is a subgroup of \(G\), there may be no subgroup \(G_2\) of \(G\) such that \(G = G_1 \oplus G_2\). [Hint: Set \(G = \mathbb{Z}\) and \(G_1 = 2\mathbb{Z}\).]
Solution
Take \[ G=\mathbb Z,\qquad G_1=2\mathbb Z. \] Suppose there were a subgroup \(G_2\le\mathbb Z\) with \[ \mathbb Z=2\mathbb Z\oplus G_2. \] Every subgroup of \(\mathbb Z\) has the form \(m\mathbb Z\) for some \(m\ge0\). If \(m=0\), then \(2\mathbb Z+G_2=2\mathbb Z\ne\mathbb Z\). If \(m>0\), then \[ 2m\in2\mathbb Z\cap m\mathbb Z \] is nonzero, contradicting directness. Therefore \(2\mathbb Z\) is a subgroup of \(\mathbb Z\) that has no complementary subgroup.