Give an example of a free abelian group \(G\) of rank \(n\) having a subgroup \(H\) of rank \(n\) for which \(H \neq G\).
Solution
Goal: Provide an explicit example of a free abelian group \(G\) of rank \(n \ge 1\) containing a proper subgroup \(H \subsetneq G\) of full rank \(n\), and prove these properties.
1. Definition of \(G\) and \(H\):Let \(G = \mathbb{Z}^n = \bigoplus_{i=1}^n \mathbb{Z} \mathbf{e}_i\), where \(\{\mathbf{e}_1, \dots, \mathbf{e}_n\}\) is the standard canonical \(\mathbb{Z}\)-basis.Define the subgroup:\[H = 2\mathbb{Z} \times \mathbb{Z} \times \dots \times \mathbb{Z} = \operatorname{span}_{\mathbb{Z}}\{2\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n\}.\]
2. Verification that \(G\) is free abelian of rank \(n\):Proof: The set \(\{\mathbf{e}_1, \dots, \mathbf{e}_n\}\) freely generates \(G\) over \(\mathbb{Z}\), so \(\operatorname{rank}(G) = n\).
3. Verification that \(H\) is free abelian of rank \(n\):Proof:
3.1. The generating set \(\mathcal{B} = \{2\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n\}\) generates \(H\) by definition.
3.2. If \(c_1 (2\mathbf{e}_1) + \sum_{i=2}^n c_i \mathbf{e}_i = \mathbf{0}\) for \(c_i \in \mathbb{Z}\), then \((2c_1, c_2, \dots, c_n) = (0, 0, \dots, 0)\), which implies \(2c_1 = 0 \implies c_1 = 0\) and \(c_2 = \dots = c_n = 0\).
3.3. Thus \(\mathcal{B}\) is \(\mathbb{Z}\)-linearly independent, making \(H\) a free abelian group with basis of size \(n\), so \(\operatorname{rank}(H) = n\).
4. Verification that \(H\) is a proper subgroup (\(H \neq G\)):Proof:
4.1. The element \(\mathbf{e}_1 = (1, 0, \dots, 0) \in G\).
4.2. Every element \((x_1, \dots, x_n) \in H\) has \(x_1 \in 2\mathbb{Z}\).
4.3. Since \(1 \notin 2\mathbb{Z}\), \(\mathbf{e}_1 \notin H\), proving \(H \subsetneq G\).
4.4. In particular, the quotient group is \(G/H \cong \mathbb{Z}/2\mathbb{Z}\) of order 2.
5. Conclusion:\(G = \mathbb{Z}^n\) and \(H = 2\mathbb{Z} \times \mathbb{Z}^{n-1}\) provide the required example of a proper full-rank subgroup. Q.E.D.