Suppose that \(G = \sum G_\alpha\). Show this sum is direct if and only if the equation
\[ x_{\alpha_1} + \dots + x_{\alpha_n} = 0 \]
implies that each \(x_{\alpha_i}\) equals 0. (Here \(x_{\alpha_i} \in G_{\alpha_i}\) and the indices \(\alpha_i\) are distinct.)
Solution
Assume first that \[ G=\bigoplus_\alpha G_\alpha. \] By definition, every element of \(G\) has a unique expression as a finite sum of elements from distinct \(G_\alpha\)’s. If \[ x_{\alpha_1}+\cdots+x_{\alpha_n}=0 \] with distinct indices, then this is one expression for \(0\), while \[ 0+\cdots+0 \] is another. Uniqueness forces \(x_{\alpha_i}=0\) for every \(i\).
Conversely, assume the stated vanishing condition. Since \(G=\sum G_\alpha\), every element has at least one finite expression. Suppose \[ x_{\alpha_1}+\cdots+x_{\alpha_r} =y_{\beta_1}+\cdots+y_{\beta_s}. \] Move all terms to one side and combine terms belonging to the same subgroup. This yields \[ z_{\gamma_1}+\cdots+z_{\gamma_t}=0 \] with distinct \(\gamma_i\) and \(z_{\gamma_i}\in G_{\gamma_i}\). By hypothesis every \(z_{\gamma_i}=0\), so the two original expressions agree coordinate by coordinate. Thus finite decompositions are unique, and the sum is direct.