If the sum is direct, given the uniqueness of representation and that $0=0+\cdots+0$ regardless of which groups are considered in the sum, we have $x_{\alpha_{i}}=0$ for all $i=1,\ldots,n$ . Vice versa, suppose that for some $x\in G$ , $x=x_{\alpha_{1}}+\cdots+x_{\alpha_{n}}=y_{\beta_{1}}+\cdots+y_{\beta_{m}}$ . Then, $x_{\alpha_{1}}+\cdots+x_{\alpha_{n}}+-y_{\beta_{1}}+\cdots+-y_{\beta_{m}}=0$ . After regrouping and combining elements of same groups, we obtain $\sum_{k}(x_{\alpha_{i_{k}}}+-y_{\beta_{j_{k}}})+\sum_{s}x_{\alpha_{i'_{s}}}+\sum_{t}(-y_{\beta_{j'_{t}}})=0$ , where for each $k$ we have a pair of elements from the same group, i.e. $\alpha_{i_{k}}=\beta_{j_{k}}$ , and the remaining elements have unique indices. This equation implies that for all $k$ , $x_{\alpha_{i_{k}}}+-y_{\beta_{j_{k}}}=0$ , and for all $s$ and $t$ , $x_{\alpha_{i'_{s}}}=0$ and $y_{\beta_{j'_{t}}}=0$ . In other words, if for some $i$ and $j$ , $\alpha_{i}=\beta_{j}$ , then $x_{\alpha_{i}}=y_{\beta_{j}}$ , and for all other unique indices $\alpha_{i}$ and $\beta_{j}$ , we have $x_{\alpha_{i}}=0$ and $y_{\beta_{j}}=0$ . Hence, the sum is direct.