Exercise 2, § 7

Let $(X_\alpha, f_{\alpha\beta})$ be an inverse system of non-empty sets such that $\varprojlim X_\alpha = \emptyset$. Let $G_\alpha$ denote the free $\mathbf{Z}$-module of formal linear combinations of elements of $X_\alpha$, with coefficients in $\mathbf{Z}$, and let $g_{\alpha\beta}$ denote the homomorphism $G_\beta \to G_\alpha$ which agrees with $f_{\alpha\beta}$ on $X_\beta$. Show that the group $G = \varprojlim G_\alpha$ consists of the identity element alone. [Argue by contradiction: if $z = (z_\alpha)$ is an element of $\varprojlim G_\alpha$, consider, for each $\alpha$, the finite set $F_\alpha$ of elements of $X_\alpha$ whose coefficient in $z_\alpha$ is not zero, and observe that $f_{\alpha\beta}(F_\beta) = F_\alpha$.] Hence construct an example of an inverse system $(G_\alpha, g_{\alpha\beta})$ of groups such that the $g_{\alpha\beta}$ are surjective, the $G_\alpha$ are infinite and $\varprojlim G_\alpha$ consists only of the identity element.