Let \(\Gamma\) be a group. We claim that \(\mathbb{Z}\) is a projective \(\mathbb{Z}\Gamma\)-module if and only if \(\Gamma\) is trivial.
One direction is obvious: if \(\Gamma\) is trivial, then \(\mathbb{Z}\Gamma \cong \mathbb{Z}\), so \(\mathbb{Z}\) is not just a projective \(\mathbb{Z}\Gamma\)-module, but a free one.
For the other direction, suppose that \(\mathbb{Z}\) is a projective \(\mathbb{Z}\Gamma\)-module. We have a short exact sequence
\(\displaystyle 0 \to K \to \mathbb{Z}\Gamma \overset{\varepsilon}{\to} \mathbb{Z} \to 0,\)
where \(\varepsilon\) is the map defined by \(\varepsilon \left( \sum a_i g_i \right) = \sum a_i\). Since \(\mathbb{Z}\) is projective, this sequence splits, i.e., there is a map \(j: \mathbb{Z} \to \mathbb{Z}\Gamma\) such that \(\varepsilon j = id_{\mathbb{Z}}\). Such a map is determined by its action on \(1 \in \mathbb{Z}\). Say \(j(1) = x = \sum_{g \in G} a_g g \in \mathbb{Z}\Gamma\). Note that \(x\) is nontrivial since \(j \) is a splitting. Since the action of \(G\) on \(\mathbb{Z}\) is trivial, we have
\(\displaystyle \sum_{g \in G} a_g g = x = hx = \sum_{g \in G} a_g hg = \sum_{g \in G} a_{h^{-1}g}g\)
for all \(h \in G\). This implies that \(a_g = a_{h^{-1}g}\). Since the action of \(G\) on \(G\) is transitive, we get that \(a_g = a_{g’}\) for all \(g, g’ \in G\). Since \(x \neq 0\), it follows that \(G\) is finite (since otherwise the sum \(x = \sum_{g \in G} a_g g\) is an infinite sum of nonzero terms). Moreover, we have \(1 = \varepsilon(x) = \sum_{g \in G} a_g = \vert G \vert \cdot a_{g_1}\). Since, \(a_{g_1} \in \mathbb{Z}\), it must be that \(\vert G \vert = 1\), and so \(G\) is trivial.