# When is $latex \mathbb{Z}$ a projective $latex \mathbb{Z}\Gamma$-module?

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.