I want to prove this statement with elementary considerations about group cohomology (started studying it today)
If $G$ is a group such that its cohomological dimension is $0$, then $G$ is the trivial group.
Recall that $\text{cd}G := \text{proj. dim}_{\mathbb{Z}G}\mathbb{Z}$. Just from the definition it's clear that in our setting $\mathbb{Z}$ is a projective $\mathbb{Z} G$-module, and therefore the canonical projection $$ \mathbb{Z}G\xrightarrow{\epsilon} \mathbb{Z}$$ splits. This means that $\mathbb{Z}G \simeq \ker(\epsilon)\oplus j(\mathbb{Z})$ where $j$ is the splitting. But I'm not able to find any contradiction yet. So I tried using this splitting in the following way: by the triviality of the action of $G$ on $\mathbb{Z}$, we would have $$j(z)=j(g.z)=g.j(z)$$ for any $g\in G$. and then this leads me to feel that I should find some contradiction here, but I'm unable to do that.
Can someone point me in the right direction?
Suppose that the augmentation $\epsilon : \mathbb{Z}[G] \to \mathbb{Z}$, $g \mapsto 1$ splits. This means there's a map of $\mathbb{Z}[G]$-modules $j : \mathbb{Z} \to \mathbb{Z}[G]$ such that $\epsilon(j(n)) = n$ for all $n \in \mathbb{Z}$. Let $z = j(1) \in \mathbb{Z}[G]$. Then $j(g \cdot 1) = j(1) = z = g \cdot j(1) = g \cdot z$ for all $g$, hence $z \in \mathbb{Z}[G]^G$ is invariant.
This is only possible if $G$ is finite, and $z$ has to be a scalar multiple of $\xi = \sum_{g \in G} g \in \mathbb{Z}[G]$ (easy to check), say $z = n \xi$ for some $n \in \mathbb{Z}$. Clearly $\epsilon(z) = n \cdot \#G$, but at the same time $z = j(1) \implies \epsilon(z) = 1$ (since $j$ is a splitting). Hence $n \cdot \#G = 1$, which can only happen if $\# G = 1$.