I will say how to compute the module of coinvariants---this will give a criterion to say if it vanishes. We work over $\mathbb{Z}$, since this is the most general setting, but the following analysis works over any commutative ring.

Given a $G$-module $M$, pick generators $x_1, \ldots, x_k \in M$. These are elements so that every other $m \in M$ is a $\mathbb{Z}$-linear combination of the elements $gx_i$ where $i \in \{1, \ldots, k\}$ and $g \in G$. This is the same as saying that the $x_i$ generate $M$ as a left $\mathbb{Z}G$-module.

If $M \cong \bigoplus_{i=1}^k \mathbb{Z}Gx_i$, then $M$ is free as a left $\mathbb{Z}G$-module, and so the coinvariant module is just $\bigoplus_{i=1}^k \mathbb{Z}x_i$, the free $\mathbb{Z}$-module on the same generators. If $M$ is not free on the $x_i$, then there must be some relations. In other words, the surjection $\bigoplus_{i=1}^k \mathbb{Z}Gx_i \twoheadrightarrow M$ must have some kernel.

Let's suppose that $G$ is finite, or otherwise ensure that the kernel of this map is finitely generated. Pick generators for the kernel $y_1, \ldots, y_l$. We obtain an exact sequence

$$
\bigoplus_{j=1}^l \mathbb{Z}Gy_j \xrightarrow{\varphi} \bigoplus_{i=1}^k \mathbb{Z}Gx_i \to M \to 0.
$$

Since the map $\varphi$ is a map from one free module to another, it is given by a $l \times k$ matrix with entries in the ring $\mathbb{Z}G$. Let $\epsilon\varphi$ be the same matrix, but where we have replaced every group element $g \in G$ with the integer $1 \in \mathbb{Z}$. So $\epsilon\varphi$ is an integer matrix. Writing $\mathbb{Z}$ for the integers with the trivial right action of $G$, we have
\begin{align}
M_G &\cong \mathbb{Z} \otimes_G M \\
&\cong \mathbb{Z} \otimes_G \mathrm{coker} \, \varphi \\
&\cong \mathrm{coker} \left( \mathbb{Z} \otimes_G \varphi \right) \\
&\cong \mathrm{coker} \left( \epsilon \varphi \right).
\end{align}
In other words, the coinvariants are still generated by the same generating set; however, any group elements appearing in the relations are ignored. For example, any relation $gx_1 = x_2$ holding in $M$ just becomes $x_1=x_2$ in $M_G$. Similarly, the relation $x_1 + gx_2 = hx_3$ becomes $x_1 + x_2 = x_3$, $x_1-gx_1=0$ becomes $x_1-x_1=0$, etc. This gives a systematic way of obtaining a presentation for the $\mathbb{Z}$-module $M_G$ from a presentation for the $\mathbb{Z}G$-module $M$.

submoduleof $M$ on which $G$ acts trivially, and $M_G$ is the largestquotientof $M$ on which $G$ acts trivially. Thus, for example, if $G$ acts as a commuting family of semisimple endomorphisms, then $M^G$ (the common $1$-eigenspace) is isomorphic to $M_G$ (the quotient by the non-$1$ eigenspaces); and, if $G$ acts by unipotent automorphisms, so that $M$ has a composition series with quotients which $G$ acts trivially, then $M^G$ is at the "bottom" of the composition series … (1/2) – LSpice Oct 11 at 2:22isby commuting semisimple automorphisms, so $M^G \cong M_G$.) – LSpice Oct 11 at 2:24