Let $G$ be a group, $M$ a $G$-module, then the group of coinvariants is the module $M_G := M/I_GM$, where $I_G$ is the kernel of the augmentation map $\epsilon : \mathbb{Z}G\rightarrow \mathbb{Z}$.

The dual notion of $G$-invariants is very simple, and for most $G$-modules, it's easy to tell whether or not for example $M^G = 0$.

On the other hand, I don't have a good intuition for $M_G$.

I'd appreciate a list of results that say something about this module of coinvariants, and criteria which might help recognize when it vanishes. Even facts in the case where $G$ is a finite, abelian, or even finite abelian group would be welcome.

References would also be helpful.

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    $\begingroup$ It may help to think in terms of the characterisation, rather than the definition: $M^G$ is the largest submodule of $M$ on which $G$ acts trivially, and $M_G$ is the largest quotient of $M$ 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) $\endgroup$
    – LSpice
    Oct 11, 2018 at 2:22
  • 4
    $\begingroup$ … and $M_G$ is at the "top", in the sense that it is the quotient of $M$ by the next-smallest term. (2/2) $\endgroup$
    – LSpice
    Oct 11, 2018 at 2:23
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    $\begingroup$ (In particular, if $G$ is finite Abelian and the module is over a field (I can't tell whether that's what you have in mind), then the action of $G$ is by commuting semisimple automorphisms, so $M^G \cong M_G$.) $\endgroup$
    – LSpice
    Oct 11, 2018 at 2:24
  • 2
    $\begingroup$ What about the modular case? Surely cyclic group of order $p$ can act by unipotent upper triangular $2\times 2$ matrices over $\mathbb{F}_p$, and tautologically this is an action by commuting non-semisimple automorphisms. $\endgroup$ Oct 11, 2018 at 3:11
  • $\begingroup$ @LSpice Thanks for your comments, though I was definitely thinking more about the non-semisimple case (ie, $M$ is a more or less arbitrary $\mathbb{Z}G$-module) $\endgroup$ Oct 11, 2018 at 4:08

1 Answer 1


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 $M$.


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