# Every $K[G]$-module is torsionless?

Let $$G$$ be a finite group and $$K$$ a field. Consider the group ring $$R:=K[G]$$. Let $$M$$ be a (left) $$R$$-module. Is it true that then there exists a set $$S$$ and an injective $$R$$-module homomorphism $$M\hookrightarrow \prod_{s\in S}R$$? In other words I am asking if any $$R$$-module can be embedded into a cofree $$R$$-module (relative to $$R$$).

For free $$R$$-modules $$M$$ this is clearly true. I tried to find a group ring $$K[G]$$ which is torsion-free as a left module over itself together with a $$K[G]$$-module with torsion — in order to disprove the statement. This does not work, however, since the group ring has non-trivial zero-divisors for any non-trivial group. Any other ideas?

• Commented Sep 3, 2023 at 11:54
• Well, Maschke only applies if the characteristic of $K$ does not divide the group order. Commented Sep 3, 2023 at 11:59
• I know. But I have no better idea. And for a counterexample, this theorem tells us where to look. Commented Sep 3, 2023 at 12:03
• That's true. Thanks anyway. For $R$ a semisimple ring, any simple $R$-module is a submodule of the regular $R$-module $R$. With Maschke (in case that the characteristic does not divide the group order) we could then find a desired embedding. Commented Sep 3, 2023 at 12:17
• @Margaret For $e \in G$, if $e.m = m$ where $M$ is an $K[G]$-module and $m \in M$, then $M$ is a $K$ vector space and a representation of $G$ and hence can be decomposed into irreducible representations and all these irreducible representations are subspace of $K[G]$ (regular representation). I am guessing, you should be able to embed $M$ into $\oplus_{i = 1}^k K[G]$. So you should look for action of $G$ on $M$ with $e.m \neq m$. Do u have an example where this is true ? Commented Sep 3, 2023 at 23:35

You'd be justified in finding this answer unsatisfactory as it relies heavily on citing results.

A $$R$$-module that embeds into $$R^I$$ for some set $$I$$ is called torsionless (not to be confused with torsionfree). $$R$$ in our case is is an injective cogenerator for the category of $$R$$-modules and thus by proposition 4 in this paper, every $$R$$-module is torsionless.

Thanks to Lukas for pointing out the goal was to show any $$K[G]$$ module was torsionless, not just torsion-free.

Let $$K$$ be a field and $$G$$ a finite group, and set $$A= K[G]$$, the group algebra of $$G$$ over $$K$$.

Definition: An $$A$$-module $$M$$ is torsionless if there is an injective homomorphism $$\phi\colon M \to A^I$$ from $$M$$ to a direct product of copies of $$A$$ (thus the indexing set $$I$$ is not required to be finite).

[I should confess I only learned the above definition today! Note that if $$M$$ is torsionless, then $$M$$ is torsion-free (since $$R^I$$ is evidently torsion-free), but the converse need not be true for a general ring $$A$$. For example it is false for $$A=\mathbb Z$$ because $$\mathbb Q$$ is torsion-free but does not embed in $$\mathbb Z^I$$ for any $$I$$ because it is divisible.]

Now for any (left) $$A$$-module $$M$$ let $$M^{\dagger} = \text{Hom}_A(M,A)$$, where the $$A$$-module structure on $$M^{\dagger}$$ is given by $$g(f)(m) = f(g^{-1}(m)), \quad \forall g \in G, f \in M^{\dagger}, m \in M.$$ The property of being torsionless can be expressed in terms of $$M^{\dagger}$$:

Claim: $$M$$ is torsionless if and only if, for every $$m \in M$$, there is an $$f\in M^{\dagger}$$ such that $$f(m) \neq 0$$, that is, if and only if the natural map from $$d\colon M\to (M^{\dagger})^{\dagger}=:M^{\ddagger}$$, given by $$d(m)(f)= f(m), (\forall m \in M, f \in M^{\dagger})$$ is injective.

Proof of claim: To see this first note that $$\theta\colon M \to R^I$$ is injective if and only if $$\ker(\theta)=\{0\}$$. Thus if $$m \in M\backslash \{0\}$$, we must have $$\theta(m) \neq 0$$. But $$\theta(m) = (\theta_i(m))_{i \in I}$$ where $$\theta_i \colon M \to R$$ is the $$i$$-th component of $$\theta$$ and $$\theta(m)\neq 0$$ if and only if there is some $$i_0 \in I$$ with $$\theta_{i_0}(m) \neq 0$$. Since $$\theta_{i_0}\in M^{\dagger}$$ it follows that $$d(m)\in M^{\ddagger}$$ is nonzero (since $$d(m)(\theta_i)\neq 0$$) and hence $$\ker(d)=\{0\}$$. For the converse, suppose that for each $$m \in M\backslash\{0\}$$ we have $$d(m)\neq 0$$. Then there is some $$\theta_m \in M^{\dagger}$$ with $$d(m)(\theta_m) = \theta_m(m) \neq 0$$. Then if we set $$I =M$$ and define $$\Theta\colon M \to R^M$$ to be the map with components $$\Theta_m=\theta_m$$ it follows immediately that $$\Theta\colon M \to R^M$$ has $$\ker(\Theta)=\{0\}$$, that is, $$\Theta$$ is injective.

Frobenius reciprocity: If $$V$$ is a $$H_1$$-representation and $$W$$ is an $$H_2$$-representation then $$\mathrm{Hom}_{H_1}(V,\text{Ind}_{H_2}^{H_1}(W)) \cong \mathrm{Hom}_{H_2}(\mathrm{Res}^{H_1}_{H_2}(V),W)$$

But now applying this reciprocity we find that $$\begin{split} M^{\dagger} &= \mathrm{Hom}_{K[G]}(M,K[G]) = \mathrm{Hom}_{K[G]}(M,\text{Ind}_{\{e\}}^G(K))\\ &= \mathrm{Hom}_K(\mathrm{Res}^G_{\{e\}}(M),K) = M^{\star}. \end{split}$$ where this identification is compatible with the standard $$A=K[G]$$-module structure on $$M^*$$.

But now it is a reasonably standard fact that the map $$d\colon M\to (M^\vee)^\vee = (M^*)^*$$ is injective (assuming the axiom of choice) for any vector space $$M$$, as this is the standard embedding of a vector space into its double dual, but for completeness, we give a proof:

Proof that $$d\colon M \to M^{\star\star}$$ is injective: The injectivity is equivalent to the assertion that, if $$m \in M \backslash\{0\}$$ then there is some $$f \in M^*$$ with $$f(m)\neq 0$$. This however follows immediately from the fact that any linearly independent subset $$X$$ of $$M$$ can be extended to a basis of $$M$$. But the set of all linearly independent subsets of $$M$$ which contain $$X$$ is partially ordered by inclusion, and since for any totally ordered subset $$\{S_{t}: t \in T\}$$ the union of the subsets $$S_t$$ is its least upper bound, Zorn's Lemma (equivalent to the axiom of choice) ensures a maximal linearly independent set exists. But any such set must span $$M$$ and hence is a basis as required. if $$L\leq M$$ is subspace of $$M$$ then any functional $$f\colon L \to K$$ extends to a functional $$\tilde{f}\colon M \to K$$. Applying this to $$X = \{m\}$$ we see that $$M$$ has a basis $$B$$ containing $$m$$, and then the function $$\delta_m\colon B \to K$$ given by $$\delta_m(b) = 0$$ if $$b\neq m$$ and $$\delta_m(m)=1$$ extends to a linear functional $$f\colon M \to K$$ with $$f(m)=1$$.

Clarification: If $$H_1$$ is a finite group and $$H_2$$ is a subgroup of $$H_1$$, then given an $$H_2$$-representation $$(W,\rho)$$ we define $$\mathrm{Ind}_{H_2}^{H_1}(W) = \{f\colon H_1 \to W: f(h_2h_1)=\rho(h_2)(f(h_1)), \forall h_1\in H_1, h_2 \in H_2\}$$ a left $$A$$-module via $$g(f)(g_1) = f(g_1g)$$. This is the right adjoint to the restriction functor from $$G$$-representations to $$H$$-representations: the counit on $$W$$ is the map of $$H_2$$-representations $$\mathrm{Res}^{H_1}_{H_2}\mathrm{Ind}_{H_2}^{H_1}(W)\to W$$ given by $$[f:G\to W]\mapsto f(e)$$. What I am calling "induction" is sometimes referred to as "coinduction", where induction is taken to be the left adjoint to restriction.

• Nice! This argument is much more reasonable than mine. Commented Sep 4, 2023 at 10:47
• @LukasHeger it was lucky really I think - for the torsionless property (apologies again for conflating a term I didn't know with one I did!) it seemed sensible to understand what an element of $M^{\dagger}$ "looked like" but then the definition is an obviously natural notion of "dual" module, so given I already knew another natural notion the eternal optimist can always hope they are the same, and thankfully once you think of it, it is easy to prove! Commented Sep 4, 2023 at 11:09
• + 1 for your answer. It seems very elegant. If for every $m \in M$ there exists an $f \in M^{\dagger}$ such that $f(m) \neq 0$, how to argue that $M$ is torsion-less i.e., it can be injectively embedded into direct product of copies of $A$ ? Commented Sep 5, 2023 at 0:39
• @Balajisb I've added a proof of the equivalence of the two characterisations of torsionless to my answer. Commented Sep 5, 2023 at 3:57

Yes, if $$G$$ is finite, every $$K[G]$$-module is torsionless.

The left regular $$R$$-module becomes a $$K$$-algebra by pullback along the embedding $$K\hookrightarrow K[G]=R$$. Denote by $$R^*$$ the $$R$$-module $$R^*:=\operatorname{Hom}_K(R,K)$$ induced by the right action of $$R$$ on itself.

Lemma 1. The $$R$$-module $$R^*$$ is injective.
Proof. For any $$R$$-module $$N$$ we have the following natural isomorphim by the tensor-hom adjunction: $$\operatorname{Hom}_R(N,R^*)=\operatorname{Hom}_R(N,\operatorname{Hom}_K(R,K))\cong \operatorname{Hom}_K(R\otimes_R N,K)\cong \operatorname{Hom}_K(N,K).$$ In the last step we use that $$R\otimes_R N\cong N$$ as $$R$$-modules and then apply the forgetful functor from $$R$$-mod to $$K$$-mod. Since $$K$$ as a vector space over itself is injective, the functor $$\operatorname{Hom}_K(-,K)\cong \operatorname{Hom}_R(-,R^*)$$ is exact. Thus, the $$R$$-module $$R^*$$ is injective, QED.

Lemma 2. Any left $$R$$-module $$M$$ embeds as an $$R$$-module into a direct product $$\prod_{M\setminus \{0\}} R^*$$.
Proof. Without loss of generality assume $$M\neq 0$$. In the proof of Lemma 1 we showed that the functors $$\operatorname{Hom}_K(-,K)$$ and $$\operatorname{Hom}_R(-,R^*)$$ are isomorphic. In particular, for any non-zero $$R$$-module $$N$$, there exists a non-zero $$R$$-module homomorphism $$N\rightarrow R^*$$. For any $$m\in M\setminus \{0\}$$, we therefore find a non-zero morphism from the cyclic module $$\langle m \rangle$$ into $$R^*$$. Since $$R^*$$ is injective, this extends to a morphism $$\phi_m\colon M\rightarrow R^*$$ with $$\phi_m(m)\neq 0$$. By the universal property of the product we obtain a morphism $$M\rightarrow \prod_{M\setminus\{0\}}R^*.$$ The map is injective by construction, QED.

Lemma 3. The $$R$$-module $$R^*$$ is isomorphic to $$R$$ as an $$R$$-module over itself.
Proof. The Frobenius map $$R^*\rightarrow R; f\mapsto \sum_{g\in G}f(g)g^{-1}$$ is an isomorphism of $$R$$-modules. Note that in the definition of the Frobenius map we used that the group $$G$$ is finite, QED.

Now, Lemma 2 and 3 together show that any $$R$$-module is torsionless.

• Alternatively for Lemma 3: for $M$ a $K$-vector space and $A$ a set, write $M^A$ for the Cartesian product of $A$ copies of $M$ and $M^{\oplus A}$ for the direct sum of $A$ copies of $M$. Thus $M^{\oplus A} \subseteq M^A$ with equality holding when $A$ is finite. Now $\text{Hom}_K(K^{\oplus A},M) = M^A$, that is, $K^{\oplus A}$ is the free vector space on $A$. Now taking $M=K$ shows $(K^{\oplus A})^\star = K^A$ and so for $A$ finite, $(K^{\oplus A})^\star = K^{\oplus A}$. By functoriality, if $A$ is a finite $G$-set ($G$ any group) it follows $K^{\oplus A}$ is a self-dual $K[G]$-module. Commented Sep 5, 2023 at 15:24

This answer shows that any $$K[G]$$-module is torsion-free, which is, a priori not the question asked, as torsion-free is strictly weaker than torsionless in general.

Let $$A = K[G]$$ be the group algebra of a finite group $$G$$ over a field $$K$$ (of arbitrary characteristic I think?) We claim that any $$A$$-module is torsion-free.

The category of (left say) $$A$$-modules is "locally finite" in that if $$M$$ is an $$A$$-module, then for any $$m\in M$$, $$m \in A.m \leq M$$ is an element of the cyclic submodule it generates, which is finite-dimensional (and a quotient of $$A$$). Thus if $$m \in M$$ is torsion and we let $$N$$ be a maximal proper submodule of $$A.m$$, then $$m+N \in A.m/N$$ is a torsion element in the simply $$A$$-module $$A.m/N$$. Hence to show that any $$A$$-module is torsion-free, it suffices to show this for simple $$A$$-modules. But this follows if we show that $$S$$ is isomorphic to a submodule of $$A$$, which we now show:

If $$S$$ is simple, then so is $$S^{\star} = \mathrm{Hom}_K(S,K)$$. Picking any $$f \in S^{\star}$$, the map $$p\colon A \to S^\star$$ given by $$p(a) = a.f$$ has image $$p(A)\leq S^{\star}$$ a submodule of $$S^{\star}$$ containing $$f$$, and hence by the simplicity of $$S^{\star}$$, the map $$p$$ is surjective. Since $$S^{\star}$$ is finite-dimensional, $$S \cong S^{\star\star}$$, and so we may view $$p^t$$ as an injective map $$p^t\colon S \to A^{\star}$$. We are therefore reduced to showing that $$A \cong A^{\star}$$ as a left module over itself.

For this, note that $$K[G]$$ has $$G$$ as a basis and $$A^\star$$ has the corresponding dual basis $$\{\delta_g: g\in G\}$$. We claim that the map $$g \mapsto \delta_g$$ extends to an isomorphism $$\theta$$ of left $$A$$-modules.

To see this, note that, for any $$g,g_0 \in G$$ we have $$g_0\theta(\sum_{g \in G} c_g.g)(h) = \sum_{g \in G} c_g \delta_g(g_0^{-1}h)= \delta_{g,g_0^{-1}h}.c_g \quad \forall h \in G,$$ (where $$\delta_{a,b}$$ is the Dirac delta function, equal to 1 if $$a=b$$ and $$0$$ otherwise). On the other hand, $$\theta(g_0\sum_{g \in G} c_g.g)(h) = \sum_{g\in G} c_g \delta_{g_0g}(h) = \delta_{g_0g,h}.c_g, \quad \forall h \in G.$$

Since $$g_0g=h$$ if and only if $$g=g_0^{-1}h$$, the result follows. (In fact the above shows that, for any finite $$G$$-set $$X$$, $$K[X]$$ the associated permutation module for $$K[G]$$ is isomorphic to its dual.)

Update: In fact, the argument that shows that a simple module is torsionless readily extends to show that every $$A$$-module is torsionless. First, we need some notation: if $$S$$ is any set, then $$A^{\oplus S}$$ is the free $$A$$-module on $$S$$, which may be thought of as the space of functions $$\{ t\colon S \to A: |S\backslash t^{-1}(0)|<\infty \},$$ that is, functions on $$S$$ taking values in $$R$$ with only finitely many elements of $$S$$ mapped to a nonzero element of $$A$$.

Given an arbitrary $$A$$-module $$M$$, pick $$S$$ a generating set for $$M^\star$$ (so that $$M^\star = \sum_{s\in S} A.s$$). Then there is an obvious map $$\theta_S \colon R^{\oplus S} \to M^{\star}$$ given by $$\theta_S((a_s)_{s \in S}) = \sum_{s \in S} a_s.s$$.

Since $$S$$ is a generating set, $$\theta_S \colon R^{\oplus S}\to M^\star$$ is surjective. Thus taking dual spaces we obtain an injective map $$\theta_S^{\dagger}\colon M^{\star\star} \to R^S$$, where $$R^S$$ denotes the Cartesian product (i.e., all functions $$(a_s)_{s \in S}$$). But the duality map $$d\colon M \to M^{\star\star}$$ is injective, and hence $$\theta_S^{\dagger}\circ d \colon M \to R^S$$ is injective, showing that $$M$$ is torsionless as required.

• Note that torsionless and torsionfree are not the same Commented Sep 4, 2023 at 7:56