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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.