Is there any difference between a $G$-module and a $KG$-module?

Question

The following is the definition of a left $R$-module:

Definition 1: Let $R$ be a ring with multiplicative identity $1$. A left $R$-module is an abelian group $M$ with a mapping from $R\times M$ to $M$, $(r,x)\mapsto rx$, such that

1. $r(x+y)=rx+ry,$
2. $(r+s)x=rx+sx,$
3. $(rs)x=r(sx),$
4. $1x=x.$

In the book The Symmetric Group the author, B. Sagan, defines a $G$-module in the following way

Definition 2: Let $V$ be a vector space and $G$ be a group with identity $\epsilon$. Then $V$ is a $G$-module if there is a multiplication, $g\mathbf{v}$, of elements of $V$ by elements of $G$ such that

1. $g\mathbf{v}\in V$,
2. $g(c\mathbf{v}+d\mathbf{w})=c(g\mathbf{v})+d(g\mathbf{w})$,
3. $(gh)\mathbf{v}=g(h\mathbf{v})$,
4. $\epsilon\mathbf{v}=\mathbf{v}$,

for all $g,h\in G;\mathbf{v},\mathbf{w}\in V$; and scalars $c,d$ of the field.

Any vector space is an abelian group and the group algebra $KG$ is a ring. We can therefore consider $V$ as a $KG$-module satisfying the conditions in Definition 1, and it is easy to see that it then also satisfies the conditions in definition 2. My question therefore is, is there any difference between a $G$-module and a $KG$-module?

Answer 1

You are correct.

As noted in the other answer, a $G$-module can be seen as a homomorphism from $G$ to the group of units of the ring $\operatorname{End}(V).$

More generally, if $R$ is a unital ring we can define a $G$ map as a map $\alpha:G\to R^{\times},$ a homomorphism from $G$ to the group of units of $R.$

Given $S$ is a unital commutative ring, and a ring homomorphism $\phi:S\to R$ with image in the center of $R$ (the elements of $R$ which commuting with all other elements,) then we get can extend $\alpha$ and $\phi$ to a ring homomorphism $SG\to R.$

In the case $R=\operatorname{End}(V),$ where $V$ is a vector space over the field $K,$ then we can choose $\phi:K\to R$ as $\phi(\lambda)=\lambda I$ and we get $KG\to \operatorname{End}(V),$ which we can see is equivalent to making $V$ a $KG$-module.

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We can define a $G$-action in any category, as a group homomorphism $G\to \operatorname{Aut}(M)\subseteq \operatorname{End}(M).$

In an abelian category, where $\operatorname{End}(M)$ is always a ring, we can define a ring action as a homomorphism from a ring $A$ to the ring of endomorphisms on some category object. Then we can do the same trick as above. But what the center of that ring is depends on the category (when there is such an $S.$)

In the category of abelian groups, there is a basic homomorphism from $\mathbb Z.$

In the category of $R$-modules, where $R$ is a commutative unital ring, the most general case will be a homomorphism from $R.$

Given a $G$-action in an abelian category, we find it is equivalent to an $SG$-action on the category, where $S$ is the most general center of the endomorphism rings in the category.

In the category of vector spaces, $S$ will be $K.$

Answer 2

The axioms in definition 2 imply that each $g$ in $G$ acts like a unit of $End(V_K)$ on $V_K$, and it does so such that the assignment is a homomorphism of groups $G\to GL(V_K)$. By the universal property of $K[G]$, this gives rise to a unique ring homomorphism $K[G]\to End(V_K)$. That is another way of specifying a left $K[G]$ module structure on $V$.

Of course, restriction of any such ring homomorphism $K[G]\to End(V_K)$ to elements of $G$ gives you a map $G\to GL(V_K)$ that satisfies definition 2. If you stare at it a bit, you'll see these two transformations constitute an isomorphism of categories between the categories of $G$-modules and $K[G]$ modules.

If categories aren't your bag, this can be taken to mean: yes, they amount to the same thing!

Answer 3

It's not really important (so this is really a long comment and not an answer) but it is a little eccentric for Sagan to call something a "G-module", because one normally reserves that term for an abelian group $M$ with a compatible action of a ring $R$ (so something with an addition and a (possibly non-commutative) multiplication. If you just have a group $G$, people normally speak of a $G$-representation (or even sometimes "linear $G$-representation", since some texts discuss "permutation representations" of a group, though "group actions" is now the more standard term.

Indeed, in something of the same spirit as Thomas Andrews' answer, if you let $\mathcal C$ be your favourite category, then a ``$\mathcal C$-representation'' of a group is a homomorphism into $\text{Aut}(M)$ for $M$ an object of $\mathcal C$. Almost always, when people say "G-representation", however, they mean "linear representation", and so are taking $\mathcal C$ to be $\textbf{Vect}$ the category of vector spaces. Nevertheless, people speak of "representation theory" as being (at least in part) a way of studying abstract objects, such as groups, by seeing how they can act on "more concrete" mathematical objects. Since "concrete" is a relative term, it seems reasonable enough to consider (someone's) "concrete objects" to be some category, such as $\textbf{Set}$ or $\textbf{Vect}$.

In this generality, depending on $\mathcal C$ there may or may not be an algebra corresponding to $KG$: there is no (or at least no obvious) such algebra when $\mathcal C$ is $\textbf{Set}$ the category of sets, but once $\mathcal C$ is additive one has at least the group ring $\mathbb ZG$.

One final minor point about the terminology: the data of a linear representation of a group is, as the other answers have already clarified, equivalent to the data of a $KG$-module, apart again from the fact that to say something is a $G$-representation does not tell you the field the vector space is over, whereas if you say $KG$-module, this corresponds to a $G$-representation on a $K$-vector space. Often the field is fixed (e.g. the complex numbers) but there are contexts where that is not the case. To give an example of such a context and how the terminology can be used differently, in "modular representation theory" of a finite group $G$ one seeks to understand how the representations of $G$ over fields of characteristic dividing the order of $G$ relate to those over fields of characteristic zero. Using module terminology I would need to say something like "how $kG$-modules and $KG$-modules are related, where $k$ has characteristic dividing $|G|$ and $K$ has characteristic zero."

[Just so that this post has some content that isn't linguistic, let $G$ be a finite group and let $(V,\sigma)$ be a representation of $G$ over a field of characteristic zero, which for simplicity I'll take to be $\mathbb Q$. Then you can chose a basis $\{e_1,\ldots,e_n\}$ for $V$ with respect to which the matrices for the action of the elements of $G$ are all integer-valued, so that $G$ preserves the free abelian group $L=\bigoplus_{i=1}^n \mathbb Z.e_i$, and hence one obtains, by reducing mod $p$, a $G$-representation on $L_p=\bigoplus_{i=1}^n \mathbb F_p.e_i$. Now while the isomorphism class of$\mathbb F_p$-representation you get this way will depend on your choice of basis, but the irreducible $\mathbb F_p$ representations which occur in $L_p$ are independent of the choice of basis, hence the $\mathbb Q$-representation $V$ of $G$ has a well-defined set of irreducible $\mathbb F_p$-representations associated to it, and indeed one can extend this construction to handle any complex representation $(W,\tau)$ of $G$ b replacing $\mathbb Q$ with a number field.]