Understanding the difference between group representations and modules

Solution 1:

There's essentially no real difference between modules and representations. Think of them as two sides of the same coin.

Given a $\mathbb{K}G$-module $V$, you have a linear action of $G$ on a $\mathbb{K}$-vector space $V$. This in turn gives you a homomorphism from $G$ to $\mathrm{GL}(V)$ (invertible $\mathbb{K}$-linear endomorphisms). Such a homomorphism is a representation. And then this can be turned around. Given a representation, you get an associated module.

Specifically, let $V$ be a $\mathbb{K}G$-module and let $g,h \in G$, $v,w\in V$, and $c\in\mathbb{K}$. Give a name to the map: $v \mapsto g\cdot v$ say: $\varphi(g):V \to V$ (so $\varphi(g)(v)=g \cdot v$). Then $\varphi(g)(v+cw)$ $=g\cdot(v+cw)$ $=g\cdot v+cg\cdot w$ $=\varphi(g)(v)+c\varphi(g)(w)$. Thus $\varphi(g)$ is $\mathbb{K}$-linear. Then because $\varphi(1)$ is the identity map ($1 \cdot v=v$) and $\varphi(g^{-1})(\varphi(g)(v))=g^{-1}\cdot g\cdot v=(g^{-1}g\cdot v=1\cdot v=v$ etc. we get $\varphi(g)$ is an invertible linear map. Therefore: $\varphi:G \to \mathrm{GL}(V)$. Moreover, $\varphi(gh)=\varphi(g)\varphi(h)$ (easy to check) so $\varphi$ is a homomorphism (which we call a representation). Without going into the details, this all reverses.

So $\mathbb{K}G$-modules = representations of $G$ on $\mathbb{K}$-vector spaces.

If you've studied group actions, you've already seen this type of correspondence. Let $G$ act on $X$. Then the map $x \mapsto g \cdot x$ turns out to be a bijection on $X$. Thus if we define $\varphi(g)(x)=g\cdot x$ for all $x\in X$, then $\varphi(g) \in S(X)$ (permutations on $X$). Moreover, $\varphi(gh)=\varphi(g)\varphi(h)$ so $\varphi : G\to S(X)$ is a group homomorphism. We call such things permutation representations. And again this can be reversed. Given a permutation representation: $\varphi:G \to S(X)$, one can define a group action $g \cdot x \equiv \varphi(g)(x)$.

So $G$-action on $X$ = permutation representation of $G$ on $X$.

If you look into other branches of algebra, you'll see this kind of thing over and over again: Lie algebra modules = Lie algebra representations etc.

It's just different points of view. You can either think of "Algebra Thing" acting on "Thing" or a homomorphism from "Algebra Thing" to Maps from "Things to Things".

Solution 2:

A left $R$-module is an abelian (and thus additive) group $V$ on which a ring $R$ acts. The behavior of these actions respects the ring structure, in other words we have linearity in both directions, unity as the identity map and associativity (which is to say the ring's multiplication is composition):

$$(r+s)\cdot v=r\cdot v+s\cdot v, \qquad r\cdot(x+y)=r\cdot x+r\cdot y,$$ $$1_R\cdot v=v, \qquad (rs)\cdot v=r\cdot(s\cdot v), \quad \forall r,s\in R,~~v,x,y\in V.$$

Since each action amounts to an endomorphism of $V$, we may identify $R$-actions with a subring of the ring of endomorphisms, $\mathrm{End}(V)$. Thus, just as $\rho:G\to\mathrm{GL}(V)$, we have $\rho:R\to\mathrm{End}(V)$ to help us "represent" how $R$ acts on $V$. (Notice a vector space $V$ is an example of the additive group.)

This is similar to the usual notion representation, except now we can add actions together instead of simply multiply (compose) them. We are admitting more structure, endomorphisms instead of just automorphisms. There is an intuitive sense in which a vector space over $k$ with representation of $G$ is essentially the same as $V$ interpreted as a $kG$-module: given one, we can easily go to the other.

Given $V$ and $G$, we can extend the actions of $G$ to actions of $kG$ by admitting linearity:

$$\left(\sum_{g\in G} a_g g\right) v:=\sum_{g\in G}a_g(g\cdot v) \qquad \forall v\in V, ~a_g\in k.$$

Conversely, $G$ is a (not necessarily proper) subgroup of $(kG)^\times$ (the group of units, or invertible elements, in the group algebra $kG$), so given $V$ as a $kG$-module, we can make it a vector space over the field $k$ by defining scalar multiplication via $av=(ae)\cdot v$, where $a\in k$ and $e\in G$ is identity, and then it is automatically a representation of $G$ by letting $g\in G$ act via $g\cdot v$ as a $kG$-action.

Solution 3:

The difference between a representation of $G$ and a $K[G]$-module is first of all a notational/vocabulary one. Every representation over $K$ can be realized as a $K[G]$-module, and vice-versa. Regarding the intuition that a representation is a way to represent the group elements as matrices, without losing the group structure: it is not too difficult to see that a $K[G]$-module has a natural structure of $K$-vector space, and so choosing a basis, you can again realize your group elements as matrices.

Now, there are differences in techniques. For example, saying that any complex representations of a finite group splits into a direct sum of irreducible representations is exactly equivalent to the statement that the ring $\mathbb{C}$ is semisimple. At this moment, non-commutative algebra comes into play, especially the Artin-Wedderburn theorem, which is a structure theorem for semisimple ring. Moreover, if you want to use (co)homological tools to analyse your representation, the "natural" language is in terms of modules.

A last comment: representation theory is a vast subject, in which it is very important to be able to grasp different view points. For example, even though module theory is well-suited for cohomological considerations, your group might have extra structure that is not encoded in your group ring (e.g. topological, geometric). On the other hand, sometimes you want to consider a special class of representations, and so you have to change you group ring to something better suited (e.g. a Hecke algebra). In being able to pass from a given point of view to a different one, you give yourself more tools to deal with the problems you encounter.