I know that we have a bijection between moduls over a group algebra and (G-linear) representations of a group but I've got problems understanding this. Is there an example (which is not trivial) where group, representations, group algebra and its modules are well-defined? Whenever I read something about this, the "rest of the details are left as an exercise". I'm quite dumb, but I really want to understand this.
Thank you in advance :)
On
Let me explain what that famous bijection really is. It comes from some common adjunction of functors!
Usually one defines a linear representation as a group homomorphism $$\rho\colon G\to \operatorname{GL} (V).$$ Note that $\operatorname{GL} (V)$ is (by definition) the group of units (invertible elements) in the $k$-algebra $\operatorname{End} (V)$ of $k$-linear endomorphisms $V\to V$.
Now comes a little bit of category theory: forming the group algebra $k [G]$ of a group $G$ is a functor $$k [-]\colon \mathit{groups} \to k\mathit{-algebras},$$ and taking the group of units is a functor $$(-)^\times\colon k\mathit{-algebras} \to \mathit{groups}.$$ It is easy to check (!) that these two functors are adjoint, in the sense that there is a natural bijection between morphisms $$\operatorname{Mor}_{k\mathit{-algebras}} (k[G], A) \cong \operatorname{Mor}_{\mathit{groups}} (G, A^\times)$$ (hint: all elements of $G$ are invertible in $k[G]$).
If we specify this to $A = \operatorname{End} (V)$, we get $$\operatorname{Mor}_{k\mathit{-algebras}} (k[G], \operatorname{End} (V)) \cong \operatorname{Mor}_{\mathit{groups}} (G, \operatorname{GL} (V)).$$ Now on the right hand side we immediately recognize representations of $G$, and on the left hand side we have $k[G]$-modules $V$.
Maybe one specific instructive example is the following: $k[G]$ is naturally a module over itself (left or right). This corresponds to what we call the (left or right) regular representation of $G$.
Other examples? Well, start from any representation of $G$, and then it gives you some $k[G]$-module. But since in this correspondence you just automatically extend the action of $G$ on $V$ to the action of $k[G]$, I think such examples are not very interesting for understanding the bijection we are talking about.
Let's get some things straight first. $G$ is a group and $\Bbbk$ a field.
Example 1. The simplest group is $G=\{1\}$, every $\Bbbk$-vector space is a linear $G$-representation, the group algebra is equal to $\Bbbk$ itself and this makes perfect sense, but is also quite boring.
Example 2. The second simplest group is $G=\Bbb Z/2\Bbb Z\cong\{1,\tau\}$ where $\tau^2=1$. Here, a linear $G$-representation is a vector space $V$ on which $\tau$ acts as an involution, i.e. $\tau$ corresponds to an invertible linear map $T\colon V\to V$ with $T\circ T =\operatorname{id}_V$. Indeed, I define the map $T$ as $T(v):=\tau.v$. Hence, we can understand the linear $G$-representations as tuples $(V,T)$ where $V$ is a $\Bbbk$-vector space and $T$ is a linear involution on $V$.
On the other hand, a $R:=\Bbbk[G]\cong\Bbbk[x]/\langle x^2-1\rangle$ (polynomial ring quotient) where $\tau$ corresponds to the image of $x$. Indeed, consider the (surjective) $\Bbbk$-algebra homomorphism $\phi:\Bbbk[x]\to R$ which maps $x\mapsto \tau$. Since $\phi(x^2-1)=\phi(x)^2-1=\tau^2-1=0$, we have $x^2-1\in\ker(\phi)$. On the other hand, $\Bbbk[x]/\ker(\phi)\cong\Bbbk[G]=\Bbbk\oplus\Bbbk\tau$ has dimension $2$ as a vector space. Hence, it is the quotient $\Bbbk[x]$ by a polynomial of degree $2$, hence it follows that $\ker(\phi)=\langle x^2-1\rangle$. Now, what are $R$-modules? Any $R$-module $V$ must be a $\Bbbk$-vector space because we can just restrict scalar multiplication from $R\supseteq\Bbbk$. Then, the only thing that remains to be said is what the scalar multiplication $\tau\cdot v$ means. because once this is defined, we have $(a+b\tau)\cdot v= a\cdot v + b\cdot(\tau\cdot v)$ and scalar multiplication by $a,b\in\Bbbk$ is already defined. Furthermore, we know that the map $T\colon V\to V$ given by $T(v):=\tau\cdot v$ must be an involution, because $$T(T(v))=\tau\cdot\tau\cdot v = \tau^2 \cdot v = 1\cdot v = v.$$ Hence, an $R$-module $V$ is also just a vector space with a fixed linear involution $T:V\to V$.
Example 3. You can work out Example 2 for any cyclic group $G=\langle \tau\rangle$ of order $n$ - You will get tuples $(V,T)$ where $V$ is a $\Bbbk$-vectr space and $T$ is an invertible endomorphism with $T^n=\operatorname{id}_V$. The group algebra is $\Bbbk[G]=\Bbbk[x]/\langle x^n-1\rangle$.
After these examples, it quickly becomes complicated and also quite interesting. It is not an easy task to understand all representations of most "interesting" groups. However, the correspondence between modules over the group algebra and representations of the group is always the same formal correspondence: Every module over $\Bbbk[G]$ is particular a $\Bbbk$-vector space, because $\Bbbk\subseteq\Bbbk[G]$. Then, every group element $g\in G$ corresponds to some linear automorphism of $V$ and these automorphisms have to satisfy the relations of the group. In other words, it all boils down to a group homomorphism $\rho:G\to\operatorname{GL}(V)$. From such a group homomorphism, you can get a $G$-action:
\begin{align*} G\times V &\longrightarrow V\\ (g,v) &\longmapsto \rho(g)(v) \end{align*}
and also a $\Bbbk[G]$-module structure:
\begin{align*} \left(\sum_{g\in G} a_g g\right) \cdot v &= \sum_{g\in G} a_g\cdot \rho(g)(v). \end{align*}
The whole thing works vice-versa in both cases: Given a $G$-action on $V$, just define $\rho(g)$ to be the map $\rho(g)(v)=g.v$ and when $V$ is a $\Bbbk[G]$-module, just define $\rho(g)$ to be the map $\rho(g)(v)=g\cdot v$.
Alright. Hope this helps.