# What is the difference between an $FG$-module and a group algebra?

I am beginning the study of representation theory, and am having trouble understanding the difference between an $$FG$$-module and a group algebra of $$G$$. This seems to be the only other question on this topic, but I don't think the answer is relevant to the definitions I have.

In the course I am taking, we have the following definition for an $$FG$$-module:

Let $$V$$ be a vector space over $$F$$ and let $$G$$ be a group. Then $$V$$ is an $$FG$$-module if a multiplication $$vg (v \in V, g \in G)$$ is defined, satisfying the following conditions for all $$u, v \in V, \lambda \in F$$ and $$g, h \in G$$ :

1. $$v g \in V$$;
2. $$v(g h)=(v g) h$$
3. $$v 1=v$$;
4. $$(\lambda v) g=\lambda(v g)$$;
5. $$(u+v) g=u g+v g$$.

Now, the definition of group algebra is given:

The vector space $$F G$$, with multiplication defined by $$\left(\sum_{g \in G} \lambda_g g\right)\left(\sum_{h \in G} \mu_h h\right) = \sum_{g, h \in G} \lambda_g \mu_h(g h)$$ $$\left(\lambda_g, \mu_h \in F\right)$$, is called the group algebra of $$G$$ over $$F$$.

In my mind these are essentially the same thing. Both are a "precise combination" of a field and the elements of a group such that multiplication is defined with closure, associativity, identity, scalars, and distributivity. I don't really see what makes them distinct.

I suppose there is some sort of "hierarchy" that goes $$\text{Group Algebra}\subset \text{FG-module}\subset\text{Vector space}?$$ Is my understanding correct? I feel pretty flaky about it.

• One is a ring and the other is a module: they are completely different things! Your question is not very different to "what is the difference between a car and a tv?" (Do not call other people's work lackluster. You can ask your question without doing that...) Feb 15 at 1:12
• If you know what a module over a ring is, and you know what an algebra over a field is, you must see that the two definitions are different. For example, you can take two elements in an algebra and multiply them to obtain a third one, but if you have a module then you cannot multiply two of its elements. Feb 15 at 1:16
• A module is like a vector space, just defined over a ring instead of a field. So first we define the group algebra $FG$ (if it's an algebra then it is in particular a ring), and then we consider modules over that ring. Now, your definition of an $FG$-module looks a bit different (more confusing), but it's equivalent to what I wrote. By your definition, the "scalar multiplication" is only defined by scalars from the group $G$. But it can be checked that this can be naturally extended to scalar multiplication with scalars from $FG$, and so we get a module over $FG$. (note that $G\subseteq FG$)
– Mark
Feb 15 at 1:24
• This is one of those situations where people will be confused at your confusion: an $\mathbb k$-algebra is something that behaves like the space of endomorphisms of a $\mathbb k$-vector space: $\operatorname{End}(V)$ is a vector space along with a product (composition of functions) which is $\mathbb k$-bilinear. Because algebras are modelled in this way, you can consider algebra maps $A\longrightarrow \operatorname{Env}(W)$ where $W$ is some vector space, and this $W$ is an $A$-module (i.e., a way of seeing $A$ as a collection of operators on $W$). They are very different things!
– Pedro
Feb 15 at 14:52
• If someone could summarize the ideas in these comments and write it in an answer, I'd be happy to accept it! Feb 15 at 15:55

A module is like a vector space, just defined over a ring instead of a field. So first we define the group algebra $$FG$$ (if it's an algebra then it is in particular a ring), and then we consider modules over that ring. Now, your definition of an $$FG$$-module looks a bit different (more confusing), but it's equivalent to what I wrote. By your definition, the "scalar multiplication" is only defined by scalars from the group $$G$$. But it can be checked that this can be naturally extended to scalar multiplication with scalars from $$FG$$, and so we get a module over $$FG$$. (note that $$G\subseteq FG$$)
This is one of those situations where people will be confused at your confusion: an $$\mathbb k$$-algebra is something that behaves like the space of endomorphisms of a $$\mathbb k$$-vector space: $$\operatorname{End}(V)$$ is a vector space along with a product (composition of functions) which is $$\mathbb k$$-bilinear. Because algebras are modelled in this way, you can consider algebra maps $$A\longrightarrow \operatorname{Env}(W)$$ where $$W$$ is some vector space, and this $$W$$ is an $$A$$-module (i.e., a way of seeing $$A$$ as a collection of operators on $$W$$). They are very different things!