Relation between $G$-modules and $KG$-modules

Question

In this question the following definitions of a ring module and a $G$-module where given:

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. $1x=x.$

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.

The OP then writes "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." It was concluded that there is no real difference between a $G$-module and a $KG$-module. My questions:

1. It is easy to see that conditions 3 and 4 in Definition 2 are identical to conditions 3 and 4 of Definition 1, but how does the second condition in Definition 2 follow from Definition 1?
2. Assume we go in the opposite direction, starting with a $G$-module and want to show that we can consider it an $R$-module with $R$ being the group algebra $KG$. We can see that condition 1 in Definition 1 follows from condition 2 in Definition 2 by letting $c=d=1$, but how does condition 2 in Definition 1 follow from Definition 2?

Answer 1

It's important to understand that these are different terms, there is just a natural correspondence between them. Of course condition $2$ in the first definition can't follow directly from the second definition, as we don't even have an addition operation in $G$, but we do in $KG$.

If $V$ is a $G$-module, then we can naturally give it a $KG$-module structure as follows:

$(\sum\limits_{g\in G} k_gg)v=\sum\limits_{g\in G}k_g(gv)$

Where $gv$ is the action of the element $g\in G$ on the vector $v\in V$. Now just check by hand that all the conditions of a $KG$-module are satisfied. In particular, condition $2$ is really straightforward.

Conversely, if $M$ is a $KG$ module then since we have a natural embedding $K\subseteq KG$, we get a $K$-vector space structure on $M$ by restricting the scalars from $KG$ to $K$. And since also $G\subseteq KG$, restricting the scalars from $KG$ to $G$ gives $M$ a structure of a $G$-module. In other words, $gx$ (for $g\in G, x\in M$) is defined as we view $g$ as an element of $KG$. Again, check that all the required conditions are satisfied. For example:

$c(gx)=(cg)x=(gc)x=g(cx), \ \ c\in K, g\in G, x\in M$

This just follows from condition $3$ in definition $1$.

Finally, it can be checked that what we defined here is indeed a bijective correspondence between representations of $G$ and $KG$-modules. It also preserves many properties. (for example, simple modules correspond to irreducible representations)

Answer 2

how does the second condition in Definition 2 follow from Definition 1?

In definition 1, from the first condition you get $r(x+y)=rx+ry$ for every $r,s\in KG$. Restricting to elements $r$ of the form $1_K\cdot g$, and identifying them with their counterparts in $G$, you clearly have the second condition in Definition 2.

Assume we go in the opposite direction, starting with a $$ -module and want to show that we can consider it an $$ -module with $$ being the group algebra $$ [...] how does condition 2 in Definition 1 follow from Definition 2?

It does not (cannot) follow directly from Definition 2 because Definition 2 makes no reference to a second operation between elements. It works because the definition of addition in $KG$ is the linear extension of that defined by the action of $G$ on $V$, and that extension satisfies condition 2 in Definition 1 by virtue of the way addition is defined (in addition to the axioms in Definition 2.)

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A more consistent way of saying it would be:

Definition 1: Let $R$ be a $K$ algebra with multiplicative identity $1$. A left $R$-module is a vector space $V$ with a mapping $B:R\times V\to V$, $(r,x)\mapsto rx$, such that

1. $B$ is bilinear (that is $B(x,-)$ is a linear transformation for any $x$ and $B(-,y)$ is a linear transformation for any $y$)
2. $B(rs,x)=B(r,B(s,x))$ for all $r,s\in R$ and $x\in V$
3. $B(1,x)=x$ for all $x\in V$

Definition 2: Let $G$ be a group with multiplicative identity $\epsilon$. A left $G$-module is a vector space $V$ with a mapping $B:G\times V\to V$, $(r,x)\mapsto rx$, such that

1. $B(g,-)$ is a linear transformation for any $g\in G$
2. $B(gh,x)=B(g,B(h,x))$ for all $g,h\in G$ and $x\in V$
3. $B(\epsilon,x)=x$ for all $x\in V$

As you can see there is very little difference between the two, and then thing that makes it work for $R=KG$ is that Conditions 1+2 of Definition 2 make the linear extension of multiplication to $KG$ work.