The Question
I use $\mathbb{K}G$-module to denote a $G$-action on a vector space over $\mathbb{K}$ (side question - is this the standard notation?).
A $\mathbb{K}[G]$-module differs in that we allow formal sums of elements of $G$ in forming the group algebra $\mathbb{K}[G]$, making sure that everything is compatible with the abelian group structures.
Suppose I work with just $\mathbb{K}G$-modules - do the standard constructions of induced modules and Frobenius reciprocity still hold?
My guess is yes, because really forming an induced module in the $R$-module setting ($R$ a ring) just seems to mainly depend on defining the rule for how elements of $R$ are compatible with "multiplication" and the tensor product (the rest is just ensuring linearity), similarly for forming the Hom$(V,W)$ objects to prove Frobenius reciprocity. So in the $\mathbb{K}G$-module case, we just don't have to prove the axioms that involve compatibility with a second binary operation on $G$, because there's no sum on $G$. But also I'm very new to this, so am worried I'm missing something.
Background for why I'd ask this
Suppose I form the span over $\mathbb{C}$ of elements of $S_n$ of a specific cycle type:
\begin{align} V_{\lambda} = \text{span}_{\mathbb{C}}\{\sigma \in S_n \,\,|\,\, \text{$\sigma$ has cycle type $\lambda$} \} \end{align}
Because conjugating by elements of $S_n$ preserves cycle type, I can define a $S_n$-action on $V_\lambda$ as
\begin{align} g \cdot \sigma = g \sigma g^{-1}, \quad g \in S_n, \, \sigma \in V_{\lambda} \end{align}
However, it doesn't seem like I can use conjugation to make this space into a module over the group algebra $\mathbb{C}[S_n]$, because $g^{-1}$ doesn't exist in general for $g \in \mathbb{C}[S_n]$. That is the following axiom can't be shown to hold:
\begin{align} (g_1 + g_2)\cdot x \stackrel{?}{=} g_1 \cdot x + g_2 \cdot x \end{align}
I need to perform calculations using induced modules and Frobenius reciprocity with objects like $V_{\lambda}$ and the action by conjugation, but only see these topics discussed in the literature using the full group algebra.
I know it's technically not my stated question, but any comment on how to get around this difficulty would of course be excellent in conjunction with my proposed solution/question.
There is a natural correspondence between $G$-actions over $\mathbb K$ and $\mathbb K[G]$-modules.
If we have a $\mathbb K[G]$-module $V$ then we have a map $f :\mathbb K[G] \to \operatorname{End}(V)$ where $f(x)(v) = x \cdot v$. Given such a map, we can restrict just to the subset $G$ and we obtain a map $\hat f : G \to \operatorname{Aut}(V)$. For an element $g$ of $G$, $f(g)$ of course has to be invertible because $f(g^{-1}) \circ f(g) = \operatorname{id}$.
Conversely, given a map $\hat f : G \to \operatorname{Aut}(V)$, we can extend $\hat f$ linearly to get a map $f :\mathbb K[G] \to \operatorname{End}(V)$. This means that
$$ f(a_1g_1 + \dots + a_r g_r) = a_1 \hat f(g_1) + \dots + a_r \hat f(g_r). $$
In fact, $\mathbb K[G]$ is exactly the $\mathbb K$-algebra you get when you want to extend group homomorphisms $G \to \operatorname{Aut}(V)$ to algebra homomorphisms $\mathbb K[G] \to \operatorname{End}(V)$.