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Let $G$ be a group. I am happy to assume niceties such as finite and abelian, but perhaps it is not necessary to answer my question.

Consider the $|G|$-dimensional vector space $V$ (over some nice field $K$, possibly even $\mathbb C$ if need be) whose basis $\{v_g\}_{g \in G}$ is labeled by the group elements of $G$. We know any subgroup $H \subset G$ has a natural action on this vector space: any $h \in H$ defines a linear map $h \cdot v_g = v_{hg}$.

I was now wondering: take a normal subgroup $N \trianglelefteq G$. Is there a natural action of $G/N$ on $V$? I would imagine something in the line of $[g'] \cdot v_g = k \; v_g$ where $k\in K$, but not sure what $k$ should be. Possibly something like ``$k = -1$ if $[g'] = [g]$ and $k=1$ otherwise''?

(For those interested in the more general context I am wondering about: given a (central) extension $1 \to A \to E \to B \to 1$, do $A$ and $B$ have a natural action on the regular representation of $E$?)

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  • $\begingroup$ Does "natural' have a precise meaning here? $\endgroup$
    – Gregory Arone
    Commented Aug 23, 2017 at 9:54
  • $\begingroup$ @GregoryArone Thanks for the question. No, it does not. More precisely, I am just relying on an intuitive notion of 'natural', but this might have a precise version from some category theory perspective, which I simply don't know enough about. $\endgroup$
    – Ruben Verresen
    Commented Aug 23, 2017 at 11:39
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    $\begingroup$ The closest I can think of is something like writing the regular rep as the sum of the $N$-fixed points (which is a submodule for $G$) and the rest (this is canonical since the first part is really the $G/N$-reps, so it is a sum of isotypic components). Then one could just define $G/N$ to act trivially on the rest (I am not sure one can do much better than that). $\endgroup$
    – Tobias Kildetoft
    Commented Aug 23, 2017 at 11:47

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