My question boils down to whether or not you can "induce" without the need for representation theory.
Let $G$ be a group, with $H$ a subgroup. Let's assume [$G:H$] = $n$ is finite to simplify the discussion (not to mention this is in the setting that is most useful to me).
Given a surjective homomorphism $\phi: H \twoheadrightarrow Q,$ where $Q$ is a finite group of size $k$ (again, to stay in a nice setting), I would like to extend the domain of $\phi$ to the group $G$ via the induced representation.
We can embed $Q$ into the matrix group $GL(F^k),$ where $F$ is any field to yield the homomorphism $\pi: H \rightarrow GL(F^k)$. Let's assume $F$ is a finite field or even $\mathbb{F}_2$ to keep $\pi$ mapping to a finite quotient. This yields a representation of $H$ where the image of $\pi$ is isomorphic to $Q.$
From this $H$-module, we can get the following induced $G$-module $$Ind^G_H\pi: G \rightarrow GL((F^k)^n).$$
Is there a way to relate the image of $Ind^G_H\pi$ to $Q$ somehow? I would like to believe the image is isomorphic to $Q^n,$ but this isn't clear to me. Does my choice of a field $F$ (possibly infinite) even matter? Is there a way that I can perform this "induction" without the need to embed $Q$ inside a matrix group (which I needed to do to be able to use representation theory tools). If the latter question is in the affirmative, what is the functor or process that I am applying to this abstract group $Q?$
Thank you very much for your time!