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This question didn't receive an answer on MathSE, so I'm asking it here.

Let $G$ be a finite group and let $k$ be an algebraically closed field of characteristic zero. Every $1$-dimensional representation of $G$ over $k$ factors through $G^{\mathrm{ab}} = G/[G, G]$, and every finite abelian group has a faithful representation over $k$. Taken together, these simple facts show that one can characterise the abelianisation $G^{\mathrm{ab}} = G/[G, G]$ of $G$ as the smallest quotient of $G$ such that every $1$-dimensional representation of $G$ factors through it.

Let $G_n$ be the smallest quotient of $G$ such that every $n$-dimensional representation of $G$ over $k$ factors through $G_n$. Clearly $G_1 = G^{\mathrm{ab}}$ and I've read that if $G$ has a generating set of size $m$ then $G_m = G$, i.e. $G$ has an $m$-dimensional faithful representation. What do the $G_n$ look like in general? Can they be characterised in a different way? What do they tell us about $G$?

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    $\begingroup$ "I've read that if $G$ has a generating set of size $m$ then $G_m=G$" shouldn't be taken too seriously. Indeed, every finite simple group has a generating pair, while $G_2\neq G$ for most finite simple groups (furthermore for every $n$, every large enough nonabelian finite simple group has no nontrivial representation of dimension $\le n$). $\endgroup$
    – YCor
    Commented Dec 18, 2023 at 9:45
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    $\begingroup$ I'd be curious of a general description of those finite groups $G$ such that $G=G_2$, i.e., residually embeddable in $\mathrm{GL}_2(\mathbf{C})$. It consists of groups whose only nonabelian Jordan-Hölder factors are isomorphic to $A_5$, but not conversely, since $A_5$ itself is not in the class (unlike the binary icosahedral group of order 120). $\endgroup$
    – YCor
    Commented Dec 18, 2023 at 9:56
  • $\begingroup$ I just wanted to stress what YCor has already pointed out: $G=G_,m$ just means residually embeddable in $GL_m(k)$ and not embeddable in $GL_m(k)$. I am not sure in which direction the implication in the last paragraph goes, but i cannot be an if and only if. $\endgroup$
    – HenrikRüping
    Commented Dec 18, 2023 at 10:21

2 Answers 2

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It may be relevant to note a 1965 Theorem of Isaacs and Passman ( Pacific Journal of Mathematics), which may be interpreted as follows: for any positive integer $m$, there are only finitely many possibilities for $G/F(G)$ if all complex irreducible characters of the finite group $G$ have degree at most $m$. Here, $F(G)$ denotes the unique largest nilpotent normal subgroup of $G$.

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  • $\begingroup$ What Isaacs and Passman prove is a little stronger. They show that $G$ has an Abelian normal subgroup $A$ such that $[G:A]$ is bounded in terms of $M$ alone if all complex irreducible characters of $G$ have degree at most $m.$ I used the Fitting subgroup to make the statement more succinct. $\endgroup$
    – Geoff Robinson
    Commented Dec 19, 2023 at 13:35
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This is not an answer but only a too long comment.

While I dont know the answer for this question, I would split it up into two parts: one containing group theory and one containing representation theory.

Given another group $H$, we can assign to any group $G$ the natural map $$ \prod_{f\in Mor(G,H)} f:G\to \prod_{Mor(G,H)} H$$ and look at its image (denoted by $G_H$). Thinking of this image as $G/\bigcap_{f\in Mor(G,H)}Ker(f)$, we see that $G\to G_H$ is natural in $G$.

Has this already been studied. For example I bet someone already looked at the case of $G=F_3$ and $H=F_2$?

In the example where $G$ is finite and $H=k^*$ where $k$ is an algebraically closed field of characteristic zero, $H$ is so large that $G_H$ is naturally isomorphic to $G_{ab}$. This works for any abelian group containing $\mathbb{Z}$ and all $\mathbb{Z}/p$'s; it is nothing special about algebraically closedness and characteristic zero.

Can representation theory be used to understand the case of $H=GL_n(k)$ for $k\ge 0$?

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    $\begingroup$ Of course this has been studied. The case $(G,H)=(F_3,F_2)$ is somewhat trivial since $F_3$ is embeddable in $F_2$. However a less trivial result is that the map remains injective if one considers only surjections $F_3\to F_2$. $\endgroup$
    – YCor
    Commented Dec 18, 2023 at 11:32
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    $\begingroup$ "containing all $Z/p$": well, not if you only allow only prime $p$. And also not enough for $G=\mathbf{Q}$, when $H=\mathbf{Z}\oplus$(torsion). Actually, for a group $H$, there is the easy equivalence (a) $G\to G_H$ is injective for every abelian group (b) $H$ contains $\mathbf{Q}/\mathbf{Z}$ as a subgroup. $\endgroup$
    – YCor
    Commented Dec 18, 2023 at 11:34
  • $\begingroup$ Yes the "In the example where...." section I thought of having the assumption that $G$ is finite also in the second sentence, so that we can use the classification of finitely generated abelian groups. But I agree its better to have this general statement that works without this assumption. Also this shows that $G_1=G_ab$ does not need the assumption that $G$ is finite, since $\mathbb{Q}/\mathbb{Z}$ embeds into $k^*$ as the subgroup generated by all roots of unity. $\endgroup$
    – HenrikRüping
    Commented Dec 18, 2023 at 11:58

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