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If $G$ is a linear group over a ring $R$, is every homomorphic image of $G$ again a linear group?

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What exactly do you mean by "linear group over a ring"? A subgroup of the endomorphisms of $R^n$? –  Bill Cook Aug 24 '12 at 20:03
    
$G$ islinear if it is a subgroup of $GL(n,R)$ for some $n$ and some ring $R$. –  user17488 Aug 24 '12 at 20:04
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You mean "commutative ring". Because every group $G$ is a subgroup of $GL_1(\mathbf{Z}G)$, where $\mathbf{Z}G$ is its group algebra. –  YCor Aug 25 '12 at 10:26

2 Answers 2

No. For example, the free groups $F_{\infty}$ on countably many elements is a subgroup of $F_2$, which is linear (over $\mathbb{C}$ for example), so $F_{\infty}$ is linear. Its quotients are all countably-generated groups. It is straightforward to write down countably-generated groups which are not linear over a particular commutative ring, but to get one not linear over any commutative ring seems to require slightly more work.

Lemma: Let $A_{\infty}$ denote the group of alternating permutations (fixing finitely many elements) of $\mathbb{N}$. Then $A_{\infty}$ is simple.

Proof. Let $N$ be a normal subgroup and let $\rho : A_{\infty} \to A_{\infty}/N$ be the corresponding quotient. $A_{\infty}$ is the union of a chain of distinguished subgroups $A_n \subset A_{\infty}$ which are simple for $n \ge 5$; consequently their image in $A_{\infty}/N$ is either trivial or all of $A_n$ for $n \ge 5$, and if $A_n$ has nontrivial image then so does $A_m$ for all $m \ge \text{max}(n, 5)$ (by simplicity) and for all $m \le \text{max}(n, 5)$ (since these are contained in a sufficiently large $A_m$), hence for all $A_m$. Hence the image of all of $A_{\infty}$ is either trivial or all of $A_{\infty}$, so $N$ is either all of $A_{\infty}$ or trivial. $\Box$

Claim: $A_{\infty}$ is not linear over any commutative ring $R$.

Proof. Let $\rho : A_{\infty} \to \text{GL}_n(R)$ be a homomorphism. For any prime ideal $P$ of $R$, the image of $A_{\infty}$ in $\text{GL}_n(R/P)$ is either trivial or all of $A_{\infty}$ by the lemma. If the image is all of $A_{\infty}$ for some prime ideal $P$, then by passing to the fraction field of $R/P$ we may assume WLOG that $R$ is a field. Suppose $R$ has characteristic $p$. Since $A_{\infty}$ contains $S_n$ as a subgroup for every positive integer $n$, it contains every finite group as a subgroup, so in particular it contains $(\mathbb{Z}/q\mathbb{Z})^n$ as a subgroup for some prime $q \neq p$ and every positive integer $n$. By passing to the algebraic closure we may simultaneously diagonalize the action of such a group, and then taking $n$ sufficiently large gives a contradiction.

If no prime ideal $P$ has the above property, then $A_{\infty}$ is contained in the intersection of the kernels of the maps $\text{GL}_n(R) \to \text{GL}_n(R/P)$; consequently it is contained in the group of matrices congruent to $I \bmod N(R)$ where $N(R)$ denotes the nilradical. But this group has a filtration by normal subgroups consisting of the matrices congruent to $I \bmod N(R)^k$ for positive integers $k$, and all of the corresponding quotients are solvable (by inspecting where the commutators land). $A_{\infty}$ necessarily has nontrivial image in one of these quotients, which is a contradiction. $\Box$

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It easy indeed to write down, but your countable direct sum doesn't work, as it is is linear over a field of characteristic 2. –  YCor Aug 25 '12 at 10:28
    
@Yves: my apologies; I'm used to "linear" meaning "linear over $\mathbb{C}$." I no longer know an easy example which works over an arbitrary commutative ring; what did you have in mind? –  Qiaochu Yuan Aug 25 '12 at 17:58
    
Your example is great, but in a different direction, any finitely generated, non-residually-finite group works (a certain Baumslag-Solitar group, for example). –  user641 Aug 25 '12 at 21:23
    
@Steve: the proof I know that a f.g. non-r.f. group is not linear works over a field. Does the result generalize to linear over an arbitrary commutative ring? –  Qiaochu Yuan Aug 25 '12 at 21:57
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Yes, it's correct that every f.g. commutative ring (not necessarily a domain) is residually finite (more generally, any noetherian ring is residually Artinian). Therefore any f.g. group linear over a commutative ring is residually finite. –  YCor Aug 28 '12 at 6:23

In a somewhat different category, a meaningful (counter-) example is the quotient of (one form of) the Heisenberg group $N$ of upper-triangular nilpotent 3-by-3 real matrices by the subgroup of those with integer entries in the upper right corner and 0's immediately superdiagonal. (The Lie algebra did not change, etc.)

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So I presume that for Lie-theoretic reasons this quotient has no faithful finite-dimensional representations over $\mathbb{C}$ as a topological group. Is it clear that it has no faithful finite-dimensional representations over an arbitrary commutative ring $R$ as an abstract group? (Or is this what you meant by "different category"?) –  Qiaochu Yuan Aug 25 '12 at 18:42
    
Right, the Lie algebra is (still) nilpotent, finite-dimensional. I've not thought about how this group might behave in other "categories" of linear groups over other rings... –  paul garrett Aug 25 '12 at 18:59
    
Indeed you don't need continuity, because $G=\text{Heis}_3(\mathbf{Q})/\mathbf{Z}$ is not linear over any field. Proof: you can suppose the field alg. closed so any rep is triangulable on a finite index subgroup of $G$; since $G$ has no proper finite index subgroup, this is $G$ itself. So the derived subgroup maps to unipotent matrices. But certainly $\mathbf{Q}/\mathbf{Z}$ has no faithful unipotent representation over any field. –  YCor Aug 28 '12 at 8:34
    
@Yves: very good. –  paul garrett Aug 28 '12 at 12:28

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