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Dear all,

I am interested in residually finite, perfect groups. Are all of them known to be residually alternating? If not, how could one construct a counterexample?

A group $G$ is residually alternating if for every $g \in G$ there exists a finite alternating quotient $G/N$ such that $g \notin N$.

By a result by Katz and Magnus free groups are known to be residually alternating. This has recently been extended by Henry Wilton.

Elisabeth

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3 Answers

up vote 11 down vote accepted

I think the question needs to be made a little more precise. One can take a finite simple group $G$ which is not any alternating group. Then $G$ is perfect, residually finite (!) and is not residually alternating.

If we ask for finitely generated infinite perfect groups which are not residually alternating, then things are a little more involved: you can take $G=SL_n({\mathbb Z})$ for $n\geq 3$. Then (by the congruence subgroup property for $SL_n({\mathbb Z})$) the only simple quotients are $SL_n({\mathbb Z}/p{\mathbb Z})$ for some prime $p$. These are not alternating groups if $n$ is large enough. [$SL_n({\mathbb Z})$ is perfect, and residually finite, as can be easily seen].

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Yes I was indeed looking for finitely generated counterexamples. Many thanks for your answer! –  Elisabeth Fink May 17 '13 at 14:57
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By the congruence subgroup property, any finite quotient of $SL_n(\mathbb{Z})$ factors through $SL_n(\mathbb{Z}/N\mathbb{Z})$ for some $N.$ But how do you conclude that any finite simple quotient factors through $SL_n(\mathbb{Z}/p\mathbb{Z})$ for some prime $p$? Also, a nitpick: it is $PSL_n(\mathbb{Z}/p\mathbb{Z})$ that is simple as an abstract group. –  Victor Protsak May 17 '13 at 17:35
    
@victor: yes this needs a proof. A finite quotient of $SL_n({\mathbb Z})$ , by CSP, is a quotient of $SL_n({\mathbb Z}/n{\mathbb Z})$. By strong approximation, the latter is the product of $SL_n({\mathbb Z}/p^e{\mathbb Z})$ for varying primes $p$ and integers $e$. The latter groups have unipotent groups and the unique simple quotient $SL_n({\mathbb Z}/p{\mathbb Z})$ (modulo centre, as you rightly say). –  Aakumadula May 18 '13 at 2:24
    
in the above comment, please change "have unipotent groups" to "have nilpotent groups as normal subgroups" –  Aakumadula May 18 '13 at 3:08
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If you allow the group to be finite, any non-alternating finite simple group is a counterexample. Otherwise you can still obtain counterexamples from wreath products of such groups with the infinite cyclic group.

To give a specific counterexample: let $$ G := (\mathbb{Z},+) \wr {\rm PSL}(2,7) \ = \ (\mathbb{Z},+)^8 \rtimes {\rm PSL}(2,7), $$ where ${\rm PSL}(2,7) \cong \langle (3,7,5)(4,8,6), (1,2,6)(3,4,8) \rangle$ acts on $(\mathbb{Z},+)^8$ by permuting the factors. Then $G'$ is perfect, it is residually finite as $(\mathbb{Z},+)$ is so, and it does not admit a surjection to a nontrivial alternating group.

The example can be constructed in GAP as follows:

gap> LoadPackage("rcwa");
gap> G := WreathProduct(CyclicGroup(IsRcwaGroupOverZ,infinity),PSL(2,7));;
gap> StructureDescription(G);
"Z wr PSL(3,2)"
gap> IsPerfect(G); # not yet (as Derek remarked) ...
false
gap> G := DerivedSubgroup(G);; # ... but now we have our example.
gap> IsPerfect(G);
true
gap> StructureDescription(G);
"(Z x Z x Z x Z x Z x Z x Z) . PSL(3,2)"
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Thanks that's really helpful! –  Elisabeth Fink May 17 '13 at 14:57
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I don't think that example is perfect. –  Derek Holt May 17 '13 at 14:57
    
@Derek: Indeed, thanks! -- One needs to take the derived subgroup of $G$ to obtain a perfect group. I have edited my answer accordingly. –  Stefan Kohl May 17 '13 at 16:45
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@Stefan: That's a nice example of the use of your RCWA package. Groups like that can hard to compute with. You can represent them as finitely presented groups or as integral matrix groups, but in both cases it can be difficult to carry out structural computations. –  Derek Holt May 18 '13 at 14:59
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I guess you are looking for finitely generated counterexamples. Perfect crystallographic space groups provide one source of examples. These are virtually abelian groups that are extensions of a finitely generated abelian group by a finite perfect group. They are residually finite, and not residually alternating - they are not even residually finite simple.

As a specific example, let $X = A_5$ given by the presentation $\langle a,b \mid a^2=b^3=(ab)^5=1 \rangle$, and let $a$ and $b$ act on ${\mathbb Z}^4$ as in the deleted permutation module. For example, their actions could be given by the integral matrices

$a \to \left(\begin{array}{rrrr}0&0&0&1\\0&1&0&0\\-1&-1&-1&-1\\1&0&0&0\end{array}\right),\ \ \ b \to \left(\begin{array}{rrrr}0&0&1&0\\1&0&0&0\\0&1&0&0\\0&0&0&1\end{array}\right).$

Now let $G$ be the semidirect product of ${\mathbb Z}^4$ with $X$ using this action. Then $G$ itself is not perfect, but its commutator subgroup $G'$ has index 5 in $G$ and is perfect. There are examples like this for all finite perfect groups $X$.

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Many thanks Derek! This is very helpful! –  Elisabeth Fink May 17 '13 at 14:58
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