Are residually finite, perfect groups residually alternating? 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
 A: 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 $X$.
A: 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)"

A: 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].  
