Re: Quotients of Sym(X)


In article <gs1niu$2fss$1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Dominic van der Zypen <dominic.zypen@xxxxxxxxx> wrote:

Which groups arise as quotients of symmetric groups?

A correct answer has been given for quotients of finite symmetric groups.
The OP clarified in email that his question concerns infinite groups,
e.g. whether there is ever a surjection from a Sym(X) onto the integers.

If C is any infinite cardinal <= card(X), then
Sym_C(X) = { f : X -> X | card( supp(f) ) ) <= C }
is easily seen to be a normal subgroup of Sym(X)
(where the supp(f) is the "support" of f : { x in X | f(x) \not = x } ).

I believe it's true that these are the only normal subgroups of Sym(X)
for infinite X, apart from the "infinite alternating group", i.e.
the derived subgroup of Sym_C(X) when C = aleph_0. (In particular,
there will be no surjections onto Z). I looked for references to
corroborate my memory on this and while I didn't quite find what I wanted,
I got some partial confirmation:

MR0140406 (25 #3826) Kent, Clement F.
Constructive analogues of the group of permutations of the natural
numbers. (Trans. Amer. Math. Soc. 104 1962 347--362.)

If $S_\infty$ is the group of all permutations of the natural numbers,
$F$ the subgroup of permutations which change only a finite number of
numbers, $A$ the subgroup of even permutations of $F$ and
$\varepsilon$ the identity permutation, then it is known that
$S_\infty\supset F\supset A\supset\{\varepsilon}$ is the unique
composition series for $S_\infty$. [...]

and

MR0167517 (29 #4789) Gaughan, Edward D.
The index problem for infinite symmetric groups.
(Proc. Amer. Math. Soc. 15 1964 527--528.)

For an infinite set $M$ with cardinal $X$, let $S(X,Y)=$
{$\sigma:\sigma$ is a permutation on $M$ such that
$|\text{spt}\,\sigma|<Y$}, where $\text{spt}\,\sigma={m\in
M:\sigma(m)\neq m}$. Generalizing the results of Onofri [Ann. Mat.
Pura Appl. (4) 7 (1929), 103--130], Higman [Publ. Math. Debrecen 3
(1954), 221--226; MR0072136 (17,234d)] and Scott [Univ. Kansas Dept.
of Math. Rep. No. 5 (1956), 1--22], the author proves the following:
If $\chi_0<Y\leq X^\ast$, where $X^\ast$ is the successor of $X$, then
$S(X,Y)$ has no proper subgroups of index less than $X$.

dave
.

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