GROUPS OF HOMEOMORPHISMS AND NORMAL SUBGROUPS OF THE GROUP OF PERMUTATIONS

In this paper, it is proved that no nontrivial proper normal subgroup of the group of permutations of a set X can be the group of homeomorphisms of (X,T) for any topology T on X.

It is proved that no nontrivial proper normal subgroup of the group of permutations of a fixed set X can be the group of homeomorphisms of (X,T) for any topology T on X.As a by-product we obtain a characterization of completely homogeneous spaces.

NOTATIONS AND RESULTS.
We denote the cardinality of a set A by IAI.S(X) denotes the group of all permutations (bijections) of a set X.If p is a permutation of a set X, then H(P) {x : X p(x) x} A(X) denotes the group of all permutations of a set X which can be written as a product of an even number of transpositions.If = Is any [nfinlte cardinal number, If (X,T) is a topological space, the group of all homeomorphlsms of (X,T) onto Itself is called the group of homeomorphisms of (X,T) and denotes by H(X,T).
First we consider the case when X is finite set.LEMMA ,[.No nontrivial proper normal subgroup of the group S(X) of permutations of a finite set X can be the group of homeomorphisms of the topological space (X,T) for any topology T on X.

PROOF.
We can directly verify the result when IXI 4. When IXI > 5, the only normal subgroups of S(X) are the trivial subgroup, A(X) and S(X) itself.Let T be a topology on X such that H(X,T) A(X).Then (X,T) is a homogeneous space since all 3- cycles are in A(X).Thus (X,T) is a product of a discrete space and an indiscrete by Ginsburg [2].Then a partition {Eili= of X forms a basis for the topology space T o X.Here some E contain at least two elements (say a and b) for otherwise T is discrete in which case H(X,T) S(X) A(X), a contradiction.Now the transposition (a,b) is a homeomorphism of (X,T).This is again a contradiction for (a,b) A(X).
Hence the result.Now we proceed to extend the result to the case when X is an infinite set.
We use the following lemma proved by Baer [3].LEMMA 2.2.The normal subgroups of the group S(X) of permutations of X are precisely the trivial subgroup, A(X), S(X) and the subgroups of S(X) of the form H for some infinite cardinal number LEMMA 2.3.Let X be any infinite set and T any topology on X such that A(X) is a subgroup of a group of homeomorphisms of (X,T).Then a) (X,T) is homogeneous.b) PROOF.
(a) Let a and b be two distinct points of X.Now choose two more distinct points c and d of X other than a and b.Consider the permutation p (a,b) (c,d).It is a homeomorphism of (X,T) since it is an element of A(X) and it maps a to b. Hence (X,T) is homogeneous. (b) Let A be a nonempty open set of (X,T) and Ac BC X.  (c,d) is a homeomorphlsm of (X,T) since p g A(X).Then } p(A) N B is open.Since (X,T) is homegeneous by (a), (X,T) is discrete.This contradicts the hypothesis.
Hence the result.
(e) If (X,T) is indiscrete, the result is obvious.Otherwise every finite subset of X ls closed by (c).If a nonempty finite subset F of X is open, then both F and X\F are open which contradicts (d).Hence the result.REMARK 2.1.Lemma 2.3 shows that if A(X) is a subgroup of H(X,T), then (X,T) is discrete or the nonempty open sets o (X,T) form a filter.LEMMA 2.4.Let (X,T) be an in[inlte topological space in which nonempty open sets form a filter.Let A be a proper closed subset of (X,T).Then every permutation of X which moves only the elements o A is a homeomorphlsm of (X,T).
PROOF.Let p be a permutation of K which moves only the elements of A. If U is a nonempty open set, p(U) U 0 (X\A) and U n (X\A) is open and nonempty by hypothesis.Thus p is" an open map.Similarly we can prove that p is a continuous map.Hence p is a homeomorphism.LEMMA 2.5.Let (X,T) be an infinite topological space which is neither discrete nor indiscrete such that the group of homeomorphisms H of (X,T) is a normal subgroup sets of (X,T) form a filter.Let p be any permutatlo of X which moves every element of K and keeps every element of X\K fixed.Then by Lemma 2.4, p is a homeomorphlsm of homeomorphlsm of (X,T) since H is normal.
suitable subset and that subset is also closed by Lemma 2.3.Now consider a permutation of X which maps K onto X\k and X\K onto K.Such a permutat[on exists since IKI IX\KI.Now t is a homeomorphlsm of iX,T)onto itself by the last paragragh.Hence t(K) X\K is closed.Now both K and X\K are open which contradicts Lemma 2.3.Hence the result.LEMMA 2.6.Let (X,T) be an infinite topological space which is neither discrete nor indiscrete such that the group of homeomorphlsms H of iX,T) is a normal subgroup of S(X) containing A(X).
Let K be a proper closed subset of (X,T).
Then every permutation p such that IM(p)l IKI is a homeomorphism and every subset M of X such PROOF.
By the previous remark, the nonempty open sets in (X,T) form a filter.
Let t be a permutation of X which moves every element of K and leaves every element of X\K fixed.
Then by Lemma 2.4, t EH and IM(t)l--IKI Then by Lemma 2.2, every permutation p of X such that IM(p)l IKI belongs to H since H is normal in S(X).Now prove that every subset M of X such that IMI IKI is closed.Without loss of generality, we may assume that IMI--IKI for otherwise we can take a suitable subset of K and the subsets of K are also closed by Lemma 2.3.Ue have IKI IMI and IX\KI IX\M since IKI < IXI by Lemma 2.5.Then there exists a permutation p of X which maps K onto M, M onto K and keeps every other element fixed.Then IM(p)l IKI + IMI.If K is finite, M is also finite and hence closed by Lemma 2.3.Therefore we may assume that K is infinite.Then IM(p)l IKI + IMI--IKI.Thus p is a homeomorphism by the first paragraph.Then M pK) is closed.LEMMA 2.7.Let (X,T) be an infinite topological space which is neither discrete nor indiscrete such that the group of homeomorphisms H of (X,T) is a nontrlvlal normal subgroup of S(X).Then T T for some infinite cardinal number such that IXl where {} 0{A-X: lEnA < =}.
PROOF.Since H is a nontrlvial normal subgroup of S(X), it contains A(X) by Lemma 2.2.Then by Lemma 2.3, every finite set is closed in (X,T).Let a inf {IBl: B c X, B is not closed in (X,T) }.
Then a is an infinite cardinal number such that Now prove that T T Here T T for otherwise there exists U T but then M is closed in (X,T) by Lemma 2.6 since X\U is closed in (X,T)and IMI IX\UI.This contradicts the definition of a Also T c T. For, if A T A * , IX\AI < , then X\A is closed in (X,T) by the definition of .Thus A T. Hence the result.
LEMMA 2.8.The group of homeomorphisms of a topological space (X,T), where T is either discrete, indiscrete or of the form T {} U {AcX IX\AI < =} If A X or A B, the the result is evident.Otherwise choose an element a of A and b of B\A.Also choose two distinct points c and d, other than a and b, both from ether A, B\A or X\B.Now the permutation p (a,b) (c,d) is a homeomorphism of (X,T) since p E A(X).Then A U {bl AUp(A)is open.Thus 8 U (AU {b}) b cB\A Is open.Hence the result.(c)Since (X,T) is not indiscrete, there exists a proper nonempty open set A of (X,T).Let bX\A.Then X\{b} is open by (b).Thus } Is closed.Then every sngleton subset of (X,T) is closed, since (X,T) Is homogeneous by(a).Hence every fnlte subset, being a finite union of sngleton subsets is closed. (d) Let A and B be two nonempty open subsets of (X,T).Prove that A N B .Otherwise choose an element a from A and b from B. Here a b.Choose two distinct points c and d other than a and b both from either A or B or X\(AU B).Now p (a,b)