@ Appl. Gen. Topol. 23, no. 2 (2022), 269-280 doi:10.4995/agt.2022.16143 © AGT, UPV, 2022 On the group of homeomorphisms on R: A revisit K. Ali Akbar a and T. Mubeena b a Department of Mathematics, School of Physical Sciences, Central University of Kerala, Periya - 671320, Kasaragod, Kerala, India (aliakbar.pkd@gmail.com, aliakbar@cukerala.ac.in) b Department of Mathematics, School of Mathematics and Computational Sciences, Univer- sity of Calicut, Thenhipalam - 673635, Malappuram, Kerala, India (mubeenatc@gmail.com, mubeenatc@uoc.ac.in) Communicated by F. Lin Abstract In this article, we prove that the group of all increasing homeomor- phisms on R has exactly five normal subgroups, and the group of all homeomorphisms on R has exactly four normal subgroups. There are several results known about the group of homeomorphisms on R and about the group of increasing homeomorphisms on R ([2], [6], [7] and [8]), but beyond this there is virtually nothing in the literature concern- ing the topological structure in the aspects of topological dynamics. In this paper, we analyze this structure in some detail. 2020 MSC: 20F38; 37B02. Keywords: group of homeomorphisms; normal subgroups; dynamical sys- tems; fixed points; conjugacy; bounded functions. 1. Introduction There have been several papers discussing about the normal subgroups of the group of homeomorphisms on various metric spaces ([2], [6], [7] and [8]). It is natural to ask: Which subsets will arise as normal subgroups of the group of homeomorphisms on a metric space? We provide a proof in the case of the group of (increasing) homeomorphisms on R. This paper contains a detailed Received 29 August 2021 – Accepted 14 March 2022 http://dx.doi.org/10.4995/agt.2022.16143 K. Ali Akbar and T. Mubeena proof that highlights the differences and similarities between our results and those given in the references. A dynamical system is simply a pair (X,f) where X is a metric space and f : X → X is a continuous function. A point x ∈ X is said to be periodic with period n if fn(x) = x for some n ∈ N, and fm(x) 6= x for 1 ≤ m < n where fn = f ◦f ◦ ...◦f is the composition of f with itself n times. If f(x) = x then we say that x is a fixed point of f. We denote the set of all fixed points of f by Fix(f), and the complement of Fix(f) by Fix(f)c. Two dynamical systems (X,f) and (Y,g) are said to be conjugate (we simply say f is conjugate to g), if there exists a homeomorphism h from X to Y such that h ◦ f = g ◦ h. Being conjugate is an equivalence relation in the class of dynamical systems. A homeomorphism h from X to X such that h◦f = f◦h is called a conjugacy of (X,f) or simply a conjugacy. Let A,B be two subgroups of a group G. Then A is invariant in B if A ⊂ B and if bAb−1 ⊂ A for every b ∈ B. A subgroup N of a group G is normal if and only if it is invariant under conjugation, if and only if it is a union of conjugacy classes of G. Since our group in this paper is a group of homeomorphisms, the algebraic notion of conjugacy here coincides with topological conjugacy in the sense of dynamical systems theory. For preliminaries from topological dynamics and group theory, the reader may refer [3], [4] and [5]. In this paper, we study the normal subgroups of the group of (increasing) homeomorphisms of R and analyze the topological structure in the aspects of topological dynamics in some detail. The proofs are different from those given in the above references. The classification here is only based on the set of fixed points of members of the normal subgroups. In articles [2], [6] and [8], the set of fixed points or the support of the homeomorphisms are used to classify normal subgroups. First we discuss the ideas of proof involved in the references [2], [6] and [8] to convince the reader that our proof is different from the known ones. For a set X, let π(X) be the group of all permutations (bijections) on X and G be a subgroup of π(X). For a topological space X, let H(X) be the group of homeomorphisms on X. Suppose F is a non-empty family of subsets of X. We define S(F,G) = {g ∈ G : Fix(g) ⊃ F for some F ∈F}. For S(F,H(X)), we shall write S(F). We say that the family F is ecliptic relative to G whenever it satisfies the following two conditions. (1) If F1,F2 ∈F, then there exists an F3 ∈F such that F3 ⊂ F1 ∩F2, (2) If F1 ∈F and g ∈ G, then there exists an F2 ∈F such that F2 ⊂ g(F1). An ecliptic family which satisfies the following additional condition will be called strictly ecliptic. (3) If F ∈ F and U ⊂ X is open (U 6= ∅), then there exists an h ∈ H(X) such that h(Fc) ⊂ U, where Fc is the complement of F in X. The objective of the reference [8] is to investigate the normal subgroups for a class of spaces which includes the n-cell Bn and the author proved that some of these normal subgroups can be defined in terms of the family of fixed point © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 270 On the group of homeomorphisms on R: A revisit sets of their elements. For a family F of subsets of X, define S(F,G) = {g ∈ G : Fix(g) ⊃ F for some F ∈ F}. It is proved that S(F,G) is a subgroup of G and if F is ecliptic relative to G, then S(F,G) is a normal subgroup of G. If X is a topological space such that for any non-empty, open set U, there is an open set V ⊂ U which is homeomorphic to an open ball in a Euclidean space of positive dimension and supposing there is a strictly ecliptic family F on X relative to H(X) and if N is a normal subgroup of H(X), then the author proved that either N ⊃ S(F) or N consists of the identity 1. They also proved that if N is a normal subgroup of H(Bn) which contains an element not in H0(Bn) = {h ∈ H(Bn) : Fix(h) ⊃ Sn−1(the boundary of Bn)}, then N ⊃ H0(Bn). The objective of the reference [2] is to analyze the algebraic structure of H(I) in some detail for some interval I. For a subgroup H of π(X), x ∈ X, let Hx := isotropy subgroup of H at x. The authors proved that every translation is a product of two involutions and every element of H(I) is a product of at most four involutions. They considered a signature theorem, which provides a useful criteria for the conjugacy in H(I). Using this idea, they enumerate completely the normal subgroups of H := H(I) ≤ π(I). Let F be the isotropy subgroup H0 of H(R). The idea of proof is as follows. For an interval I, we denote I0 for the interior of I. For a map f : I → I, S(f,x) :=sign(f(x) −x). An element t ∈ F is a translation if it does not have interior points as fixed points and let T denote the set of all translations. Also T+ := {t ∈ T : S(t,x) > 0} and T− := {t ∈ T : S(t,x) < 0}. Now, let S be a semi-group which is invariant in F, Qa = {f ∈ H : f(x) = x for all x in some neighborhood Nf (a)} for a ∈ I and Q = Q0 ∩ Q1. The authors proved that if S 6⊂ Q0, then S contains an element with at most one interior fixed point. Also T+ and T− are complete conjugacy classes in F, T is a complete conjugacy in H and F = TT. Using these ideas, the authors also proved that if N is an invariant subgroup of F , then N ⊂ Q0 or N ⊂ Q1 or N = F. If N is an invariant subgroup of H then either N = H, N = F or N ⊂ Q. Now the only subgroups of H are H,F,Q and {1} since Q is simple. If N is normal in Q0 (respectively in Q1), then either N = {1}, N = Q or N = Q0 (respectively in Q1). Hence the only normal subgroups of F are F,Q0,Q1,Q and {1}. The reference [6] is an expository paper, the author provides a relatively complete but concise account of the classification of H := H(I), in terms of a suitable topological signature concept. For φ ∈ H, the author first associated the function s(φ) : R → S = {−1, 0, 1} defined by s(φ)(x) = sign(φ(x) − x). For s ∈ Σ := {h : R → S : h is continuous}, let Spt(s) = R \ int(s−1(0)), Ha = {φ ∈ H : Spt(φ) is bounded above}, Hb = {φ ∈ H : Spt(φ) is bounded below} and Hc = {φ ∈ H : Spt(φ) is bounded}. The author first observed the following facts: (1) s(φ−1) = −s(φ) (2) For φ1,φ2 ∈ H+ (the set of all increasing homeomorphisms) with s(φ1) ≥ 0 and s(φ2) ≥ 0, it holds s(φ1 ◦φ2) ≥ max{s(φ1),s(φ2)}. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 271 K. Ali Akbar and T. Mubeena These facts provide a one-to-one correspondence between the collection of normal subgroups N of H (resp. H+) and Σ(N), the family of s-functions closed under the operation (s(φ),s(ψ)) → s(φ−1 ◦ ψ) and closed under topo- logical equivalence. Consider the group H(R) = {f : R → R : f is a homeomorphism} under composition of functions, and its subgroups IH(R) = {f ∈ H(R) : f is increasing}, Hl = {f ∈ IH(R) : Fix(f)c is bounded above} and Hr = {f ∈ IH(R) : Fix(f)c is bounded below}. Our main results prove that: (1) The group IH(R) has exactly five normal subgroups. They are: (a) The whole group IH(R) (b) The trivial group {1} (c) Hl (d) Hr (e) H = Hl ∩Hr. (2) For H(R) there are exactly four normal subgroups. They are: (a) The whole group H(R) (b) The trivial group {1} (c) H = {f ∈ IH(R) : Fix(f)c is bounded} (d) IH(R). 2. Main results Let IH([a,b]) denote the group (under composition of functions) of all in- creasing homeomorphisms on the closed interval [a,b] and let H([a,b]) denote the group (under composition of functions) of all homeomorphisms on the closed interval [a,b]. In fact H(R) and H([a,b]) are topological groups with respect to compact-open topology. This happens since the homeomorphism group on a locally connected and locally compact second countable space is a topological group (see [1]). Consider R∗ = R ∪{−∞,∞} with order topology. Any closed interval [a,b] in R is homeomorphic to R∗ = R∪{−∞,∞}, and the groups IH([a,b]) and IH(R∗) are isomorphic. We write: (1) CA([a,b]) = {f : [a,b] → [a,b] : f is a homeomorphism such that f(t) > t ∀ t ∈ (a,b)} (2) CB([a,b]) = {f : [a,b] → [a,b] : f is a homeomorphism such that f(t) < t ∀ t ∈ (a,b)}. (3) H = Hl ∩Hr = {f ∈ IH(R) : Fix(f)c is bounded} For f ∈ CA([a,b])∪CB([a,b]), we have f(a) = a and f(b) = b. Hence CA([a,b]) and CB([a,b]) are subsets of IH([a,b]). We define CA([a,b]) ◦ CB([a,b]) := {f ◦g : f ∈ CA([a,b]),g ∈ CB([a,b])}. Two continuous maps f,g : R → R are said to be order conjugate if there exists an increasing homeomorphism h on R such that h◦f = g◦h. The maps f,g : R → R defined by f(x) = x + 1 and g(x) = x−1 are conjugate to each other through h(x) = −x+ 1 2 ∈ H(R). But f and g are not order conjugate. Contrary to assume there is an h ∈ IH(R) such © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 272 On the group of homeomorphisms on R: A revisit that h◦f = g◦h. Then h(x+ 1) = h(x)−1. That is, h(x+ 1)−h(x) = −1 < 0. Which is a contradiction to the assumption that h ∈ IH(R) and hence the maps f and g are not order conjugate. Lemma 2.1. (1) Assume that f,g ∈ IH(R) are such that Fix(f) = Fix(g) (a) f is conjugate to g; (b) If for every t ∈ R it holds (f(t)−t)(g(t)−t) ≥ 0 then f and g are order conjugate. (2) Assume that f,g ∈ IH([a,b]) are such that Fix(f) = Fix(g) (a) f is conjugate to g; (b) If for every t ∈ [a,b] it holds (f(t) − t)(g(t) − t) ≥ 0 then f and g are order conjugate. Proof. (1) Assume that f,g ∈ IH(R) are such that Fix(f) = Fix(g) (a) Case 1: Fix(f) = Fix(g) = ∅ Assume that f(0) > 0. For n ∈ N, inductively define f−n = f−n+1 ◦f−1. Since f is increasing (fn(0)) increases and thus di- verges to ∞, and (f−n(0)) decreases and diverges to −∞. More- over, for t ∈ R there exists unique n ∈ Z such that, fn(0) ≤ t < fn+1(0). Consider this unique n, and define k : (−∞,f(0)) → (−∞, 1) by k(t) = t f(0) and h : R → R by h(t) = k(f−n(t)) + n for t ∈ R. Note that h(f(0)) = 1 and h is a homeomorphism of R. Then h◦f(t) = h(t) + 1 ∀ t ∈ R. This h gives a conjugacy from f to x + 1. Similarly we can prove that, f is conjugate to x− 1 if f(0) < 0. The maps x + 1 and x− 1 are conjugate to each other. Hence the proof. Case 2: Fix(f) = Fix(g) 6= ∅ In this case, define f̃ : R∗ → R∗ by f̃|R (the restriction map f̃ to R)= f, f̃(−∞) = −∞ and f̃(∞) = ∞. Similarly define g̃ also. Let Fix(f̃)c = Fix(g̃)c = ∪α(aα,bα) (disjoint union of open intervals). The restriction maps f̃|[aα,bα] : [aα,bα] → [aα,bα] and g̃|[aα,bα] : [aα,bα] → [aα,bα] are increasing with aα = f̃(aα) = g̃(aα) and bα = f̃(bα) = g̃(bα) and Fix(f̃|(a,b)) = Fix(g̃|(a,b)) = ∅. Let hα be a conjugacy from f̃|[aα,bα] to g̃|[aα,bα] for every α. By Case 1, this conjugacy hα exists for every α. Define h : R∗ → R∗ as h(x) = { hα(x) if x ∈ (aα,bα) x otherwise . Then h is a conjugacy from f̃ to g̃. Hence h|R is a conjugacy from f to g. (b) Suppose (f(t)−t)(g(t)−t) ≥ 0 for all t ∈ R. If Fix(f) = Fix(g) = ∅ then f(0)g(0) > 0. This implies either both f(0) and g(0) are positive or both f(0) and g(0) are negative. Hence both f and g are either order conjugate to x+1 or to x−1. If Fix(f) = Fix(g) 6= ∅ then consider the maps f̃, g̃ as in Case 2 of (1) (a) in the proof of Lemma 2.1. If Fix(f̃)c = Fix(g̃)c = ∪α(aα,bα) (disjoint union © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 273 K. Ali Akbar and T. Mubeena of open intervals) then the restriction maps f̃|[aα,bα] : [aα,bα] → [aα,bα] and g̃|[aα,bα] : [aα,bα] → [aα,bα] are increasing with aα = f̃(aα) = g̃(aα) and bα = f̃(bα) = g̃(bα), and (f̃(t)−t)(g̃(t)−t) > 0 for all t ∈ (aα,bα). If hα is an order conjugacy from f̃|[aα,bα] to g̃|[aα,bα] for every α then the map h : R ∗ → R∗ defined by h(x) = { hα(x) if x ∈ (aα,bα) x otherwise is an order conjugacy from f to g. Hence the proof follows. (2) Assume that f,g ∈ IH([a,b]) are such that Fix(f) = Fix(g). Without loss of generality, we can assume that f,g ∈ IH(R∗). Then f(−∞) = −∞ and f(∞) = ∞. Hence Fix(f|R) = Fix(g|R). (a) By (1) (a) of Lemma 2.1, f|R is conjugate to g|R and hence f is conjugate to g; (b) By (2) (a) of Lemma 2.1, if for every t ∈ R it holds (f(t)−t)(g(t)− t) ≥ 0 then f|R and g|R are order conjugate and hence f is order conjugate to g. � For a map f : R → R, if t1, t2 ∈ Fix(f) and s /∈ Fix(f) for all s ∈ (t1, t2) then we say that t1 and t2 are adjacent. Lemma 2.2. Let f ∈ IH(R) and let {{aα,bα}}α be the pairs of adjacent fixed points. Define g : R → R by g(x) = { x+f(x) 2 if aα < x < bα for some α f(x) otherwise Then f is order conjugate to g. Proof. The proof follows from Lemma 2.1. � Proposition 2.3. CA([a,b]) ◦CB([a,b]) = IH([a,b]). Proof. Let h ∈ IH([a,b]). Define hA(x) := { h(x) if h(x) > x x otherwise and hB(x) :={ h(x) if h(x) < x x otherwise . Then h = hA ◦ hB. But hA /∈ CA([a,b]) and hB /∈ CB([a,b]). Now consider g(x) = x + 1. Then g(x) > x for all x ∈ R. Let h′A = hA ◦ g and h ′ B = g −1 ◦ hB. Then h ◦ g(x) > h(x) for all x. Therefore h ◦ g(x) > x whenever h(x) > x. Hence h′A(x) > x if h(x) > x. If h(x) ≤ x then h′A(x) = g(x) > x. Hence h ′ A ∈ CA([a,b]). Similarly we can prove that h′B(x) ∈ CB([a,b]). Hence the proof follows since h = h ′ A ◦h ′ B. � Corollary 2.4. If N is a normal subgroup of IH([a,b]) that contains an ele- ment of either CA([a,b]) or CB([a,b]) then N = IH([a,b]). Proof. By Lemma 2.1, the sets CA([a,b]) and CB([a,b]) are exactly the con- jugacy classes of IH([a,b]), and CA([a,b]) = {f−1 : f ∈ CB([a,b])}. So if subgroup N is normal and intersect either CA([a,b]) or CB([a,b]), then it au- tomatically contains these sets. Hence N = IH([a,b]) by Proposition 2.3. � © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 274 On the group of homeomorphisms on R: A revisit We introduce the following notation: For f ∈ IH(R) and t0 ∈ Fix(f), we denote ft0 (x) := { x+f(x) 2 if x ≥ t0 f(x) if x < t0 , f∗t0 (x) := { f(x) if x ≥ t0 x if x < t0 , and f∗∗t0 (x) :={ x if x ≥ t0 f(x) if x < t0 . Lemma 2.5. Let f ∈ IH(R) and let t0 ∈ Fix(f). Then f∗t0 is order conjugate to f−1t0 ◦f. Proof. For f ∈ IH(R), first observe that ft0|[t0,∞) is x+f(x) 2 and f∗t0|[t0,∞) is f(x). Hence by Lemma 2.1, ft0|[t0,∞) is order conjugate to f ∗ t0 |[t0,∞). For t ∈ [t0,∞), first suppose that f(t) − t ≥ 0. Then f(t) ≥ t+f(t) 2 . Which implies f−1t0 (f(t)) ≥ f −1 t0 ( t+f(t) 2 ) = t. If f(t) − t ≤ 0 then we can prove that f−1t0 (f(t)) ≤ t. Hence (f(t) − t)((f −1 t0 ◦ f)(t) − t) ≥ 0 for all t ∈ [t0,∞). By Lemma 2.1, f∗t0 is order conjugate to f −1 t0 ◦f on [t0,∞). Also f−1t0 ◦f|(−∞,t0) is the identity function. Hence the proof follows. � Corollary 2.6. Let N be a normal subgroup of IH(R). Let f ∈ N and let t0 ∈ Fix(f). Then f∗t0 ∈ N. Proof. By Lemma 2.1, ft0 is order conjugate to f and by Lemma 2.5, f ∗ t0 is order conjugate to f−1t0 ◦f. Hence the proof follows. � Proposition 2.7. Let N be a normal subgroup of IH(R). If there exists an f ∈ N with Fix(f) 6= ∅ is bounded above then Hr ⊂ N. Proof. Let N be a normal subgroup of IH(R) and let f ∈ N such that Fix(f) is bounded above and let t0 = Sup Fix(f), the supremum of Fix(f). If g ∈ N with g(t0) = t0 and h ∈ IH(R) with h(t0) = t0 then h|[t0,∞)◦g|[t0,∞)◦h −1|[t0,∞) is the same as h◦g◦h−1|[t0,∞). Hence N|[t0,∞)= {g|[t0,∞) : g ∈ N, g(t0) = t0} is a normal subgroup of IH([t0,∞)). Since t0 = Sup Fix(f), either f(t) > t or f(t) < t on (t0,∞). That is, f|[t0,∞] ∈ CA([t0,∞]) or CB([t0,∞]). Then by Corollary 2.1, it follows that N|[t0,∞) = IH([t0,∞)). Now let φ ∈ Hr. Choose a fixed point s0 of φ such that every number less than s0 is also a fixed point of φ. Consider χ(t) = t0 −s0 + φ(t−t0 + s0). If τ(t) = t−t0 + s0 then φ◦τ = τ ◦χ. Hence φ is order conjugate to χ. Observe that χ = χ∗t0 . Hence the order conjugate χ of φ is the identity outside [t0,∞). Now χ|[t0,∞) ∈ IH([t0,∞)). Hence there exists χ̃ ∈ N such that χ̃ = χ on [t0,∞). Thus χ̃∗t0 = χ, and hence χ ∈ N. Then φ ∈ N since N is normal. Hence Hr ⊂ N. � Remark 2.8. Let N be a normal subgroup of IH(R). If there exists f ∈ N such that Fix(f) 6= ∅ is bounded below then by considering analogues arguments involved in the proof of Proposition 2.7, we have Hl ⊂ N. Remark 2.9. Let N be a subgroup of IH(R∗). If Fix(f) = {−∞,∞} for all f ∈ N then either f ∈ CA([−∞,∞]) or f ∈ CB([−∞,∞]). Hence by © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 275 K. Ali Akbar and T. Mubeena Proposition 2.3, N = IH(R∗). From this it follows that, if N is a subgroup of IH(R) with Fix(f) = ∅ then N = IH(R). Remark 2.10. If N be a normal subgroup of IH(R) with f ∈ N and t0 ∈ Fix(f), then analogues to Corollary 2.6, we can prove that f∗∗t0 ∈ N. Corollary 2.11. Let t1 < t2 be adjacent fixed points of some f ∈ N where N is a normal subgroup of IH(R). If g(x) = { f(x) if t1 < x < t2 x otherwise then g ∈ N. Proof. Observe that g = f∗t1 ◦f ∗ t2 −1. Hence g ∈ N by Corollary 2.6. � The following two lemmas are important to prove our main theorem. We consider these lemmas before considering our main theorem. We first make a back ground to complete the proof of following lemma. Let f : R → R ∈ IH(R) with unique fixed point a. Define g : R → R by g(t) = f(a + t) − a for t ∈ R. Then g(0) = 0 if and only if f(a) = a, and g is order conjugate to f by the order conjugacy h(t) = a + t for t ∈ R. By Lemma 2.1, there are only 3 elements in IH(R) with a unique fixed point upto order conjugacy. Let f(x) = { 2x if x ≥ 0 x 2 if x < 0 for x ∈ R and g(x) :=   2x if x ≥ 1 1 2 (3x + 1) if − 1 ≤ x ≤ 1 x−1 2 if x ≤−1 . Observe that f has a unique fixed point at 0 and g has unique fixed point at −1. Then the map g◦f has no fixed points. By Corollary 2.6 and Remark 2.10, it follows that, if N is a non-trivial normal subgroup of IH(R) and contains an element with a unique fixed point then it contains an element without fixed points. Next, let f,g : R → R be such that f(x) = { x2 + 1 if 0 ≤ x ≤ 1 x + 1 otherwise and g(x) = x − 1. Then f and g has no fixed points, and g ◦ f has only two adjacent fixed points 0 and 1. Lemma 2.12. The group H is the smallest non-trivial proper normal subgroup of IH(R). Proof. Let N be a non-trivial normal subgroup of IH(R). Suppose there exists f ∈ N with adjacent fixed points t1 < t2. If φ is an element of IH([t1, t2]) such that φ = f|[t1,t2] then by Corollary 2.11, the extension φ̃ : R → R defined by φ̃ = { φ(x) if t1 < x < t2 x otherwise is also in N. Let h ∈ H = {f ∈ IH(R) : Fix(f)c is bounded}. Without loss of generality assume that t1 be the infimum of Fix(h)c and t2 be the supremum of Fix(h) c. Then t1, t2 ∈ Fix(h). By Lemma 2.2, there exists an order conjugate ĥ of h such that φ̃◦ĥ−1 ∈ N. Then ĥ ∈ N since ĥ ◦ φ̃ ◦ ĥ−1 ∈ N. Which implies h ∈ N. This proves H ⊂ N whenever there exists f ∈ N with adjacent fixed points. By Lemma 2.1, any homeomorphisms on R without fixed points is either order conjugate to x + 1 or to x−1. If N contains an element with a unique fixed point then it contains an element without fixed points. Therefore it follows that N always contains an element with atleast two adjacent fixed points. Hence the proof. � © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 276 On the group of homeomorphisms on R: A revisit Lemma 2.13. Let N be a normal subgroup of IH(R). If f ∈ N such that both Fix(f) and Fix(f)c are not bounded above, then N contains an element such that its set of all fixed points is bounded above. Proof. Let f ∈ N and t0 ∈ R be a fixed point of f. By Corollary 2.6, f∗t0 ∈ N. Let {(aα,aα+1)}α be the collection of all intervals of R such that f∗t0 (aα) = aα, f∗t0 (aα+1) = aα+1 and either f ∗ t0 (t) > t or f∗t0 (t) < t for all t ∈ (aα,aα+1). Consider a collection of intervals {(bα,bα+1)}α of R such that bα < aα < bα+1 < aα+1 for all α and an increasing homeomorphism g on R which is order conjugate to f∗t0 such that g(bα) = bα, g(bα+1) = bα+1 and either g(t) > t or g(t) < t for all t ∈ (bα,bα+1). This is possible since IH([a,b]) is isomorphic to IH([c,d]) for any intervals [a,b] and [c,d]. Without loss of generality we can assume that f∗t0 and g do not have common fixed points which are greater than t0. Then f ∗ t0 ◦g ∈ N and Fix(f∗t0 ◦g) is bounded above. Hence the proof. � Remark 2.14. Let N be a normal subgroup of IH(R). (1) If there exists f ∈ N such that Fix(f) and Fix(f)c are not bounded above. Then Hr ⊂ N by Proposition 2.7 and Lemma 2.13. (2) If there exists f ∈ N with Fix(f) and Fix(f)c are not bounded be- low then Hl ⊂ N. This follows by considering analogous arguments involved in the proof of Lemma 2.13 and by Remark 2.8. Now we are ready to prove our main theorems: Theorem 2.15. The group IH(R) has exactly five normal subgroups. They are: (1) The whole group IH(R) (2) The trivial group {1} (3) Hl (4) Hr (5) H = Hl ∩Hr. Proof. Let N be a non-trivial normal subgroup of the group IH(R). Suppose that the function x+1 is in N. We claim that N = IH(R). Consider a function f ∈ IH(R), and let g = (f∨(x−1))∧(x+ 1), where ∨, ∧ denote the maximum, and minimum of functions respectively. Then g ∈ IH(R) and it is at a distance ≤ 1 from the diagonal. Then Fix(f) = Fix(g), and f(x)−x and g(x)−x have the same sign between any two adjacent fixed points. Hence g is order conjugate to f. Now it is enough to prove g ∈ N. As g(x) ≥ x−1, ∀x ∈ R, we have g(x) + 2 > x ∀x ∈ R and therefore the function g(x) + 2 is order conjugate to the function φ(x) = x+ 1, since any two elements of IH(R) whose graphs are above the diagonal are order conjugate. Then g + 2 ∈ N and hence φ−1 ◦φ−1 ◦ (g + 2) = g ∈ N, where (g + 2)(x) = g(x) + 2 for x ∈ R. But f is order conjugate to g. Therefore f ∈ N. Thus N = IH(R). Similarly, if x− 1 is in N then also we can prove that N = IH(R). If there is an element f in N without fixed point then N = IH(R). This is because f is order conjugate to either x− 1 or x + 1 by Lemma 2.1. © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 277 K. Ali Akbar and T. Mubeena Now, let φ ∈ N be such that it has arbitrarily large fixed points and arbitrar- ily large non-fixed points (that is, Fix(φ) and Fix(φ)c are not bounded above). Then by Lemma 2.13 and by Remark 2.14, Hr ⊂ N. Analogues to the above claim, if there exists φ ∈ N such that Fix(φ) and Fix(φ)c are not bounded below then Hl ⊂ N by Remark 2.14. Now suppose Fix(φ)c is bounded below for every φ ∈ N. Then N ⊂ Hr. Also if Fix(φ)c is bounded above for every φ ∈ N then N ⊂ Hl. Therefore from the following table we conclude that either N ⊂ Hl or Hr ⊂ N. (1) There exists ψ ∈ N such that Then Hr ⊂ N Fix(ψ) is bounded above (2) There exists φ ∈ N such that Then (1) by Lemma 2.13 neither Fix(φ) nor Fix(φ)c is bounded above (3) For every φ ∈ N, Fix(φ)c is bounded above Then N ⊂ Hl Similarly by considering the following table analogues to the above table, we can show that N ⊂ Hr or Hl ⊂ N. (1) There exists ψ ∈ N such that Then Hl ⊂ N Fix(ψ) is bounded below (2) There exists φ ∈ N such that Then (1) by Remark 2.14 neither Fix(φ) nor Fix(φ)c is bounded below (3) For every φ ∈ N, Fix(φ)c is bounded be- low Then N ⊂ Hr Now there are only four possibilities for a non-trivial normal subgroup N of IH(R) : Case: 1 N ⊂ Hl and N ⊂ Hr In Case: 1, N ⊂ Hl ∩Hr = H. Therefore N = H. Case: 2 N = Hl Case: 3 N = Hr Case: 4 Hr ⊂ N and Hl ⊂ N In Case 4, Hr ∪ Hl ⊂ N. Let f(x) := { x if x ≥ 0 1 2 x if x < 0 and g(x) :=  2x if x ≥ 1 3x+1 2 if − 1 ≤ x ≤ 1 x if x ≤−1 . Then f ∈ Hl and g ∈ Hr. Hence g ◦f ∈ N and it has no fixed points. Therefore x + 1 ∈ N. Hence N = IH(R). © AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 278 On the group of homeomorphisms on R: A revisit By Lemma 2.12, H is the smallest normal subgroup contained in IH(R). Hence the other two cases Hr ⊂ N ⊂ Hl and Hl ⊂ N ⊂ Hr are not possible. This completes the proof. � Remark 2.16. Let N be a non-trivial normal subgroup of the group IH(R) and suppose that there is an element in N such that it has only two fixed points. Let φ be in N such that it has only two fixed points namely a and b, a < b. Consider R∗ = R ∪{−∞,∞} with order topology. If φ ∈ IH(R∗) fixes only −∞ and ∞ then φ|R ∈ IH(R) and it has no other fixed points. Let N∗ be a normal subgroup of IH(R∗) containing a map, fixing only −∞ and ∞. Consider N∗ = {f|R : f ∈ N∗}. Then N∗ is a normal subgroup of IH(R) and φ|R ∈ N∗. Then φ|R has no fixed points and hence N∗ = IH(R). Therefore N∗ = IH(R∗). Then N contains an element without fixed points since IH(R), IH(R∗) and IH([a,b]) are isomorphic. Hence N = IH(R). In this case, note that Hl ∪Hr ⊂ N, and therefore N becomes IH(R) since there is an element in N without fixed points. For a subgroup H of a group G, we denote H ≤ G. Theorem 2.17. For H(R) there are exactly four normal subgroups. They are: (1) The whole group H(R). (2) The trivial group {1}. (3) H = {f ∈ IH(R) : Fix(f)c is bounded}. (4) IH(R). Proof. Let G be a group and K ≤ N ≤ G. If K is normal in G then K is normal in N also. The subgroups Hl and Hr are not normal in H(R). Hence by Theorem 2.15, if N = IH(R) and G = H(R), then either K = {1} or K = H or K = IH(R). Next suppose IH(R) ≤ K ≤ H(R). Then the index [H(R) : K] ≤ [H(R) : IH(R)] = 2. Therefore either K = H(R) or K = IH(R). Finally, suppose there is a normal subgroup N of H(R) such that N = A∪B with A is a proper subgroup of IH(R) and ∅ 6= B ⊂ H(R) \ IH(R). Which implies IH(R)∪B is a normal subgroup of H(R). But there is no such normal subgroup for H(R) other than H(R) itself. Hence the proof. � Acknowledgements. The authors are very thankful to the referee for giving valuable suggestions. The first author acknowledges SERB-MATRICS Grant No. MTR/2018/000256 for financial support. The second author acknowl- edges University of Calicut, Seed Money (U.O. 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