American Mathematical Monthly Problem 11395 by M. Farrokhi D.G., Tsukuba, Ibakari, Japan.
Let G be the group of all continuous bijections of [0,1] to itself. If H is a finite subgroup of G, then prove that |H|= 1 or |H|= 2.
Solution by Darij Grinberg.
We begin with a fact that should be rather well-known:
Lemma 1. Let f be a continuous bijection of [0,1] to itself. Then, either f is strictly increasing and satisfies f(0) = 0 and f(1) = 1, orf is strictly decreasing and satisfies f(0) = 1 and f(1) = 0.
Proof of Lemma 1. Define a map R: G→G (the letter R stands for ”reflection”) as follows: For every f ∈G, define the function R(f) : [0,1]→[0,1] by (R(f)) (x) = f(1−x) for every x∈[0,1].Obviously, R is a bijection fromG toG.(Of course,R is not a group homomorphism.)
WLOG we have f−1(0) < f−1(1) (otherwise, replace f byR(f)). By the interme- diate value theorem, for anyx∈[0,1] there exists some u∈[f−1(0), f−1(1)] such that f(u) = x(in fact,x∈[0,1] yields 0≤x≤1,so that f(f−1(0)) ≤x≤f(f−1(1))). In other words, [0,1]⊆f−1([f−1(0), f−1(1)]). Hence, f([0,1])⊆ [f−1(0), f−1(1)]. But f being a bijection, we havef([0,1]) = [0,1],so this becomes [0,1]⊆[f−1(0), f−1(1)]. Sincef−1(0) andf−1(1) lie in the interval [0,1] and satisfyf−1(0) < f−1(1),this can only mean f−1(0) = 0 and f−1(1) = 1, so thatf(0) = 0 and f(1) = 1.
Now take two points a, b∈[0,1] such thata < b. Assume thatf(a)> f (b).Then, f(0) ≤ f(b) < f(a) (in fact, f(0) ≤ f(b) since f(0) = 0 ≤ f(b)), so that by the intermediate value theorem, there exists somec∈[0, a] satisfying f(c) =f(b). Thus, c =b (since f is a bijection), contradicting c ≤ a < b. This contradiction shows that our assumption that f(a) > f(b) was wrong. Since f(a) = f(b) cannot hold either (again by the bijectivity of f, since a < b), we conclude that f(a)< f(b) for any two points a, b∈ [0,1] such that a < b. Thus, f is strictly increasing. Altogether, Lemma 1 is proven.
According to Lemma 1, we can define a map Φ :G→Z(2Z) as follows:
Φ (f) =
0, if f is strictly increasing and satisfies f(0) = 0 andf(1) = 1;
1, if f is strictly decreasing and satisfies f(0) = 1 andf(1) = 0 (where 0 and 1 mean the residue classes of the integers 0 and 1 modulo 2). This map Φ is easily seen to be a group homomorphism. Now comes the main (but surprisingly easy) step of our proof:
Lemma 2. The only finite subgroup of Ker Φ is the trivial group {id}.
Proof of Lemma 2. Assume, for the sake of contradiction, that there exists a non- trivial finite subgroupU of Ker Φ. Then, this subgroupU contains an elementf 6= id. Since U is finite, we have f|U| = id. Since f 6= id, there exists a point x ∈ [0,1]
such that f(x) 6= x. We WLOG assume that f(x) < x (since the case f(x) > x can be handled in exactly the same way). Noticing that f is strictly increasing (since f ∈U ⊆Ker Φ,so that Φ (f) = 0), this yieldsf(f(x))< f (x),so thatf2(x)< f (x). Since f is strictly increasing, this, in turn, leads to f(f2(x)) < f(f(x)), or, equiv- alently, f3(x) < f2(x). By induction, we thus obtain fk+1(x) < fk(x) for every
1
k ∈ {0,1,2, ...}. Thus, f|U|(x) < f|U|−1(x) < ... < f1(x) < f0(x) = id (x), contra- dicting f|U| = id. This contradiction concludes the proof of Lemma 2.
Now the problem becomes trivial: LetHbe a finite subgroup ofG.Then,H∩Ker Φ is a finite subgroup of Ker Φ, so that H∩Ker Φ = {id} by Lemma 2, what rewrites as Ker (Φ|H) = {id}, so that Φ |H: H → Z(2Z) is an injective map. Thus, |H| ≤
|Z(2Z)|= 2,so that |H|= 1 or |H|= 2, qed.
2
