Let a 1 , . . . , a n be a permutation of [n]. Two disjoint order-isomorphic subsequences are called twins. We show that every permutation of [n] contains twins of length Ω(n 3/5 ) improving the trivial bound of Ω(n 1/2 ). We also show that a random permutation contains twins of length Ω(n 2/3 ), which is sharp.In this paper we regard permutations as sequences of symbols, devoid of any group-theoretic meaning. So, for us a permutation on a finite set Σ is a sequence of elements of Σ in which each element of Σ appears exactly once. We call a subsequence of a permutation subpermutation. For instance, 135642 is a permutation of [6], and 1562 is a subpermutation inside, which itself is a permutation of {1, 2, 5, 6}. We denote permutations by bold letters.Throughout the paper we consider only the permutations of finite sets of natural numbers. We say that permutations a = (a 1 , . . . , a L ) and b = (b 1 , . . . , b L ) are order-isomorphic if (a i < a j ) ⇐⇒ (b i < b j ). For example, 1562 is order-isomorphic to 1342.We call a pair of subpermutation a, b of c twins if a and b are order-isomorphic and disjoint (do not contain the same symbol). For example, 152 and 364 are twins in 135642 of length 3. We denote by t(n) the largest integer such that every permutation of [n] contains a pair of twins of length t(n).The problem of estimating t(n) was raised by Gawron [5], who observed that t(n) ≥ (n 1/2 − 1)/2 follows from the Erdős-Szekeres theorem, and that t(n) = O(n 2/3 ) follows from the first moment method. He further conjectured that t(n) = Ω(n 2/3 ). This is not known even for random permutations: the best result is due to Dudek, Grytczuk, Ruciński [4] who showed that a random permutation almost surely contains twins of length Ω(n 2/3 / log 1/3 n).In this short note, we give a first non-trivial lower bound on t(n), and remove the logarithmic factor from the Dudek-Grytczuk-Ruciński result.Theorem 1. For n ≥ 2, every permutation of [n] contains twins of length at least 1 8 n 3/5 . Theorem 2. A random permutation of [n] almost surely contains twins of length at least 1 80 n 2/3 , as n → ∞.In view of Gawron's result, Theorem 2 is sharp up to the constant factor.
