Groups with Identical k-Profiles

We show that for 1≤ k ≤ √ 2log3 n− (5/2), the multiset of isomorphism types of k-generated subgroups does not determine a group of order at most n. This answers a question raised by Tim Gowers in connection with the Group Isomorphism problem. ACM Classification: F.2.2 AMS Classification: 68Q17, 20D15, 20F69, 68Q25


Introduction
We say that a group is k-generated if it has a set of at most k generators.Let G k be the set of isomorphism types 1 of all k-generated finite groups.Let G be a finite group.Following Gowers [3], we say that the k-profile of G is the function f G : G k → N defined by letting f G (H) be the number of subgroups of G isomorphic to H (H ∈ G k ).
Tim Gowers raised the question [3], for which k does the k-profile determine a group of order n ?Such a k yields a simple isomorphism test 2 in time n O(k) for groups of order n given by their Cayley tables (see Section 3). 1 Two groups belong to the same isomorphism type if and only if they are isomorphic. 2Regarding the significance of the Group Isomorphism problem to the Graph Isomorphism problem we refer the reader to Section 13 of [1] and especially to footnote 9 in that section.
Theorem 1.1.If p is an odd prime, k and n are positive integers, and then there exist nonisomorphic p-groups of order at most n with identical k-profiles.
Remark 1.2.In particular, setting p = 3, we see that if k and n are positive integers such that 1 ≤ k ≤ 2 log 3 n − (5/2), then there exist nonisomorphic groups of order at most n with identical k-profiles.
Our examples are p-groups of class 2 and exponent p.
Theorem 1.3.For any odd prime p and positive integer k there exist nonisomorphic p-groups of class 2, exponent p, and order p N , where N = (k + 2)(k + 3)/2, with identical k-profiles.

The proof
Recall that a nilpotent group  Proof.Let x = m − k − 2, so x ≥ 0 and we wish to show that f (x) ≥ 0 where But then f (x) = x 2 + 3x ≥ 0, as desired.
Fact 2.2.For an odd prime p and a positive integer k we have Proof.
(i) p is an odd prime, (ii) m is a positive integer, and (iii) P is a relatively free group with m generators, class two, and exponent p. Note.This is false for k = 2 and m = k + 1 = 3.
Proof.In this situation, P = Z(P), |P/P | = p m , and |P | = p m(m−1)/2 .We claim that for every k-generated subgroup Q of P, there exists a k-generated subgroup Indeed, let Q be a k-generated subgroup of P and The number of distinct subgroups of the form RP is the same as the number of k-dimensional subspaces of an m-dimensional vector space over the prime field F p .Call this number N(m, k).Then Clearly, the numerator of N(m, k) is less than p mk .By Fact 2.2, the denominator is greater than Now we count the elements of P that lie in Q for some k-generated subgroup Q of P. Each such element lies in (RP ) for some subgroup RP as above.So we obtain the upper bound for e = (k We saw above that |P | = p m(m−1)/2 .Fact 2.1 shows that This gives the desired conclusion.To obtain (c), let s 1 , . . ., s d be d elements of P. Set R = s 1 , . . ., s d .Then there exist unique elements u 1 , . . ., u d of P 1 such that u −1 i s i ∈ w for each i, and R = Q where Q = u 1 , . . ., u d .By the choice of v, we see that v ∈ R .As R ≤ P 1 , we have R ∩ v, w = 1.
Lemma 2.6.Assume the hypothesis and notation of Lemma 2.5.Then there exists a bijection between the set of all d-generated subgroups of P/ v and the set of all d-generated subgroups of P/ w such that corresponding subgroups are isomorphic.
Proof.Consider a d-generated subgroup Recall that v and w are in Z(P).So By Lemma 2.5 we infer For a d-generated subgroup R of P/ w , we obtain analogous subgroups R * , R 0 , R * * of P. Note that Q and R uniquely determine Q * * and R * * .Now consider the family of all subgroups S of P such that (i) v and w are in S, and (ii) S = S 0 , v, w for some d-generated subgroup S 0 of S.
The analysis above shows that to prove Lemma 2.6, it suffices to obtain, for each subgroup S as above, a bijection between • the set of all d-generated subgroups Q of P/ v for which Q * * = S and • the set of all d-generated subgroups R of P/ w for which R * * = S such that corresponding subgroups Q and R are isomorphic.
For each subgroup S, we have S ∩ v, w = S 0 ∩ v, w = 1 by Lemma 2.5.Since P has exponent p and S/S is abelian, there exists a complement S 1 /S to S , v, w /S in S/S .Since S , v, and w are central, we have S = S 1 × v, w .Therefore, there exists a unique automorphism of S that induces the identity on S 1 and switches v and w.This establishes the desired bijection.
The groups P/ v and P/ w have order |P|/p.
By Theorem 1.3, there exist nonisomorphic groups of order p N with identical k-profiles.
Remark 2.7.We comment on the case k = 1.It is obvious that p-groups of exponent p of equal order have the same 1-profile.In particular, for every odd prime p there exist nonisomorphic p-groups of order p 3 with the same 1-profile.Moreover, for all primes p there exists a nonabelian group of order p 4 with a cyclic subgroup of order p 3 called M 4 (p), which has the same 1-profile as the direct product of a cyclic group of order p 3 and the cyclic group of order p. (For the definition of M 4 (p) see the classification of p-groups with a cyclic subgroup of index p in [2, pp. 192-193].)In particular, M 4 (2) has order 16, improving Remark 1.2 for k = 1.

The isomorphism test
We describe the isomorphism test based on k-profiles suggested by Gowers [3].Proposition 3.1.Let k, n be positive integers and suppose the groups of order n are determined, up to isomorphism, by their k-profiles.Then isomorphism of two groups of order n, given by their Cayley tables, can be decided in time n 2k+O (1) .
Proof.Let G, H be two groups of order n.By our assumption, G and H are isomorphic if and only if their k-profiles agree, so we only need to show how to compare the k-profiles of the two groups.This can be done by computing the following equivalence relation on the disjoint union X := G k ∪ H k .We say that two k-tuples (x 1 , . . ., x k ) ∈ X and (y 1 , . . ., y k ) ∈ X are equivalent if the correspondence x i → y i extends to an isomorphism of the subgroups generated by these k-tuples.This can be checked in polynomial time per instance, so n 2k+O(1) total time.Now the k-profiles of G and H agree if and only if each equivalence class is evenly divided between G k and H k .Remark 3.2.While our result shows that the comparison of k-profiles alone will not solve the Group Isomorphism problem in polynomial time, it does not rule out a role for this algorithm in improving the state of the art in this area.Indeed, Group Isomorphism is not currently known to be testable in time n o(log n) (cf.[4,6,5,7]).Therefore, if our bound on k is not very far from being tight, say the result stated in Remark 1.2 would fail if we replace 2 log 3 n by O((log n) 0.99 ), this would mean progress on the complexity of the Group Isomorphism problem.ŁUKASZ GRABOWSKI graduated from the University of Göttingen in 2011 under the supervision of Andreas Thom and Thomas Schick.He is currently a research associate at the University of Warwick, working in combinatorics and group theory.His aim for this year is to be a little bit like Stephen Curry.
THEORY OF COMPUTING, Volume 11 (15), 2015, pp.395-401 where G denotes the commutator subgroup G = [G, G] and Z(G) denotes the center of G.For an odd prime p, a relatively free p-group P of class 2 and exponent p with m generators can be obtained from a free group with m generators by factoring out all elements u p and all commutators [[u, v], w].

Lemma 2 . 5 .
Assume Hypothesis 2.3 for a group P 1 in place of P. Let d be a positive integer such that m ≥ d + 2. Let P 2 = w be a cyclic group of order p and P = P 1 × P 2 .Then there exists an element v of P 1 such that (a) | v, w | = p 2 , (b) P/ v is not isomorphic to P/ w , and (c) for every d-generated subgroup Q of P we have Q ∩ v, w = 1.Proof.By Lemma 2.4, P 1 has an element v that does not lie in Q for any d-generated subgroup Q of P. Then (a) is obvious.We obtain (b) because(P/ v ) = P 1 / v and (P/ w ) ∼ = P 1 .(3)THEORYOF COMPUTING, Volume 11 (15), 2015, pp.395-401