Abstract 1 Introduction 2 Background on Finite Abelian Groups

Testing Isomorphism of Boolean Functions over Finite Abelian Groups

Swarnalipa Datta ORCID Indian Statistical Institute, Kolkata, India Arijit Ghosh ORCID Indian Statistical Institute, Kolkata, India Chandrima Kayal ORCID Université Paris Cité, CNRS, IRIF, France Manaswi Paraashar ORCID University of Copenhagen, Denmark Manmatha Roy ORCID Indian Statistical Institute, Kolkata, India
Abstract

Let f and g be Boolean functions over a finite Abelian group 𝒢, where g is fully known and f is accessible via queries; that is, given any x𝒢, we can obtain the value f(x). We study the problem of tolerant isomorphism testing: given parameters ϵ0 and τ>0, the goal is to determine, using as few queries as possible, whether there exists an automorphism σ of 𝒢 such that the fractional Hamming distance between fσ and g is at most ϵ, or whether for every automorphism σ, the distance is at least ϵ+τ.

We design an efficient tolerant property testing algorithm for this problem over finite Abelian groups with constant exponent. The exponent of a finite group refers to the largest order of any element in the group. The query complexity of our algorithm is polynomial in s and 1/τ, where s bounds the spectral norm of the function g, and τ is the tolerance parameter. In addition, we present an improved algorithm in the case where g is Fourier sparse, meaning that its Fourier expansion contains only a small number of nonzero coefficients.

Our approach draws on key ideas from Abelian group theory and Fourier analysis, including the annihilator of a subgroup, Pontryagin duality, and a pseudo inner product defined over finite Abelian groups. We believe that these techniques will be useful more broadly in the design of property testing algorithms.

Keywords and phrases:
Analysis of Boolean functions, Abelian groups, Automorphism group, Function isomorphism, Spectral norm
Category:
RANDOM
Funding:
Arijit Ghosh: A. G. is partially supported by the Science & Engineering Research Board of the DST, India, through the MATRICS grant MTR/2023/001527.
Chandrima Kayal: C.K. is supported by French PEPR integrated project EPiQ (ANR-22-PETQ-0007).
Manaswi Paraashar: M.P. is supported by ERC grant (QInteract, Grant Agreement No 101078107).
Copyright and License:
[Uncaptioned image] © Swarnalipa Datta, Arijit Ghosh, Chandrima Kayal, Manaswi Paraashar, and Manmatha Roy; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation Boolean function learning
Related Version:
Full Version: https://arxiv.org/abs/2507.07654 [20]
Editors:
Alina Ene and Eshan Chattopadhyay

1 Introduction

Let 𝒢 be a finite Abelian group, and let 𝒢^ be its dual group, consisting of all characters, that is, homomorphisms from 𝒢 to ×;={x|x|=1}. The Fourier coefficient for a function f:𝒢 corresponding to a character χ𝒢 is denoted by f^(χ)111Note that Fourier coefficients are complex numbers., and the spectral norm of f is defined as

f^1:=χ𝒢^|f^(χ)|.

For more details refer to Section 2. The spectral norm is a key parameter in the analysis of Boolean functions, with wide-ranging applications in property testing, learning theory, cryptography, pseudorandomness, and quantum computing [43, 38, 46, 28, 44, 39, 13, 42, 27, 41, 45, 4]. Moreover, it plays an important role in understanding the coset complexity of subsets of Abelian groups [17, 31, 32, 15]. We will show that, for the isomorphism testing of Boolean functions over finite Abelian groups, the query complexity is essentially controlled by the spectral norm.

The area of property testing is concerned with determining whether a given object satisfies a fixed property or is “far” from satisfying it, where “far” is typically defined using a suitable distance measure. Over nearly three decades of research, symmetry has played a crucial role in property testing, appearing in various forms.

To illustrate the setup and the role of symmetry, we consider property testing of Boolean functions, which are functions of the form f:{1,1}n{1,1}. The fractional Hamming distance between two Boolean functions f and g on n-bit inputs is defined as

δ(f,g)=Prx{1,1}n[f(x)g(x)],

that is, the fraction of inputs on which they differ.

Let Sn be the group of all permutations of [n]:={1,,n}. For σSn, we define the permuted function fσ by

(fσ)(x1,,xn):=f(xσ(1),,xσ(n))

for all (x1,,xn){1,1}n. A fundamental question in this setting is the following.

Question 1 (Function Isomorphism Testing over {1,1}n).

Given a known Boolean function f and query access to an unknown Boolean function g, determine if there exists a permutation σ𝒢 such that fσ=g, or for all σ𝒢, is the Hamming distance between fσ and g at least ϵ, under the promise that one of these is the case?

The above question was first investigated by Fischer et al. [26] and was followed by a long line of work [1, 8, 11, 10, 2, 9].

The domain of Boolean functions discussed so far is the set {1,1}n. It is of great interest, as we discuss later, to study Boolean functions whose domain 𝒟 has some algebraic structure associated with it, like 2n. In these settings, it is natural to study the isomorphism problem for Boolean functions under the group of symmetry of 𝒟 that preserves its underlying algebraic structure.

An extremely important example is that of 𝒟=2n, that is, functions of the form f:2n{1,1} defined over the vector space 2n. These functions arise in numerous areas of computer science and mathematics, see [40] and the references therein for several examples. A natural group of symmetry for 2n is the group of all non-singular n×n matrices over 2(see [22, Chapter 3]). We now describe the question of Linear Isomorphism Testing of Boolean Functions over 2n. For A2n×n, let fA:2n{1,1} be the function fA(x)=f(Ax) for all x2n. The Linear Isomorphism Distance between two functions f:2n{1,1} and g:2n{1,1} is defined as

dist2n(f,g)=minA2n×n:A is non-singularδ(fA,g). (1)

Assume that f and g satisfy the promise that either dist2n(f,g)=0 or dist2n(f,g)ϵ, the question of Linear Isomorphism Testing is that of deciding which is the case.

In this work, we study Boolean functions over a finite Abelian group 𝒢. The natural symmetries in this setting arise from the automorphism group of 𝒢, denoted Aut(𝒢) (see Definition 31). For functions f,g:𝒢{0,1}, we define their automorphism distance as

dist𝒢(f,g)=minσAut(𝒢)δ(fσ,g),

where fσ(x)=f(σ(x)) for all x𝒢. We investigate the following general question.

Question 2 (Testing Isomorphism of Boolean Functions over Finite Abelian Groups).

Let 𝒢 be a finite Abelian group, and let f,g:𝒢{1,1} be Boolean functions such that f is known and given query access to g, determine if f and g are isomorphic under the automorphism group Aut(𝒢) of 𝒢 (that is, dist𝒢(f,g)=0) or is the automorphism distance between them at least ϵ (that is, dist𝒢(f,g)ϵ), under the promise that one of these is the case?

We now give an equivalent problem in the setting of finite Abelian groups.

Definition 3 (Subset Equivalence Problem).

Given two subsets A,B𝒢 of a finite Abelian group 𝒢, does there exist an automorphism σAut(𝒢) such that A=σ(B)?

To formulate a testing version of this problem, we introduce the following definitions. For subsets A,B𝒢, the distance between A and B is defined by

dist(A,B):=|AB|+|BA||𝒢|. (2)

We define the automorphism distance between A and B as

dist𝒢(A,B):=minσAut(𝒢)dist(A,σ(B)). (3)

We assume membership query access to a subset of 𝒢, meaning that we can query whether any given element of 𝒢 belongs to the subset. The testing version of the Subset Equivalence Problem is formulated below:

Question 4 (Testing Automorphism Distance Between Subsets of Finite Abelian Groups).

Let ϵ>0 and 𝒢 be a finite Abelian group. Let A𝒢 be a known subset, and suppose we have membership query access to an unknown subset B𝒢. Under the promise that either dist𝒢(A,B)=0 or dist𝒢(A,B)ϵ, the goal is to determine which of the two cases holds.

Question 4 is equivalent to the problem of testing can be viewed as a special case of Boolean function isomorphism testing as follows. Define the indicator function 𝟙A:𝒢{1,1} of the set A by

𝟙A(x)={1if xA,1otherwise.

Similarly, define 𝟙B for the set B. Then Question 4 reduces to testing whether the Boolean functions 𝟙A and 𝟙B are isomorphic under the action of Aut(𝒢).

Previous works on property testing of Boolean functions over 2n critically exploit the Fourier analysis framework, where the vector space structure of the domain is essential to the proofs. However, since we are working with arbitrary finite Abelian groups, these established approaches that rely on the vector space structure do not apply. As a result, we had to develop new ideas, drawing on key concepts from the theory of finite Abelian groups. We outline some of these ideas below:

  • We have used a notion of “orthogonal” subgroup H for a subgroup H of a finite Abelian group 𝒢 which is not naturally defined in the context of groups. To address this, we used the concept of annihilator of a subgroup. This concept is described in more detail in Section 3.

  • For any group automorphism A:𝒢𝒢 and any function f:𝒢{1,+1}, we needed a natural way to associate the Fourier coefficients {fA^(χ):χ𝒢^} for the function fA with the Fourier coefficients {f^(χ):χ𝒢^} for f. We use ‘Pontryagin duality’ to deal with such scenario, which is of independent interest. More details can be found in Lemma 34 and Lemma 38 from Section 2.

  • The vector space 2n can be naturally partitioned into cosets of a subspace. This property is essential in approaches like “coset hashing”, which is used in algorithms for property testing of Boolean functions over such domains (see [25, 30, 47]). However, for a finite Abelian group 𝒢, this partitioning is not generally possible, as 𝒢 may not form a vector space. To address this, we instead use normal subgroups and partition 𝒢 using the cosets of these normal subgroups. For details, see Section 2.

  • We use a new notion of inner product, called pseudo-inner product (see Definition 22) to analyze our algorithms.

  • We also introduce a notion of independence for finite Abelian groups, which we call pseudo-independence (see Definition 32), to analyze our algorithms.

  • Note that the characters of 2n are the 2nd roots of unity, meaning they take the values 1 or +1 at each point in 2n. This property is fundamental to the analysis of Boolean functions over 2n, see, for example, Gopalan et al. [30]. Since 𝒢 is a finite Abelian group, it can be written as 𝒢=p1m1××pnmn, where pi are primes for all i[n], which may not necessarily be distinct. In the case of 𝒢, the characters are complex numbers, as they correspond to the -th roots of unity at each point in 𝒢, where is the exponent of 𝒢. Observe that =LCM{p1m1,,pnmn}. Consequently, there is no natural concept of sign, which we address using a completely different approach. For more details, see [20], which we believe offers independent interest in the study of Boolean functions on more general domains.

1.1 Related work

Fischer et al. [26] were the first to investigate the problem of function isomorphism testing, a line of research that has since been extended by several subsequent works [1, 8, 11]. The problem of characterizing the testability222See the definition of testability from Bhattacharyya et al. [6]. of linearly invariant properties of functions has been a central focus in property testing. This question has seen significant progress through various works [33, 37, 7]. Also, Bhattacharyya et al. [6] provided a characterization of linearly invariant properties that are constant-query testable with one-sided error. Testing linear isomorphism of Boolean functions over 2n was studied by Wimmer and Yoshida [47], whose algorithm has polynomial query complexity in terms of the spectral norm of the functions. Grigorescu, Wimmer and Xie [35] studied the same problem in the communication setting. Linear isomorphism of Boolean functions has also been explored in the context of combinatorial circuit design [16, 5, 48], error-correcting codes [21, 24, 36], and cryptography [18, 14, 23].

1.2 Our results

We study Question 2 in the following setting: the function g:𝒢{1,1} is known, while the function f:𝒢{1,1} is unknown. We are promised that either dist𝒢(f,g)ϵ or dist𝒢(f,g)ϵ+τ. Additionally, we assume that the exponent of the finite Abelian group 𝒢 is a constant. For the remainder of the paper, we denote the exponent of 𝒢 by (𝒢), or simply . Observe that if 𝒢=p1m1××pnmn, then

(𝒢)=LCM{p1m1,,pnmn}.

The goal is to design a randomized algorithm that, with probability at least 2/3 over its internal randomness, determines which case holds while making as few queries as possible to the unknown function f. Each query allows the algorithm to obtain the value f(x) for a chosen x𝒢.

Our results are developed in the context of tolerant testing, specifically in the (ϵ,τ)-tolerant testing model. This generalizes the setting of Question 2, which corresponds to the special case of (0,τ)-tolerant testing.

We give an efficient algorithm for the above problem. The algorithm’s query complexity is polynomial in (1/τ) (independent of ϵ) and the spectral norm (see Definition 20) of the known function.

Theorem 5 (Main Result: (ϵ,τ)-Tolerant Isomorphism Testing of Boolean Functions over Finite Abelian Groups).

Let 𝒢 be a finite Abelian group, with

𝒢=p1m1××pnmn,

where each pi is a prime (not necessarily distinct), and let ϵ0 and τ(0,1/2]. Suppose we are given a known function g:𝒢{1,1} satisfying g^1s, and have query access to an unknown function f:𝒢{1,1}.

Assuming that the exponent of 𝒢 is constant, there exists a randomized algorithm that decides whether

dist𝒢(f,g)ϵordist𝒢(f,g)ϵ+τ

with probability at least 2/3, using O~(s8τ8) queries to f, where the O~() notation hides factors that are polynomial in logs and log(1/τ).

Note that Wimmer and Yoshida [47] proved Ω(logg^1)333Wimmer and Yoshida claimed an Ω(g^1) lower bound for the case when ϵ=0 and τ>0. However, their proof applies only to the special case τ=1g^1. Even if their argument can be extended to general values of τ, the lower bound established by Datta et al. [19] is quadratically stronger. lower bound for the special case of 𝒢=2n. Recently, Datta et al. [19] have been able to show a lower bound of Ω(g^12) for the above problem when 𝒢=2n.

The proof of the above theorem generalizes the work of Wimmer and Yoshida [47], who solved the (ϵ/3,2ϵ/3)-tolerant linear isomorphism problem for Boolean functions over 2n. Their testing algorithm and analysis crucially rely on the fact that 2n is a vector space over the field 2. However, finite Abelian groups do not necessarily have a vector space structure. Due to this, we had to introduce several new ideas and incorporate key concepts from the theory of finite Abelian groups. For more details, see Section 1.3 and Section 1.4.

Finite Abelian groups 𝒢 which are of the form G1×G2××Gn have been used in various works, namely [34, 3]. We give the following two corollaries, which immediately follow from Theorem 5.

Corollary 6.

Let 𝒢 be a finite Abelian group such that 𝒢 can be expressed as G×G××G where the product takes place n-times and G itself is a finite Abelian group of constant size. Also, let ϵ,τ(0,1/2]. Given a known function g:𝒢{1,1} with g^1s, and query access to a unknown function f:𝒢{1,1}, the query complexity of deciding whether dist𝒢(f,g)ϵ or dist𝒢(f,g)ϵ+τ with probability at least 2/3, is O~(poly(s,1/τ)), where O~() hides multiplicative factors that are polynomial in logs and log(1/τ).

More generally we get the following:

Corollary 7.

Let 𝒢 be a finite Abelian group such that 𝒢 can be expressed as G1×G2××Gn where i[n] and |Gi|=O(1) for all i[n]. Also let ϵ,τ(0,1/2]. Given a known function g:𝒢{1,1} with g^1s, and query access to a unknown function f:𝒢{1,1}, the query complexity of deciding whether dist𝒢(f,g)ϵ or dist𝒢(f,g)ϵ+τ with probability at least 2/3, is O~(poly(s,1/τ)), where O~() hides factors that are polynomial in logs and log(1/τ).

The main technical result we require is the following algorithm to estimate the values of the large Fourier coefficients of f:𝒢{1,1}, given query access to f, without prior knowledge of the coefficients themselves. The set of large Fourier coefficients is defined using a threshold (see Theorem 8). Note that this set is unknown, as the function f is not known.

Theorem 8 (Generalized Implicit Sieve for Abelian groups).

Let 𝒢 be a finite Abelian group with constant exponent. Also, let θ>0 be a threshold and ={xi}i=1m~𝒢 be a set of independently and uniformly chosen m~ random points from 𝒢. There exists an algorithm, Generalized Implicit Sieve Algorithm (see [20]), that takes θ and as input, makes O(ln(m~/θ16)θ8+m~ln(𝒩m~)θ2+ln𝒩θ8) queries to the truth table of a Boolean valued function f:𝒢{1,1}, and returns, with probability at least 94100, a labeled set of m~ examples, and the value of the function f at xi for each i[m~], of the form {χα1(xi),,χα𝒩(xi),f(xi)xi}, where 𝒮={α1,,α𝒩} is some set that satisfies the following two properties:

  1. 1.

    α𝒢 with |f^(χα)|θ then α𝒮, and

  2. 2.

    α𝒮, |f^(χα)|θ/2.

The output of this algorithm can be seen as a m~×(𝒩+1) matrix Q with entries in powers of ω, where ω is a primitive th root of unity with being the exponent of 𝒢.

 Remark 9 (Comments on the query complexity in Theorem 8).

Note that the query complexity of the algorithm in Theorem 5 satisfies the following:

  1. 1.

    Query complexity is independent of |𝒢|, and is assumed to be constant, and

  2. 2.

    it is polynomial in m~, 1/θ and ln𝒩.

Wimmer and Yoshida [47] proved a special case of the above theorem, which they referred to as the Implicit Sieve Algorithm for functions over 2n and they used for testing linear isomorphism testing. Theorem 8 generalizes the Implicit Sieve algorithm of Wimmer and Yoshida to finite Abelian groups. Note that the Implicit Sieve Algorithm of Wimmer and Yoshida [47] is itself a generalization of the celebrated and widely applicable Goldreich-Levin algorithm [29].

Using Theorem 5 we have the following result for the case of subset isomorphism problem.

Theorem 10.

Let 𝒢 be a finite Abelian group with constant exponent, and A,B𝒢. Also let ϵ,τ(0,1/2]. Given the known set A𝒢 with 𝟙A^1s, and membership query access to an unknown set BG, the query complexity of deciding whether dist𝒢(A,B)ϵ or dist𝒢(A,B)ϵ+τ with probability at least 2/3, is O~(s8τ8), where O~() hides multiplicative factors polynomial in logs and log(1/τ).

The above result is part of a broader research program exploring the interplay between the spectral norm of indicator functions and the structure of subsets of finite Abelian groups. In additive combinatorics and additive number theory, the spectral norm serves as a key tool for measuring the non-uniformity or pseudorandomness of functions defined on finite Abelian groups or their vector spaces. Notable breakthroughs in this area include the quantitative version of Cohen’s Idempotent Theorem due to Green and Sanders [32], as well as Chang’s Lemma [12] and its various extensions.

We also prove that for Fourier sparse functions, testing can be performed more efficiently. The proof follows a similar approach to the theorem on tolerant isomorphism testing for Finite Abelian Groups (Theorem 5), with a key observation: when both functions are Fourier s-sparse, the number of nonzero Fourier coefficients is at most s, which significantly simplifies the learning process in the implicit sieve algorithm (Theorem 8). In particular, we do not need to estimate wt4 (see [20]), which is required in the general case.

Theorem 11 ((ϵ,τ)-Tolerant Isomorphism Testing of Sparse Boolean Functions over Finite Abelian Groups).

Let ϵ0 and τ(0,1/2], and let 𝒢 be a finite Abelian group of constant exponent. Given a known function g:𝒢{1,1} with sparsity s and query access to a unknown s-sparse Boolean valued function f:𝒢{1,1}, the query complexity of deciding whether dist𝒢(f,g)ϵ or dist𝒢(f,g)ϵ+τ with probability at least 2/3, is O~(s4τ4), where O~() notation hides multiplicative factors polynomial in logs and loglog(1/τ).

1.3 Proof sketch

We start by giving the ideas behind and a sketch of the proof of our main result, Theorem 5. Throughout this section 𝒢 is assumed to be a finite Abelian group.

Proof sketch of Theorem 5 (Isomorphism Testing for finite Abelian groups).

Given a known function g:𝒢{1,1} and an unknown function f:𝒢{1,1}, where either

dist𝒢(f,g)ϵordist𝒢(f,g)ϵ+τ.

The goal is to determine which case holds with probability at least 2/3. The algorithm has query access to f, meaning it can evaluate f(x) for any x𝒢.

  • Our algorithm (see [20]) uses the Generalized Implicit Sieve Algorithm (see [20] and Theorem 8) as a subroutine. Given query access to f, m~ random elements x1,,xm~𝒢 and a parameter θ, the Generalized Implicit Sieve Algorithm returns an m~×(𝒩+1) matrix Q which satisfies the following with high probability (ignoring the last column of Q for simplicity):

    1. 1.

      For some unknown set 𝒮={r1,,r𝒩}, the (i,j)th entry of Q is χrj(xi). Thus the jth column of Q contains the evaluations of the characters corresponding to rj on m~ random points in 𝒢.

    2. 2.

      If r𝒮 then |f^(χr)|θ/2 and if r satisfies |f^(χr)|θ, then r𝒮.

  • Since

    f^(χri)=𝔼x𝒢[f(x)χri(x)]

    for all ri𝒮 (see Section 2), the output of the Generalized Implicit Sieve Algorithm can be used to estimate the Fourier coefficients f^(χri), for i[𝒩], with suitably chosen accuracy and high probability, provided that m~ is large enough. This follows from an application of the Hoeffding bound. However, note that the set 𝒮 remains unknown.

  • We now describe how we use these estimates to construct a function f~:𝒢 such that for some unknown group automorphism A, the correlation between f~A and f is large. In order to do this, we need the idea of pseudo-independence over finite Abelian groups (see Definition 32).

  • For any set of elements r1,,rk𝒢, if there exists λ1,,λk{0}, such that at least one λj is invertible in , where =LCM{p1m1,,pnmn}, and i=1kλiri=0, then we call r1,,rk to be pseudo-dependent (Note that any subset has a pseudo-independent spanning set, see Definition 32). If there does not exist any such λi’s, then r1,,rk are pseudo-independent. The idea is, if there exists such λ1,,λk with λj invertible in , then rj can be written as a combination of the other elements. That is,

    rj=λj1(i[k],ijλiri), (4)

    where λj1 is the inverse of λj in . Observe that the above definition generalized the notion of linear dependence and independence over vector spaces. Furthermore, whenever we define a group automorphism on the elements r1,,rj1,rj+1,,rk, by the property of homomorphism it automatically gets defined on rj. That is,

    ψ(rj)=λj1(i[k],ijλiψ(ri)).

    Recall that λj is invertible in and λj1 is the inverse of λj in .

  • To construct f~ we first identify a R𝒮 (without knowing R) such that the elements in R are pseudo-independent and 𝒮 is a subset of the subgroup generated by R (which can be checked by a brute-force algorithm, see [20]). We prove that such an R can be found with high probability. This will allow us to relabel the elements in R as e1,,e|R|𝒢^ and the elements in 𝒮R as a combination of e1,,e|R|, using Equation (4), for some automorphism A of 𝒢. Relabel the columns of Q by {S1,,S𝒩} where S1=e1,S|B~|=e|B~| such that for all j{|B~|+1,,𝒩}, Sj can be written as

    λ1(i[|B~|]λiSi),

    where λ,λ1,,λ|B~|{0}, and λ is invertible in . Note that the columns S1,,S|B~| correspond to those in B~.

  • We will create f~ such that for all j[𝒩], f~^(χSj)=rj and f~^(χS)=0 if S{S1,,S𝒩} where

    rj=i=1m~f(x(i))Q[i,j]m~.
  • Thus at this point we have constructed f~ that has high correlation with f under some unknown group automorphism A, with high probability. We show that to decide whether dist𝒢(f,g)ϵ or dist𝒢(f,g)ϵ+τ, it is sufficient to check the correlation between f~ and g under all possible group automorphisms on 𝒢 (see [20]) and this can be done without making queries to f.

We refer the reader to [20] for the details.

As evident, the Generalized Implicit Sieve Algorithm (see [20]) plays an important role in the proof of our main result.

Proof idea of Theorem 8. (Generalized Implicit Sieve Algorithm).

Before we outline the proof of Theorem 8, let us first explain the goal of the Generalized Implicit Sieve Algorithm (see [20]).

  • The algorithm takes a Boolean valued function f:𝒢{1,1} as input, where 𝒢=p1m1××pnmn, and pi,i[n] are primes, and not necessarily distinct. It is also given a threshold θ and a set ={x1,,xm~} of uniformly random points from 𝒢, where m~ is sufficiently large. The algorithm considers a randomly permuted coset structure (H,𝒞) such that the subgroup H has codimension t in 𝒢 (see Definition 28), where tlog(1004m~4θ32), where =LCM{p1m1,,pnmn}. Then wt2 (see [20]) is estimated for each bucket with error ±θ24 and confidence 11100t. If the estimated weight wt2~ is 3θ24, then the algorithm keeps the bucket. Otherwise, it discards the bucket. For each surviving bucket C, wt4 (see [20]) is estimated with error ±θ48wt2~(C) and confidence 11100t. If the estimated weight wt4~ is 3θ44, then the algorithm keeps the bucket. Otherwise, it discards the bucket. Then it randomly picks ||=m~ points {y1,,ym~} from 𝒢, and calculates PCf(yi)PCf(yixi)¯ for all i[m~]. Note that PC is the projection operator on functions f:𝒢, which is defined as follows (C𝒢)

    PCf(x):=βCf^(β)χβ(x).

    The Generalized Implicit Sieve Algorithm makes O(lnm~θ16θ8+m~ln(𝒩m~)θ2+ln𝒩θ8) many queries.

We now outline the proof of Theorem 8.

  • The first step in the proof of the theorem is to partition 𝒢 into cosets of a random subgroup. Since 𝒢 is Abelian, all its subgroups are normal. Define a random subgroup H by sampling β1,,βk independently and uniformly from 𝒢 and using the pseudo-inner product operator on 𝒢 to obtain

    H={α𝒢:αβi=0i[k]}.

    By Definition 22 and Definition 24, H forms a subgroup of 𝒢. Its cosets have the form

    C(b)={α𝒢:αβi=bii[k]},

    where b=(b1,,bk)k. We refer to the cosets of H as “buckets”.

  • We show that with high probability, all the big Fourier coefficients, that is, the coefficients with weight θ24 belong to different buckets (see [20]).

  • Moreover, we show that the weight of a bucket Cα is determined by the big coefficient χα(C) contained in Cα with high probability, assuming that the big coefficients are in different buckets with high probability. We use the second moment method to show this. Let for each γ𝒢,

    Yγ,α={1,if γCα0,otherwise.

    And,

    Yα=γSYγ,α.

    Also, let

    Xγ,Cα={|f^(χγ)|2if γCα0otherwise.

    And,

    XCα=γSXγ,Cα.

    The b=0 case (see [20]) is problematic, as the Cov[Yγ1,αYγ2,α]>0 in this case and hence finding an upper bound for Pr[|XCα𝔼[XCα]|ϵ] becomes difficult using Chebyshev. We use random shift (see [20]) to avoid this, and study case by case to show that Cov[Yγ1,αYγ2,α]=0. This gives us an upper bound on the variance of XCα, which in turn helps us to show that the weight of a bucket Cα is determined by the big coefficient χα(C) contained in Cα with high probability. This implies that with high probability, the Fourier coefficients other than the big coefficient χα(C) does not contribute much to the bucket Cα.

  • We then show that the algorithm does not discard any bucket C with |f^(χγ)|θ, but it discards any bucket C with |f^(χγ)|<θ24, for all γC. Furthermore, we establish that |f^(χα(C))|θ2 for every bucket C=Cα (where α(C) is the dominating element of Cα) among those that the algorithm has not discarded.

  • Additionally, we show that the small weight coefficients do not contribute much to the weight of a bucket. That is, we show that Prx[|rCα{χα(C)}f^(χr)χr(x)|θ24] is very small, where χα(C) is the dominating Fourier coefficient of the bucket Cα.

  • Finally, our target is to output a matrix Q (see [20]) with a small error, whose entries are characters evaluated at the points xi,i[m~], where xi are picked independently and uniformly at random. To show that, we pick yi independently and uniformly at random, and then prove that the distance between PCαf(yi)PCαf(yixi)¯ and |f^(χα(C))|2χα(C)(xi) is 9θ416 with high probability. Then we construct a matrix Q (see [20]), and estimate the Fourier coefficients for all the large buckets. We then show that, up to some small error, the algorithm outputs the matrix Q (see [20]) whose entries are given by χα(Ci)(xj), i[𝒩],j[m~], where 𝒩 denotes the number of survived buckets.

1.4 Challenges with finite Abelian Groups

We outline the main challenges encountered in proving our results for finite Abelian groups 𝒢. Recall that 𝒢 can be written as 𝒢=p1m1××pnmn, where pi,i[n] are primes, and not necessarily distinct.

  • While estimating wt2 and wt4, it is required to find H for a subgroup H of 𝒢. But due to the lack of vector space structure, H cannot be defined in terms of “basis”, which is possible when 𝒢=2n, as 2n is a vector space. This problem had to be taken care of in a different way. Given a subgroup H of 𝒢, we define H by

    H:={x𝒢:xy=0yH},

    where is a pseudo inner product defined by

    xy:=(i=1Tpimix(i)y(i))(mod),

    and =LCM{p1m1,,pnmn}. It is to be noted that H is also a subgroup of 𝒢.

    This is extremely useful in proving the fact that, given a bucket Cα, wt2(Cα) and wt4(Cα) can be written as an average of some quantity, which in turn helps to reduce the query complexity of wt2 and wt4 estimation. More precisely, it follows that,

    wt2(𝒞α)=𝔼x𝒢,zH[f(x)f(x+z)χr(z)]

    and

    wt4(Cα)=𝔼x,z1,y1𝒢,z,yH[f(z1)f(xz1z)f(y1)f(xy1y)χr(zy)]

    respectively (see [20] for more details). Applying a concentration inequality the query complexity becomes dependent on θ and m~ only, where θ and m~ are as defined in Theorem 8, and it is not dependent on the order of the group 𝒢.

  • To prove our main theorem, we also need to show that, for any automorphism A:𝒢𝒢 and any function f:𝒢{1,+1}, the Fourier coefficients fA^(χr),r𝒢 of fA are same as the Fourier coefficients of f upto some permutation. In case of 2n, it was quite easy, as 2n is a vector space. As an Abelian group is not a vector space in general, we needed to use Pontryagin duality and dual automorphisms. For more details, see Lemma 34 and Lemma 38 in Section 2.

  • Due to the lack of vector space structure, we cannot partition 𝒢 into subspaces. Instead, we define a normal subgroup V0,r1,,rk (see Lemma 26), and partition 𝒢, and hence 𝒢^ into its cosets. For details see Section 2.

  • While testing isomorphism, the notion of linear independence is required. But when there is no vector space structure, linear independence does not make any sense. A different notion of “independence” needed to be defined in order to tackle this problem.

    For any set of elements r1,,rk𝒢, if there exists λ1,,λk{0}, with at least one λj an unit in (that is, λj is invertible in ), =LCM{p1m1,,pnmn}, such that i=1kλiri=0, then we call r1,,rk to be dependent. If there does not exist any such λi’s, then r1,,rk are independent. The idea is, if there exists such λ1,,λk with λj an unit, then rj can be written as a combination of the other elements. That is,

    rj=λj1(i[k],ijλiri).

    So, whenever we define a group automorphism on the elements r1,,rj1,rj+1,,rk, by the property of homomorphism it automatically gets defined on rj. This plays an important role in the Algorithm (see [20] for more details).

  • We know that the characters of a Boolean function from 2n to {0,1} takes values 1 or +1 at each point of 2n. So, to determine the dominating character in a bucket at a point xiM with high probability, determining the sign of PCαf(yi)PCαf(yi+xi) is sufficient, where χα(C) being the dominating character of the bucket Cα.

    But in our case, the characters of a finite Abelian group 𝒢 are complex-valued at each point of 𝒢. So sign does not make any sense here. The problem needed to be tackled differently. We first calculate PCαf(yi)PCαf(yixi)¯, and then we show that PCαf(yi)PCαf(yixi)¯|f^(χα(C))|2χα(C)(xi) is small with high probability, which helps us to understand the dominating character in the bucket Cα at the point xiM. For details see [20].

2 Background on Finite Abelian Groups

Throughout this paper, 𝒢 denotes a finite Abelian group p1m1××pnmn, where pi are primes for all i[n] and may not be distinct, and ω denotes a primitive th root of unity, where =LCM{p1m1,,pnmn}.

Let us define the characters of 𝒢 first.

Definition 12 (Character).

For 𝒢=p1m1××pnmn, where pi,i[n] are primes, a character of the group 𝒢 is a homomorphism χ:𝒢× of 𝒢, that is, χ satisfies the following: for all x,y𝒢 we have χ(x+y)=χ(x)χ(y).
Equivalently, a character of 𝒢 is of the form

χ(x(1),,x(n))=χr(1)(x(1))χr(n)(x(n)),

where r(i)pimi and χr(i) is a character of pimi and is defined by

χr(i)(x(i))=ωpimir(i)x(i).

Thus for any (r(1),,r(n))𝒢 we can define a corresponding character of 𝒢 as χr(1),,r(n) that on input x=(x(1),,x(n)) is defined as

χ(r(1),,r(n))(x(1),,x(n)) =ωp1m1r(1)x(1)ωpnmnr(n)x(n) (5)
=ωi=1Tr(i)x(i)npimi(mod), (6)

where =LCM{p1m1,,pnmn}.

Now let us look at some properties of characters.

Lemma 13.

Let χ be a character of 𝒢. Then,

  1. 1.

    χ0(x)=1 for all x𝒢.

  2. 2.

    χ(x)=χ(x)1=χ(x)¯ for all x𝒢.

  3. 3.

    For any character χ of 𝒢, where χχ0, x𝒢χ(x)=0.

  4. 4.

    |χ(x)|=1 for all x𝒢.

Now let us define the dual group of 𝒢.

Definition 14 (Dual group).

The set of characters of 𝒢 forms a group under the operation (χψ)(x)=χ(x)ψ(x) and is denoted by 𝒢^, where χ and ψ are characters of 𝒢. 𝒢^ is called the dual group of 𝒢.

The following theorem states that 𝒢 is isomorphic to its dual group.

Theorem 15.

𝒢^𝒢, that is 𝒢 and 𝒢^ are isomorphic to each other.

Let us look at the definition of Fourier transform for functions on 𝒢.

Definition 16 (Fourier transform).

For any 𝒢=p1m1××pnmn, with pi,i[n] primes, and any function f:𝒢, the Fourier transform f^:𝒢^ is

f^(χr(1),,r(T))=1|𝒢|x𝒢f(x)ωp1m1r(1)x(1)ωpnmnr(n)x(n),

where x=(x(1),,x(n)).

 Remark 17.

The Fourier transform of a function f:𝒢 is defined by

f^(χ)=1|𝒢|x𝒢f(x)χ(x)¯,

where χ(x)¯ is the conjugate of χ(x). The Definition 16 follows from this, as χ=χr(1),,r(n) for some r(i)pimi,i{1,,n}.

The following theorem states that any function from 𝒢 to can be written as a linear combination of characters of 𝒢.

Theorem 18 (Fourier inversion formula).

Any function f:𝒢 can be uniquely written as a linear combination of characters of 𝒢, that is,

f(x)=χr(1),,r(n)𝒢^f^(χr(1),,r(n))χr(1),,r(n)(x), (7)

where x=(x(1),,x(n)).

Theorem 19 (Parseval).

For any two functions f,g:𝒢,

𝔼x𝒢[f(x)g(x)¯]=χ𝒢^f^(χ)g^(χ)¯.

More specifically, if f:𝒢{1,1} is a Boolean-valued function then

χ𝒢^|f^(χ)|2=1.

Now let us define the Fourier sparsity sf of a function f on 𝒢.

Definition 20 (Fourier Spectral Norm).

The Fourier Spectral Norm or the Spectral norm of f:𝒢, denoted by f1 is defined as the sum of the absolute value of its Fourier coefficients. That is,

f^1=r𝒢|f^(χr)|.
Definition 21 (Sparsity and Fourier Support).
  • Fourier support supp(f^) of a function f:𝒢 denotes the set {χf^(χ)0}.

  • The Fourier sparsity sf of a function f:𝒢 is defined to be the number of non-zero Fourier coefficients in the Fourier expansion of f (Theorem 18). Alternately, Fourier sparsity can be defined as the size of the Fourier support. In this paper, by sparsity of a function, we mean the Fourier sparsity of the function. Moreover, by s-sparse function we mean functions with Fourier Sparsity s.

We know that since 2n is also a vector space over the field 2, it has a linear algebraic structure. The characters are in one-to-one correspondence with 2-linear forms. However, since a general Abelian group is not a vector space, these do not hold here. But one can define something along similar lines.

An element r of 𝒢 is of the form (r(1),,r(n)), where r(i) is an element of pimi.

Definition 22 (Pseudo inner product).

For x,r𝒢 we denote by the following pseudo inner product.

rx:=(i=1npimir(i)x(i))(mod),

where =LCM{p1m1,,pnmn}.

Observation 23.

rx0χr(x)1.

Now, we partition 𝒢 with the help of a normal subgroup.

Definition 24.

Let r1,,rk𝒢 such that rj=(rj(1),,rj(n)), where rj(i)pimi for all i[n] and j[k]. Let b=(b1,,bk)k, where =LCM{p1m1,,pnmn}. We define the set Vb,r1,,rk by

Vb,r1,,rk={x𝒢:rjx=bj(mod)j[k]}. (8)

That is,

Vb,r1,,rk ={x𝒢:(i=1Tpimirj(i)x(i))=bj(mod)j[k]}.

If b1=b2==0 then we use V0,r1,,rk to denote Vb,r1,,rk.

 Remark 25.

When k=1, then b, and we have only one element r𝒢, so this set becomes Vb,r={rx=b(mod)}.

Lemma 26.

V0,r1,,rk is a normal subgroup of 𝒢 and for any b,r1,,rk either Vb,r1,,rk is a coset of V0,r1,,rk or Vb,r1,,rk=.

Proof.

Clearly, 0V0,r1,,rk, where 0 is the identity element of 𝒢. Let x,yV0,r1,,rk. Then, since rj(xy)=i=1Tpimirj(i)x(i)i=1Tpimirj(i)y(i)=0(mod) for all j[k], so xyV0,r1,,rk. So V0,r1,,rk is a subgroup of 𝒢, and hence a normal subgroup of 𝒢 since 𝒢 is Abelian.

Since V0,r1,,rk is a normal subgroup of 𝒢 we can now consider its cosets. For each coset B of V0,r1,,rk, let a(B) be a fixed coset representative (we choose a coset representative and fix it all through the proof). Then, if yB, then y=a(B)+x, where xV0,r1,,rk. Now, for each rj,

rjy=rj(a(B)+x)=rja(B),

since rjx=0 for all rj,j[k]. So, rjy is fixed for each coset B. If we assume rja(B)=bj(mod) for each j[k], then we have Vb,r1,,rk as a coset of V0,r1,,rk, where b=(b1,,bk).

If for a b=(b1,,bk)k, there does not exist any a(B) such that rja(B)=bjj[k], Vb,r1,,rk=.

 Remark 27.

Note that since 𝒢 is an Abelian group, any subgroup of 𝒢 is a normal subgroup of 𝒢.

Definition 28 (Codimension).

Let Vb,r1,,rk be as defined in Definition 24. Then the codimension of Vb,r1,,rk is given by

Codim( Vb,r1,,rk)=mink{k:r1,,rk{r1,,rk} such that
Vb,r1,,rk={x𝒢:rjx=bj(mod)j[k]}},

where b=(b1,,bk)k, and =LCM{p1m1,,pnmn}.

Definition 29 (Random coset structure).

Let us consider β1,,βk𝒢, which are chosen independently and uniformly at random from 𝒢. Also let H={α𝒢:αβi=0i[k]}. So H is a subgroup, hence a normal subgroup of 𝒢. Then, we define the sets C(b) by C(b)={α𝒢:αβi=bii[k]}, for all b=(b1,,bk)k.

 Remark 30.

Observe that the nonempty C(b) are cosets of H. We will often refer to them as buckets.

Definition 31.

Let G be a group, with group operation . A group automorphism on G is a bijective function ψ:GG which satisfies the following.

ψ(xy)=ψ(x)ψ(y),x,yG.

The set of all automorphisms of G forms a group under the composition of functions and is denoted by Aut(G).

Definition 32.

Let r1,,rk be elements in 𝒢. We call r1,,rk to be dependent if there exists λ1,,λk with at least one λj invertible in such that i=1kλiri=0.

If there does not exist any such λ1,,λk with at least one λj invertible in such that i=1kλiri=0, then we call r1,,rk to be independent.

For example, consider the group 4. The element 3 is dependent on the element 1, but it is independent of the element 2. Whereas, 2 is dependent on 3, as 2=23(mod4). So the set {2,3} is dependent.

Lemma 33 (Hoeffding’s Inequality).

Let X1,,Xk be real independent random variables, each taking value in [-1, 1]. Then

Pr[|i=1kXi𝔼i=1kXi|ϵ]2exp(ϵ22k).

The following structural result about an Abelian group and its dual will play a key role in our analysis. For a first reading, the proof of this lemma may be skipped without affecting the readability of the rest of the paper.

Lemma 34.

Let A be a map from 𝒢 to 𝒢. Let us define a map A^:𝒢^𝒢^ by A^(χr)=χrA. Also, let us define another map A^^:𝒢𝒢 by χA^^(r)=A^(χr). Then A is an automorphism if and only if A^ is an automorphism if and only if A^^ is an automorphism.

Proof.

Proof of the part [A is an automorphism if and only if A^ is an automorphism]
When A is an automorphism.

First let us show that A^ is also a homomorphism. Observe

A^(χr1χr2)(x) =A^(χr1+r2)(x)
=χr1+r2A(x)
=χr1(A(x))χr2(A(x))
=A^(χr1)(x)A^(χr2)(x),

which shows that A^ is indeed a homomorphism.

To show injectivity, let A^(χr)=1. Then,

A^(χr)(x)=1x𝒢 χr(A(x))=1x𝒢
χr(y)=1y𝒢, where y=A(x)
χr=χ0,

where the equality in the second last line holds because A is an automorphism. Since χ0 is the identity element of G^, so A^ is injective.

Since the domain and the range of A^ are same, so A^ is bijective, and hence, it is an automorphism on G^.

When A^ is an automorphism.

To show that A is a homomorphism, we observe that, for all r𝒢,

A^(χr)(x+y)=A^(χr)(x)A^(χr)(y)=χr(A(x))χr(A(y))=χr(A(x)+A(y))
χr(A(x+y))=χr(A(x)+A(y))
χr(A(x+y)A(x)A(y))=0.

Therefore, A(x+y)=A(x)+A(y), hence A is a homomorphism.

Now let us show that A is injective. To do that, let A(x)=0. Then,

A^(χr)(x)=χr(A(x))=χr(0)=1r𝒢,

which is only possible if x=0, since A^ is an automorphism and maps characters to characters. So, A is injective.

Since the domain and the range of A are same, so A is bijective, and hence, A is an automorphism.

Proof of the part [A^ is an automorphism if and only if A^^ is an automorphism]
When A^ is an automorphism.

Then, for each x𝒢,
χA^^(r1+r2)(x)=A^(χr1+r2)(x)=χr1+r2(A(x))=χr1(A(x))χr2(A(x))=A^(χr1)(x)A^(χr2)(x) χA^^(r1+r2)(x)=χA^^(r1)(x)χA^^(r2)(x) χ{A^^(r1+r2)A^^(r1)A^^(r2)}(x)=0,

which implies that A^^(r1+r2)=A^^(r1)+A^^(r2) for all r1,r2𝒢. So A^^ is a homomorphism.

Now, let A^^(r)=0. Then, for all x𝒢,

χA^^(r)(x)=1A^(χr)(x)=1χr(A(x))=1.

Since A^ is an automorphism, so A is an automorphism by the first part of the proof, therefore χr(y)=1 for all y𝒢, where y=A(x). So r=0, hence A^^ is injective.

Since the domain and the range of A^^ is same, so A^^ is bijective, and hence, it is an automorphism.

When A^^ is an automorphism.

Then, for each x𝒢,

A^(χr1+r2)(x)=χA^^(r1+r2)(x)=χA^^(r1)+A^^(r2)(x)=χA^^(r1)(x)χA^^(r2)(x)
A^(χr1χr2)(x)=A^(χr1)(x)A^(χr2)(x),

which shows that A^ is a homomorphism.

Now, let A^(χr)(x)=1 for all x𝒢. Then, for all x𝒢,

χA^^(r)(x)=1A^^(r)=0r=0,

since A^^ is an automorphism. Therefore, A^ is injective.

Since the domain and the range of A^ is same, so A^ is bijective. Hence, A^ is an automorphism.

Definition 35 (Boolean function isomorphism).

Let f,g:𝒢{1,+1} be Boolean valued functions. Then f is isomorphic to g if there exists an automorphism A on 𝒢 such that f=gA.

Definition 36 (Automorphism distance).

Let f,g:𝒢{1,+1} be Boolean valued functions. The automorphism distance between f and g is defined by

dist𝒢(f,g)=minAAut(𝒢)δ(f,gA),

where δ(f,gA) is the fractional Hamming distance between f and gA.

Definition 37 (ϵ-close and ϵ-far from isomorphic).

Let f,g:𝒢{1,+1} be two Boolean valued functions. Then f is said to be ϵ-close (ϵ-far) from being isomorphic to g if dist𝒢(f,g)ϵ (dist𝒢(f,g)ϵ). That is, f is ϵ-close to being isomorphic to g if there exists an automorphism A on 𝒢 such that δ(f,gA)ϵ; and f is ϵ-far from being isomorphic to g if for all automorphism A on 𝒢, δ(f,gA)ϵ.

Lemma 38.

For any automorphism A:𝒢𝒢, we have

fA^(χr)=f^(χA1^^(r)).

Proof.

Observe

fA^(χr) =1|𝒢|x𝒢fA(x)χr(x)
=1|𝒢|x𝒢f(A(x))χr(x)
=1|𝒢|y𝒢f(y)χr(A1(y)) where y=A(x)
=1|𝒢|y𝒢f(y)A1^(χr)(y) by Lemma 34
=1|𝒢|y𝒢f(y)χA1^^(r)(y) by Lemma 34
=f^(χA1^^(r)).

3 The subgroup 𝑯

Throughout this section, 𝒢 denotes the finite Abelian group p1m1××pnmn, where pi are primes for all i[n] and not necessarily distinct, x denotes the subgroup generated by x, and ω denotes a primitive th root of unity and =LCM{p1m1,,pnmn}. Let H be a subgroup of 𝒢.

Let us look at the definition of H.

Definition 39 (Definition 1).

Let H be a subgroup of 𝒢. Then H is the subgroup given by

H:={x𝒢:xy=0yH},

where

xy:=(i=1Tpimix(i)y(i))(mod),

and =LCM{p1m1pnmn}.

 Remark 40.

Observe that H is also a subgroup of 𝒢.

Claim 41.

For zH,

βHωβz=|H|,

where ω is a primitive th root of unity of order , and |H| denotes the order of the subgroup H.

Also, for zH,

βHωβz=0.

Proof.

Case 1: When zH.

Then zx=0 for all xH. Hence

βHωβz=|H|.
Case 2: When zH.

Let βHχβ(z)=A. Since zH, so, by Definition 39, there exists γH such that γz0, that is, χγ(z)1. (see Observation 23) Then,

χγ(z)×A =χγ(z)βHχβ(z)
=βHχβ+γ(z)
=γHχγ(z), γ=β+γ
=A,

which implies that

A(χγ(z)1)=0A=0,

since χγ(z)1.

We need the following isomorphism result:

Lemma 42.

H is isomorphic to the quotient group 𝒢/H.

Proof.

We will show that H^ is isomorphic to 𝒢/H^, which implies that H is isomorphic to the quotient group 𝒢/H, as GG^ for any group G.

The set of characters of H is given by

Ann𝒢(H)={χ𝒢^:χ(y)=1yH},

where Ann𝒢(H) is known as the annihilator of the subgroup H in 𝒢. From Definition 39, observe that for x𝒢, χx(y)=ωxy=1 if and only if xy=0xH.

Let us define a group homomorphism :𝒢/H^Ann𝒢(H) by (ζ)=ζq, where q:𝒢𝒢/H is the quotient is the quotient group homomorphism defined by q(r)=r+H.

  • ( is injective) Let ζ𝒢/H^ such that ζq=0~, where 0~=(0,,0)[T times] is the identity element of 𝒢. So, ζ(r+H)=ζq(r)=1 for all r𝒢, which implies ζ is the identity element of 𝒢/H. Therefore, is injective.

  • ( is surjective) Let ψAnn𝒢(H). Let ζ:𝒢/H by ζ(r+H)=ψ(r) for all r𝒢. Since χr+H(x)=ωrx+Hx, so any character χr+H of 𝒢 is a character of 𝒢/H if Hx=0 for all x𝒢 (as then, the value of ωrx+Hx will be determined only by the coset representatives). Since ψ(r+H)=ψ(r)ψ(H)=ψ(r) and H is the identity element of G/H, so ζ is a character of 𝒢/H. Also, ψ=ζq. Therefore, (ζ)=ζq=ψ. Hence, is surjective.

Therefore, is an isomorphism, which implies that H is isomorphic to the quotient group 𝒢/H.

Corollary 43.

|H|×|H|=|𝒢|.

Proof.

Follows from the fact that |H|=|𝒢||H|, since H is isomorphic to the quotient group 𝒢/H by Lemma 42, and |𝒢/H|=|𝒢||H| by Lagrange’s theorem.

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