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Suppose that {A \subset B} are two subgroups of some ambient group {G}, with the index {K := [B:A]} of {A} in {B} being finite. Then {B} is the union of {K} left cosets of {A}, thus {B = SA} for some set {S \subset B} of cardinality {K}. The elements {s} of {S} are not entirely arbitrary with regards to {A}. For instance, if {A} is a normal subgroup of {B}, then for each {s \in S}, the conjugation map {g \mapsto s^{-1} g s} preserves {A}. In particular, if we write {A^s := s^{-1} A s} for the conjugate of {A} by {s}, then

\displaystyle  A = A^s.

Even if {A} is not normal in {B}, it turns out that the conjugation map {g \mapsto s^{-1} g s} approximately preserves {A}, if {K} is bounded. To quantify this, let us call two subgroups {A,B} {K}-commensurate for some {K \geq 1} if one has

\displaystyle  [A : A \cap B], [B : A \cap B] \leq K.

Proposition 1 Let {A \subset B} be groups, with finite index {K = [B:A]}. Then for every {s \in B}, the groups {A} and {A^s} are {K}-commensurate, in fact

\displaystyle  [A : A \cap A^s ] = [A^s : A \cap A^s ] \leq K.

Proof: One can partition {B} into {K} left translates {xA} of {A}, as well as {K} left translates {yA^s} of {A^s}. Combining the partitions, we see that {B} can be partitioned into at most {K^2} non-empty sets of the form {xA \cap yA^s}. Each of these sets is easily seen to be a left translate of the subgroup {A \cap A^s}, thus {[B: A \cap A^s] \leq K^2}. Since

\displaystyle [B: A \cap A^s] = [B:A] [A: A \cap A^s] = [B:A^s] [A^s: A \cap A^s]

and {[B:A] = [B:A^s]=K}, we obtain the claim. \Box

One can replace the inclusion {A \subset B} by commensurability, at the cost of some worsening of the constants:

Corollary 2 Let {A, B} be {K}-commensurate subgroups of {G}. Then for every {s \in B}, the groups {A} and {A^s} are {K^2}-commensurate.

Proof: Applying the previous proposition with {A} replaced by {A \cap B}, we see that for every {s \in B}, {A \cap B} and {(A \cap B)^s} are {K}-commensurate. Since {A \cap B} and {(A \cap B)^s} have index at most {K} in {A} and {A^s} respectively, the claim follows. \Box

It turns out that a similar phenomenon holds for the more general concept of an approximate group, and gives a “classification” of all the approximate groups {B} containing a given approximate group {A} as a “bounded index approximate subgroup”. Recall that a {K}-approximate group {A} in a group {G} for some {K \geq 1} is a symmetric subset of {G} containing the identity, such that the product set {A^2 := \{ a_1 a_2: a_1,a_2 \in A\}} can be covered by at most {K} left translates of {A} (and thus also {K} right translates, by the symmetry of {A}). For simplicity we will restrict attention to finite approximate groups {A} so that we can use their cardinality {A} as a measure of size. We call finite two approximate groups {A,B} {K}-commensurate if one has

\displaystyle  |A^2 \cap B^2| \geq \frac{1}{K} |A|, \frac{1}{K} |B|;

note that this is consistent with the previous notion of commensurability for genuine groups.

Theorem 3 Let {G} be a group, and let {K_1,K_2,K_3 \geq 1} be real numbers. Let {A} be a finite {K_1}-approximate group, and let {B} be a symmetric subset of {G} that contains {A}.

  • (i) If {B} is a {K_2}-approximate group with {|B| \leq K_3 |A|}, then one has {B \subset SA} for some set {S} of cardinality at most {K_1 K_2 K_3}. Furthermore, for each {s \in S}, the approximate groups {A} and {A^s} are {K_1 K_2^5 K_3}-commensurate.
  • (ii) Conversely, if {B \subset SA} for some set {S} of cardinality at most {K_3}, and {A} and {A^s} are {K_2}-commensurate for all {s \in S}, then {|B| \leq K_3 |A|}, and {B} is a {K_1^6 K_2 K_3^2}-approximate group.

Informally, the assertion that {B} is an approximate group containing {A} as a “bounded index approximate subgroup” is equivalent to {B} being covered by a bounded number of shifts {sA} of {A}, where {s} approximately normalises {A^2} in the sense that {A^2} and {(A^2)^s} have large intersection. Thus, to classify all such {B}, the problem essentially reduces to that of classifying those {s} that approximately normalise {A^2}.

To prove the theorem, we recall some standard lemmas from arithmetic combinatorics, which are the foundation stones of the “Ruzsa calculus” that we will use to establish our results:

Lemma 4 (Ruzsa covering lemma) If {A} and {B} are finite non-empty subsets of {G}, then one has {B \subset SAA^{-1}} for some set {S \subset B} with cardinality {|S| \leq \frac{|BA|}{|A|}}.

Proof: We take {S} to be a subset of {B} with the property that the translates {sA, s \in S} are disjoint in {BA}, and such that {S} is maximal with respect to set inclusion. The required properties of {S} are then easily verified. \Box

Lemma 5 (Ruzsa triangle inequality) If {A,B,C} are finite non-empty subsets of {G}, then

\displaystyle  |A C^{-1}| \leq |A B^{-1}| |B C^{-1}| / |B|.

Proof: If {ac^{-1}} is an element of {AC^{-1}} with {a \in A} and {c \in C}, then from the identity {ac^{-1} = (ab^{-1}) (bc^{-1})} we see that {ac^{-1}} can be written as the product of an element of {AB^{-1}} and an element of {BC^{-1}} in at least {|B|} distinct ways. The claim follows. \Box

Now we can prove (i). By the Ruzsa covering lemma, {B} can be covered by at most

\displaystyle \frac{|BA|}{|A|} \leq \frac{|B^2|}{|A|} \leq \frac{K_2 |B|}{|A|} \leq K_2 K_3

left-translates of {A^2}, and hence by at most {K_1 K_2 K_3} left-translates of {A}, thus {B \subset SA} for some {|S| \leq K_1 K_2 K_3}. Since {sA} only intersects {B} if {s \in BA}, we may assume that

\displaystyle  S \subset BA \subset B^2

and hence for any {s \in S}

\displaystyle  |A^s A| \leq |B^2 A B^2 A| \leq |B^6|

\displaystyle  \leq K_2^5 |B| \leq K_2^5 K_3 |A|.

By the Ruzsa covering lemma again, this implies that {A^s} can be covered by at most {K_2^5 K_3} left-translates of {A^2}, and hence by at most {K_1 K_2^5 K_3} left-translates of {A}. By the pigeonhole principle, there thus exists a group element {g} with

\displaystyle  |A^s \cap gA| \geq \frac{1}{K_1 K_2^5 K_3} |A|.

Since

\displaystyle  |A^s \cap gA| \leq | (A^s \cap gA)^{-1} (A^s \cap gA)|

and

\displaystyle  (A^s \cap gA)^{-1} (A^s \cap gA) \subset A^2 \cap (A^s)^2

the claim follows.

Now we prove (ii). Clearly

\displaystyle  |B| \leq |S| |A| \leq K_3 |A|.

Now we control the size of {B^2 A}. We have

\displaystyle  |B^2 A| \leq |SA SA^2| \leq K_3^2 \sup_{s \in S} |A s A^2| = K_3^2 \sup_{s \in S} |A^s A^2|.

From the Ruzsa triangle inequality and symmetry we have

\displaystyle  |A^s A^2| \leq \frac{ |A^s (A^2 \cap (A^2)^s)| |(A^2 \cap (A^2)^s) A^2|}{|A^2 \cap (A^2)^s|}

\displaystyle  \leq \frac{ |(A^3)^s| |A^4| }{|A|/K_2}

\displaystyle  \leq K_2 \frac{ |A^3| |A^4| }{|A|}

\displaystyle  \leq K_1^5 K_2 |A|

and thus

\displaystyle  |B^2 A| \leq K_1^5 K_2 K_3^2 |A|.

By the Ruzsa covering lemma, this implies that {B^2} is covered by at most {K_1^5 K_2 K_3^2} left-translates of {A^2}, hence by at most {K_1^6 K_2 K_3^2} left-translates of {A}. Since {A \subset B}, the claim follows.

We now establish some auxiliary propositions about commensurability of approximate groups. The first claim is that commensurability is approximately transitive:

Proposition 6 Let {A} be a {K_1}-approximate group, {B} be a {K_2}-approximate group, and {C} be a {K_3}-approximate group. If {A} and {B} are {K_4}-commensurate, and {B} and {C} are {K_5}-commensurate, then {A} and {C} are {K_1^2 K_2^3 K_2^3 K_4 K_5 \max(K_1,K_3)}-commensurate.

Proof: From two applications of the Ruzsa triangle inequality we have

\displaystyle  |AC| \leq \frac{|A (A^2 \cap B^2)| |(A^2 \cap B^2) (B^2 \cap C^2)| |(B^2 \cap C^2) C|}{|A^2 \cap B^2| |B^2 \cap C^2|}

\displaystyle  \leq \frac{|A^3| |B^4| |C^3|}{ (|A|/K_4) (|B|/K_5)}

\displaystyle  \leq K_4 K_5 \frac{K_1^2 |A| K_2^3 |B| K_3^2 |C|}{ |A| |B| }

\displaystyle  = K_1^2 K_2^3 K_3^2 K_4 K_5 |C|.

By the Ruzsa covering lemma, we may thus cover {A} by at most {K_1^2 K_2^3 K_3^2 K_4 K_5} left-translates of {C^2}, and hence by {K_1^2 K_2^3 K_3^3 K_4 K_5} left-translates of {C}. By the pigeonhole principle, there thus exists a group element {g} such that

\displaystyle  |A \cap gC| \geq \frac{1}{K_1^2 K_2^3 K_3^3 K_4 K_5} |A|,

and so by arguing as in the proof of part (i) of the theorem we have

\displaystyle  |A^2 \cap C^2| \geq \frac{1}{K_1^2 K_2^3 K_3^3 K_4 K_5} |A|

and similarly

\displaystyle  |A^2 \cap C^2| \geq \frac{1}{K_1^3 K_2^3 K_3^2 K_4 K_5} |C|

and the claim follows. \Box

The next proposition asserts that the union and (modified) intersection of two commensurate approximate groups is again an approximate group:

Proposition 7 Let {A} be a {K_1}-approximate group, {B} be a {K_2}-approximate group, and suppose that {A} and {B} are {K_3}-commensurate. Then {A \cup B} is a {K_1 + K_2 + K_1^2 K_2^4 K_3 + K_1^4 K_2^2 K_3}-approximate subgroup, and {A^2 \cap B^2} is a {K_1^6 K_2^3 K_3}-approximate subgroup.

Using this proposition, one may obtain a variant of the previous theorem where the containment {A \subset B} is replaced by commensurability; we leave the details to the interested reader.

Proof: We begin with {A \cup B}. Clearly {A \cup B} is symmetric and contains the identity. We have {(A \cup B)^2 = A^2 \cup AB \cup BA \cup B^2}. The set {A^2} is already covered by {K_1} left translates of {A}, and hence of {A \cup B}; similarly {B^2} is covered by {K_2} left translates of {A \cup B}. As for {AB}, we observe from the Ruzsa triangle inequality that

\displaystyle  |AB^2| \leq \frac{|A (A^2 \cap B^2)| |(A^2 \cap B^2) B^2|}{|A^2 \cap B^2|}

\displaystyle  \leq \frac{|A^3| |B^4|}{|A|/K_3}

\displaystyle  \leq K_1^2 K_2^3 K_3 |B|

and hence by the Ruzsa covering lemma, {AB} is covered by at most {K_1^2 K_2^3 K_3} left translates of {B^2}, and hence by {K_1^2 K_2^4 K_3} left translates of {B}, and hence of {A \cup B}. Similarly {BA} is covered by at most {K_1^4 K_2^2 K_3} left translates of {B}. The claim follows.

Now we consider {A^2 \cap B^2}. Again, this is clearly symmetric and contains the identity. Repeating the previous arguments, we see that {A} is covered by at most {K_1^2 K_2^3 K_3} left-translates of {B}, and hence there exists a group element {g} with

\displaystyle  |A \cap gB| \geq \frac{1}{K_1^2 K_2^3 K_3} |A|.

Now observe that

\displaystyle  |(A^2 \cap B^2)^2 (A \cap gB)| \leq |A^5| \leq K_1^4 |A|

and so by the Ruzsa covering lemma, {(A^2 \cap B^2)^2} can be covered by at most {K_1^6 K_2^3 K_3} left-translates of {(A \cap gB) (A \cap gB)^{-1}}. But this latter set is (as observed previously) contained in {A^2 \cap B^2}, and the claim follows. \Box

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