I am asked whether or not there exists an $r$ such that given any subset $A$ of any group (abelian or not) then:
$$|A\cdot A| \leq K|A| \Rightarrow |A\cdot A\cdot A| \leq K^r |A|$$
(where $|A \cdot A| = \{a \cdot b | a,b \in A\}$)
I suspect that this result is false. However, since we have proved that this result is true when the group is abelian, I am hoping to construct a counter example using non-abelian groups.
Specifically I am hoping to find a family of groups $G_n$ and subsets $A_n \subset G_n$ such that $|A_n \cdot A_n| \leq K|A_n|$ but $\forall r, \exists n \in \mathbb N$ such that $|A_n \cdot A_n \cdot A_n| > K^r|A_n|$.
Sadly, I have been unable to construct such a family.
Is it possible to find such a counter example or is this result in fact true?
Let $F$ be the free group on two generators $a$ and $b$. If $A_n = \{1, a, a^2, \dots, a^n, b\}$, then $|A_n| = n + 2$. Moreover $$ A_nA_n = \{1, a, \dots, a^{2n}\} \cup \{b, ba, ba^2, \dots, ba^n\} \cup \{ab, a^2b, \dots, a^nb\} \cup \{b^2\} $$ and hence $|A_nA_n| = (2n+1) + (n+1) + n + 1 = 4n + 3$. Thus $|A_nA_n| \leqslant 4|A_n|$. On the other hand, $A_nA_nA_n$ contains $A_nA_n$ and the set $$ C = \{a^iba^j \mid 0 \leqslant i, j \leqslant n\} $$ Since $A_nA_n \cap C = \emptyset$, one gets $|A_nA_nA_n| \geqslant (4n + 3) + (n+1)^2 \geqslant (n+2)^2 = |A_n|^2$. Therefore, if $n > 4^r$, the relation $|A_nA_nA_n| < 4^r|A_n|$ does not hold.