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If a finite subset $A$ of a group satisfies $|A^3|\le K|A|$, then repeated use of the Noncommutative Ruzsa triangle inequality gives

The Noncommutative Ruzsa triangle inequality says that nonempty finite subsets $B,C,D$ of a group satisfyThe Ruzsa covering lemma says that if $|CD|\le K|D|$, then some $X\subseteq C$ with $|X|\le K$ satisfies

Now let $A$ be a symmetric subset of a group, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\ge 4$,The hypothesis therefore gives $|A^m|\le K|A^{m-1}|$. Starting from $|A^3|\le K|A|$ proves

In particular $|A^5|\le K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\le K^3$ such thatThe set $A^2$ is symmetric and contains the identity element, so this inclusion is exactly the covering condition showing that

Small doubling alone is insufficient in a noncommutative group. Let $H$ be a finite group, let $G=H\ast ⟨x⟩$ be its free product with an infinite cyclic group, and putThen $A$ is symmetric and $|A^2|\le 5|H|+4=O(|A|)$, while $A^3$ contains the double coset $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_{1}x{h}_2$, so $|HxH|=|H{|}^2$. Letting $|H|\to \infty$ proves the small doubling does not control tripling in a noncommutative group phenomenon.

For every $x\in B{C}^{-1}$, choose one representation $x=b_x{c}_x^{-1}$. DefineThe image determinesand the chosen representative then determines $a=(a{b}_x^{-1})b_x$. Thus $F$ is injective, and counting its domain and codomain proves the Noncommutative Ruzsa triangle inequality