Let *R* be a commutative ring with a unit 1.

We consider the following group of matrices over *R*
$$M=\{A=\left(\begin{array}{c}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z\\ 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y\\ 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\end{array}\right):x,y,z\in R\}.$$

An element of *M* can be identified with a triple (*x*, *y*, *z*) of elements of *R*. Then the composition (the product) of the triples is defined by (*x*_{1}, *y*_{1}, *z*_{1})(*x*, *y*, *z*) = (*x*_{1} + *x*, *y*_{1} + *y*, *z*_{1} + *z* + *x*_{1}*y*).

Later we shall also perform the addition of the triples as elements of *R* ⊕ *R* ⊕ *R*.

The elements (0,0, *z*) lie in the center of *M* and form an abelian normal subgroup *C* of *M*. The quotient *M*/*C* is abelian so the group *M* is solvable and it is known that solvable groups are uniformly amenable. We want to find a growth function *q*(*t*, *k*) for this group, which will also give a direct proof of the fact, that *M* is uniformly amenable.

This example was inspired by a paper [10] by J. Wysoczański. He considered a group *G* which is a direct sum of groups *M* over all prime fields *F*_{p} = ℤ/ _{p}ℤ. If we take for the ring *R* the direct sum of the fields *F*_{p} and we adjoin a unit - the element with each coordinate equal to 1 in the corresponding field we get a ring *R*_{1} with a unit. The group *G* is a subgroup of the group *M* over *R*_{1}.

We now consider the group *M* over an arbitrary commutative ring *R* with a unit. Let *A* = {*a*_{0}, *a*_{1}, …, *a*_{k−1}} ⊂ *M* where *a*_{i} = (−*x*_{i}, −*y*_{i}, −*z*_{i}) for *i* = 0, 1, …, *k* − 1.

The minus sign is for the convenience of the subsequent description where the negative coefficients play a special role.

We want to estimate the number |*AU*|. By Lemma 2.1 we may assume that *a*_{0} = (0, 0, 0) is the unit of the group *M*. For a fixed *n* ∈ ℕ we define *U*_{n} ⊂ *M*,
$${U}_{n}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\{u\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}({u}_{1},{u}_{2},\phantom{\rule{thinmathspace}{0ex}}{u}_{3}):{u}_{1}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Sigma}}_{i}{t}_{i}{x}_{i},\phantom{\rule{thinmathspace}{0ex}}{u}_{2}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Sigma}}_{i}{t}_{i}{y}_{i},\phantom{\rule{thinmathspace}{0ex}}{u}_{3}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Sigma}}_{i}{t}_{i}{z}_{i}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Sigma}}_{i,j}{t}_{i,j}{x}_{i}{y}_{j}\}$$

where *i*, *j* ∈ {1, 2, …, *k* − 1}, *t*_{i} ∈ {0, 1, …, *n* − 1}, *t*_{i,j} ∈ {0, 1, …, *n*^{2} − 1}.

An element of *U*_{n} is determined by (*k* − 1)^{2} + (*k* − 1) integer coefficients. The set *J* of indices of these coefficients consists of integers and pairs of integers *J* = {*i* : 1 ≤ *i* ≤ *k* − 1, (*i*, *j*) : 1 ≤ *i*, *j* ≤ *k* − 1}.

We shall denote an element *u* ∈ *U* by *u* = (*t*_{α})_{α ∈ J}. Then, by the definition of the composition law in *M*, *w* = *a*_{s}*u* has a similar form, *w* = (*f*_{α}), where *f*_{s} = *t*_{s} − 1, *f*_{s,j} = *t*_{s,j} − *t*_{j} and *f*_{α} = *t*_{α} for other *α* ∈ *J*. In particular *f*_{α} may be negative if *α* = *s* or *α* = (*s*, *j*) and *f*_{s} ≥ − 1 and *f*_{s,j} ≥ 1 − *n*.

#### Lemma 3.1

*For any* *s* ∈ {1, 2, …, *k* − 1} *we have*
$|{a}_{s}{U}_{n}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus}\phantom{\rule{thinmathspace}{0ex}}{U}_{n}|<(\frac{{2}^{k}}{n})|{U}_{n}|.$

#### Proof

We fix an integer *s* ∈ {1, 2, …, *k* − 1}. We consider elements of *a*_{s}*U*_{n} \ *U*_{n} and for each such element we choose one representative *w* = (*f*_{α}). For any non-empty subset of indices *I* ⊂ {*s*,(*s*, 1), (*s*, 2), …, (*s*, *k* − 1)} we consider the set *B*_{I} of all chosen representatives *w* = (*f*_{α}) of the elements in *a*_{s}*U*_{n} \ *U*_{n} which have the negative coefficients *f*_{α} exactly for *α* ∈ *I*. Suppose that *B*_{I} = {*w*_{1}, *w*_{2}, *w*_{d}}.

Let *e*_{I} = (*b*_{1}, *b*_{2}, *b*_{3}) ∈ *R* ⊕ *R* ⊕ *R* where *b*_{1} = *x*_{s}, *b*_{2} = *y*_{s}, *b*_{3} = *z*_{s} + *n*∑_{(s, j) ∈ I}*x*_{s}*y*_{j} if *s* ∈ *I* and *b*_{1} = 0, *b*_{2} = 0, *b*_{3} = *n* ∑_{(s, j)∈I}*x*_{s}*y*_{j} if *s* ∉ *I*.

We shall prove that the elements *w*_{i} + *me*_{I}, *i* = 1, 2, …, *d*, *m* = 1, 2, …,*n* belong to *U*_{n} and are distinct. Only the coefficients *f*_{α}, *α* ∈ *I* of *w* change. If *s* ∈ *I* then *f*_{s} = −1 and it increases by *m* when we add *me*_{I} where *1* ≤ *m* ≤ *n* so *f*_{s} falls into the proper range for the elements in *U*_{n}. If (*s*, *j*) ∈ *I* then *1* − *n* ≤ *f*_{s, j} ≤ −1 and *f*_{s, j} increases by *mn* so it also falls into the proper range for the elements in *U*_{n}.

Suppose that *w*_{i} + *m**e*_{I} = *w*_{j} + *r**e*_{I} for different pairs (*i,m*) and (*j*, *r*). If *m* = *r* then *w*_{i} = *w*_{j} which contradicts the assumptions. If *m* ≠ *r* we may assume that *m* > *r*. Then *w*_{j} = *w*_{j} + *r**e*_{I} − *r**e*_{I} = *w*_{i} + (*m* − *r*)*e*_{I} ∈ *U*_{n} which contradicts our assumptions. Therefore |*U*_{n}| ≥ *nd* and
$|{B}_{I}|\le \frac{1}{n}|{U}_{n}|.$ There are (2^{k} − 1) non-empty subsets of {*s*, (*s*, 1), (*s*, 2), …, (*s*, *k* − 1)} and each element of *a*_{s}*U*_{n} \ *U*_{n} lies in one of them so |*a*_{s}*U*_{n} \ *U*_{n}| <
$(\frac{{2}^{k}}{n})|{U}_{n}|.$ □

#### Corollary 3.2

*For every commutative ring* *R* *with unit the group* *M* *is uniformly amenable with a growth function*
$$q(\u03f5,k)=(1+\frac{(k-1)({2}^{k})}{\u03f5}{)}^{2(k-1{)}^{2}+k-1}$$

#### Proof

For a given *k* and *ϵ* choose
$n=[\frac{(k-1){2}^{k}}{\u03f5}]+1.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{(k-1){2}^{k}}{\u03f5}<n\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{(k-1){2}^{k}}{n}<\u03f5.$ Let *A* = {*a*_{0} = (0, 0, 0), *a*_{1}, …, *a*_{k−1}} ⊂ *M*. We choose the set *U*_{n} as in the previous lemma. Then |*U*_{n}| ≤ *n*^{2(k − 1)2+k − 1} ≤
$(1+\frac{(k-1){2}^{k}}{\u03f5}{)}^{2(k-1{)}^{2}+k-1}=q(\u03f5,k).$ It suffices to prove that |*A**U*_{n}| < (1 + *ϵ*)|*U*_{n}|. We have *a*_{0}*U*_{n} = *U*_{n} and by the previous lemma
$|A{U}_{n}|\le |{U}_{n}|+(k-1)(\frac{{2}^{k}}{n})|{U}_{n}|<(1+\u03f5)|{U}_{n}|.$ □

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