Metacyclic $p$-groups of order $p^{4k+4}$ with unbounded class and nonordinary mixed varieties

Problem

For a fixed prime $p$, does sufficiently large nilpotency class force the mixed variety generated by a finite $p$-group to be an ordinary variety?

$$\exists c_p\quad\forall G\text{ finite }p\text{-group},\quad \operatorname{cl}(G)\ge c_p \Longrightarrow\operatorname{mvar}(G)\text{ ordinary}\;?$$

The prime must remain fixed while the nilpotency classes grow.

Mixed identities and the class threshold

Let $X=\{x_0,x_1,\ldots\}$ and write $G[X]=G*F(X)$. For a homomorphism $\phi:G\to H$ and a map $\xi:X\to H$, denote by

$$\operatorname{ev}_{\phi,\xi}:G[X]\longrightarrow H$$

the homomorphism extending $\phi$ and $\xi$. Put

$$I_G[X]=\{w\in G[X]:\operatorname{ev}_{1_G,\xi}(w)=1 \text{ for every }\xi:X\to G\}.$$

Following Anashin (1), the mixed variety generated by $G$ is defined by

$$\tag{3} H\in\operatorname{mvar}(G) \quad\Longleftrightarrow\quad \exists\phi:G\to H\quad \operatorname{ev}_{\phi,\xi}(w)=1 \quad(w\in I_G[X],\ \xi:X\to H).$$

The same coefficient homomorphism must work for all identities, and it need not be injective. An ordinary variety is a class defined by a set of words in $F(X)$, without coefficients.

Problem 10.4 of the Kourovka Notebook (2) asks whether, for a fixed prime $p$, the mixed variety generated by an arbitrary finite $p$-group of sufficiently large nilpotency class is an ordinary variety. Anashin (1, p. 176) already gives the class-two example

$$\langle a,b\mid a^p=b^{p^2}=1,\ a^{-1}ba=b^{1+p}\rangle \qquad(p\text{ odd}),$$

which satisfies $x^p[x,a^{-1}]=1$ but generates a nonordinary mixed variety. We use a principal unit congruent to $1$ modulo $p^2$ to obtain unbounded nilpotency class for each fixed prime, including $p=2$. Throughout,

$$[x,y]=x^{-1}y^{-1}xy, \qquad \gamma_1(G)=G,\qquad \gamma_{j+1}(G)=[\gamma_j(G),G].$$

Theorem 1. For every prime $p$ and every integer $k\geq0$, there is a finite $p$-group $G_{p,k}$ such that

$$|G_{p,k}|=p^{4k+4},\qquad \operatorname{cl}(G_{p,k})\geq k+2, \qquad \operatorname{mvar}(G_{p,k})\text{ is not ordinary}.$$

The group is a split extension of a cyclic group of order $p^{2k+3}$ by a cyclic group of order $p^{2k+1}$.

Taking $k$ at least any proposed threshold gives a negative answer to Problem 10.4. The prime remains fixed throughout the construction.

Lemma 2. Suppose that $G$ contains an abelian subgroup $A$, and that $r\geq1$ and $d\in G$ satisfy

$$x^r[x,d]=1\qquad(x\in G).$$

If some $a\in A$ satisfies $a^r\ne1$, then $\operatorname{mvar}(G)$ is not ordinary.

Principal units and geometric sums

Write

$$S_m(u)=\sum_{i=0}^{m-1}u^i,\qquad S_0(u)=0.$$

We will use

$$\tag{10} S_{m+n}(u)=S_m(u)+u^mS_n(u),\qquad S_{mn}(u)=S_m(u)S_n(u^m).$$

These identities follow by splitting the sum into consecutive intervals of equal length.

Lemma 3. For every natural number $p$ and every $h\geq0$, there is an integer $a_h$ such that

$$\tag{11} (1+p^2)^{p^h}=1+p^{h+2}(1+pa_h).$$

Moreover, if $R$ is a commutative ring and $u=1+p^2z\in R$, then for every $h\geq0$ there is $w_h\in R$ such that

$$\tag{12} S_{p^h}(u)=p^h(1+pw_h).$$

Here integer coefficients in the second identity are interpreted in $R$.

The metacyclic group and its mixed identity

Fix a prime $p$ and $k\geq0$, and put

$$N=p^{2k+3},\qquad M=p^{2k+1},\qquad r=p^{2k+2},\qquad t=p^{2k},\qquad R=\mathbb Z/N\mathbb Z.$$

Let $q=1+p^2\in R$. Since $\gcd(1+p^2,p)=1$, the element $q$ is a unit. Equation (11), with $h=2k+1$, gives $q^M=1$. Therefore the additive cyclic group $C=\mathbb Z/M\mathbb Z$ acts on the additive group of $R$ by

$$j\cdot z=q^{-j}z.$$

Define $G=G_{p,k}=R\rtimes C$. Its multiplication and inverse are

$$\tag{19} (z,j)(w,\ell)=(z+q^{-j}w,j+\ell), \qquad (z,j)^{-1}=(-q^jz,-j).$$

In particular,

$$|G|=NM=p^{4k+4}.$$

Thus $G$ is a finite $p$-group and is nilpotent. Put

$$a=(0,1),\qquad b=(1,0),\qquad d=a^{-t}=(0,-t).$$

All subsequent calculations involving $q$ and $r$ take place in $R$. We have

$$\tag{22} rp=0,\qquad r^2=0,\qquad q^t=1+r,\qquad q^{-t}=1-r.$$

The first two equalities follow from the modulus $N$. For the third, use (11) at $h=2k$ and $rp=0$. The last follows from $(1+r)(1-r)=1$.

Every power of $q$ or $q^{-1}$ has the form $1+p^2z$. For the inverse, this follows directly from

$$q^{-1}=1-p^2q^{-1};$$

the set of elements of the form $1+p^2z$ is closed under multiplication. It follows that, for every $j\in C$,

$$\tag{24} rq^j=r,\qquad S_r(q^{-j})=r.$$

Indeed, the first assertion uses $rp=0$, and the second follows from (12) with $h=2k+2$, again using $rp=0$.

For $x=(z,j)$, induction using (19) gives

$$x^m=(S_m(q^{-j})z,mj).$$

Since $M\mid r$, equations (24) yield

$$\tag{26} x^r=(rz,0).$$

Also,

$$d^{-1}xd=(q^{-t}z,j),$$

and hence

$$\tag{28} [x,d]=\bigl(q^j(q^{-t}-1)z,0\bigr)=(-rz,0).$$

Combining (26) and (28) proves

$$\tag{29} x^r[x,d]=1\qquad(x\in G).$$

The subgroup $A=R\times\{0\}$ is abelian. Its element $b=(1,0)$ satisfies

$$b^r=(r,0)\ne(0,0),$$

because $0<r<N$. Lemma 2 now shows that $\operatorname{mvar}(G)$ is not ordinary.

The lower-central-series bound

For $j\geq0$, put $v_j=(p^{2j},0)\in G$. The coordinate multiplication gives

$$[(z,0),a]=(p^2z,0)\qquad(z\in R), \qquad [v_j,a]=v_{j+1}.$$

Since $v_0\in G=\gamma_1(G)$, induction gives $v_j\in\gamma_{j+1}(G)$. In particular,

$$v_{k+1}=(p^{2k+2},0)\ne1, \qquad v_{k+1}\in\gamma_{k+2}(G).$$

Thus $\operatorname{cl}(G)\geq k+2$. Together with the order calculation and (29), this proves Theorem 1.

References

Preprint · Lean (GitHub)

1. V. S. Anashin, Mixed identities and mixed varieties of groups, Math. USSR-Sb. 57 (1987), no. 1, 171–182, doi:10.1070/SM1987v057n01ABEH003062.

2. E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved problems in group theory, 21st ed., September 2026 update, Problem 10.4, arXiv:1401.0300v46.