Non-Abelian group of exponent $3$ and Nilpotent class $2$ .

Question

Let $G$ be a non-Abelian group such that $G^3 = 1$ and $G$ is a nilpotent class $2$ group, with order $3^{32}$. Our task is to determine the structure of the group $G$ or identify any information about the type of structure it possesses.

I only know that its structure will be something like a semi-direct product of cyclic groups of orders $3$, but I don’t know how many factors will be there. Like $C^{20}_3\rtimes C^{12}_3$ or something like $C^{10}_3\rtimes C^{10}_3\rtimes C^{12}_3.$

One more question is

Can I say that group $\mathbb Z^{k_{1}}_p\rtimes\mathbb Z^{k_2}_p\cdots\rtimes\mathbb Z^{k_r}_p$ ($p$is a fixed prime, $k_i\geq 2$) has nilpotent $r$-class with non-trivial action?

Please help to resolve this problem. Thank you in advance.

Answer 1

Since nobody has attempted to answer the main question, I will do so. The number of isomorphism classes of groups satisfying the conditions is huge, and I don't believe that any kind of enumeration of these isomorphism classes is computationally feasible. It is possible to give a general description of these groups - the difficult problem would be to decide when two such groups are isomorphic.

For a group $G$ of this type, let $|G'|=3^m$ and $n=32-m$. Then $G$ can be generated by $n$ elements $x_1,\ldots,x_n$. To define $G$ we need to specify the value of the $n(n-1)/2$ commutators $[x_i,x_j]$ with $1 \le i < j \le n$. Each such commutator is an element $z_{ij}$ of $G'$, where the $z_{ij}$ are required to generate $G'$.

Each such specification of the $z_{ij}$ defines a group in the class. If we write each $z_{ij}$ as a word in a generating set $y_1,\ldots,y_m$ of $G'$, then $G$ is defined by the presentation $$\langle x_i\,(1 \le i \le n),\ y_j\,(1 \le j \le m) \mid x_i^3=y_j^3=[x_i,y_j]=1\, \forall i,j,\ [x_i,x_j]=z_{ij}\,\forall i<j\rangle.$$ The requirement that the $z_{ij}$ generate $G'$ implies that $n \ge 8$, and we could consider the groups with $8 \le n \le 31$ as $24$ separate classes.

As I said above, deciding which of these groups are isomorphic is difficult and probably not computationally feasible because in principle we need to consider the actions of ${\rm GL}(n,3)$ and ${\rm GL}(m,3)$ on $G/G'$ and $G'$.

But we can do some rough estimates (as was done by G. Higman in his proofs of lower bounds on the number of isomorphism classes of $p$-groups). The number of possible $z_{ij}$ without considering the generating $G'$ requirement is $3^{mn(n-1)/2}$, but the vast majority of such choices will generate $G'$.

The number of such groups in any single isomorphism class is at most $|{\rm GL}(n,3)||{\rm GL}(m,3)| \le 3^{m^2+n^2}$, so we get at least $3^{mn(n-1)/2-m^2-n^2}$ isomorphism classes which, for example with $n=21,\,m=10$, gives at least $3^{1748}$ isomorphism classes.

Answer 2

This should probably just be a comment but it is too complicated to write in the comment boxes.

Your second question has a special case: Has the group $\mathbb{Z}_p^k\rtimes \mathbb{Z}_p$ nilpotency class $2$?

The answer to this is "No".

As an example, let $p\geqslant 3$, and $k=3$. Consider the semi-direct product $\mathbb{Z}_p^3\rtimes \mathbb{Z}_p$ in the concrete form

$$ \big\langle \begin{pmatrix}I & 0\\v &1 \end{pmatrix},\begin{pmatrix}A & 0\\0 &1 \end{pmatrix} \mid v\in\mathbb{F}_p^3\big\rangle $$ where $A=\begin{pmatrix} 1 & 1 & 0\\0& 1 &1\\0& 0 &1\end{pmatrix}$.

The nilpotency class is $3$. Replacing $3$ by any $k$ with $k\leqslant p$, and suitably adapting $A$ you will find examples of nilpotency class as great as $p$.

Answer 3

This is not really a serious answer but mainly an illustration of how useless is the idea of "some kind of semidirect product of groups of order 3".

Every group of exponent $3$ of order $3^{32}$ has the form $C_3 \rtimes C_3 \rtimes \cdots \rtimes C_3$ ($32$ factors). Indeed, every finite group of exponent $p$ has the form $C_p \rtimes \cdots \rtimes C_p$.

Proof: Let $G$ be a finite group of exponent $p$. Since $G$ is nilpotent, its abelianization is nontrivial, so there is a nontrivial homomorphism $G \to C_p$. Since $G$ has exponent $p$, this map splits, so $G \cong H \rtimes C_p$ for some group $H$, and the claim follows by induction.