9
$\begingroup$

Let $G_d$ be the group with the following presentation $$\langle x,y \mid x^{2^{d+1}}=1, x^4=y^2, [x,y,x]=x^{2^{d}}, [x,y,y]=1\rangle,$$ where $d>2$ is an integer. It is clear that $G_d$ is a finite $2$-group of nilpotency class at most $3$. It is easy to see that $[x,y]^2=1$ and since the quaternion group $Q_8$ of order $8$ is a quotient of $G_d$, $[x,y]$ has order $2$. So the nilpotency class of $G_d$ is $2$ or $3$.

The computation with GAP shows that $G_d$ is nilpotent of class exactly $3$, whenever $d=3,4,5,6,7,8,9$.

Question: Is the nilpotency class of $G$ $3$?

$\endgroup$
  • $\begingroup$ Do you know the order of $G$ (at least for d=3,4,5,6,7,8,9)? $\endgroup$ – Boris Novikov Nov 7 '11 at 13:34
  • $\begingroup$ I do not know the order of $G_d$ for an arbitraty integer $d>2$. But the order of $G_d$ for $d=3,4,5,6,7,8,9,10,11,12,13,14$ is $2^{d+3}$. $\endgroup$ – Alireza Abdollahi Nov 7 '11 at 15:05
9
$\begingroup$

To show $w$ is nontrivial in Max's presentation, note first that the group $H_d$ formed by the "subpresentation" $\langle y,z,w \mid y^{2^{d-1}} = w, w^2=z^2=1, z^y=z, w^z=w \rangle$ is equal to the abelian group $\langle y \rangle \times \langle z \rangle$ of order $2^{d+1}$, and $w$ is certainly a nontrivial element of $H_d$.

Now note that the map $y \to yz$, $z \to zw$ defines an automorphism of $H_d$ of order 4. Let $X = \langle x \rangle$ be cyclic of order $2^{d+1}$. Then we can form the semidirect product $X \ltimes H_d$ where conjugation by $x$ induces this automorphism. So $y^x=yz$, $z^x = zw$. In this semidirect product, $x^4$ and $y^2$ are both central elements of order $2^{d-1}$. So we can factor out the cyclic subgroup $\langle x^4y^{-2} \rangle$, which intersects $H_d$ trivially, and hence gives the group with Max's presentation in which $w$ is nontrivial.

| cite | improve this answer | |
$\endgroup$
11
$\begingroup$

The nilpotency class of $G_d$ is indeed always 3. One way to see this is to rewrite the presentation of $G_d$ in such a way to exhibit that it is a polycyclic group. For this purpose, let $z:=[x,y]$ and $w:=y^{2^{d-1}}=x^{2^d}$. Clearly $w$ lies in the center of $G_d$. With a little more effort we see that $$ G_d \cong \langle x, y, z, w\mid x^4 = y^2, y^{2^{d-1}} = w, z^2 = w^2 = 1 ; y^x = yz, z^x = zw, z^y = z, w^x=w^y=w^z=w \rangle $$

This is indeed a polycyclic presentation, with relative order $4, 2^{d-1}, 2, 2$. (Thus the group has order $4* 2^{d-1}* 2* 2=2^{d+3}$).

But now it is easy to read off that $[G_d,G_d] = \langle z, w\rangle$, and thus $[[G_d,G_d],G_d]=\langle w \rangle$, which is central. Hence the nilpotency class is 3.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ PS: If you don't know about polycyclic presentations, here is a short writeup: icm.tu-bs.de/ag_algebra/software/polycyclic/htm/CHAP002.htm $\endgroup$ – Max Horn Nov 7 '11 at 15:42
  • 2
    $\begingroup$ Why $w$ is non-trivial? $\endgroup$ – Alireza Abdollahi Nov 7 '11 at 16:26
  • $\begingroup$ Because the presentation I gave is a consistent polycyclic presentation (verifying that is straight forward), and so the order of each generator is a multiple of its relative order. So the order of $w$ is a multiple of 2 (it's relative order). Since we also know that $w^2=1$, it has in fact exactly order 2. Of course you can also use Derek's argument :). $\endgroup$ – Max Horn Nov 8 '11 at 16:04
0
$\begingroup$

If the nilpotency class is $2$, then we must have $[a,b,c]=1$ for all $a,b,c\in G_d$. But we have $[x,y,x]=x^{2^d}\neq 1$ from the presentation.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @RDK: From where you know $x^{2^d}$ is non-trivial? Note that the relation $x^{2^{d+1}}=1$ cannot solely imply that the order of $x$ is $2^{d+1}$. The answer of Derek Holt can help you to understand the main difficulty. $\endgroup$ – Alireza Abdollahi Apr 6 '13 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.