1
$\begingroup$

Is there a finite $p$-group $G$ such that :

(a) $G= \langle A,x,y \rangle$, with $G/Z(G)$ has exponent $p$, $A$ is a maximal abelian normal subgroup of $G$, and $G/A$ has order $p^2$ (thus it is elementary abelian of rank $2$);

(b) the class of $G$ is equal to $p$;

(c) the subgroup $\langle B,y \rangle$ has class exactly $p$, where $B$ denotes the center of $\langle A,x \rangle$.

Thanks in advance

Improve this question
$\endgroup$
3
  • $\begingroup$ As it stands, you do not exclude the possibility that $x \in A$ and $A = B$. I think the group $G = C_{p} \wr C_{p}$ satisfies these conditions with $A = B$ elementary Abelian of rank $p$ and any $x \in A$, any $y \in G \backslash A$. $\endgroup$
    – Geoff Robinson
    Commented Nov 24, 2013 at 20:47
  • $\begingroup$ This is exactly the case that I want to avoid. The question is not very well formulated, I will edit it. $\endgroup$
    – Yassine Guerboussa
    Commented Nov 24, 2013 at 20:50
  • $\begingroup$ In the first Geoff Robinson's example (there $A=\Phi(G)$), the class of B is p-1<p and so his group does not satisfy (c). $\endgroup$
    – Yakov
    Commented Dec 22, 2015 at 17:46

1 Answer 1

Reset to default
6
$\begingroup$

I think the same example works that I gave to your question https://math.stackexchange.com/questions/571949.

Let $H = C_p \wr C_p$, and $G = H_1 \times H_2$ with $H_1 \cong H_2 \cong H$. So $|H|=p^{p+1}$, $|G|=p^{2(p+1)}$.

The maximal abelian normal subgroup $A$ is the direct product of the base groups of $H_1$ and $H_2$, and is elementary abelian of order $p^{2p}$.

Now $H$ has class $p$ and hence so does $G$, which is property (b).

The centre of $H$ has order $p$ and $H/Z(H)$ has class $p-1$ and is generated by two elements of order $p$, so $H/Z(H)$ has exponent $p$ and hence the same applies to $G/Z(G)$, which gives property (a).

We take $x \in H_1$ and $y \in H_2$ to be elements outside of $A$. So $B=Z(\langle A,x \rangle)$ includes the base group of $H_2$, and hence $\langle B,y \rangle$ contains $H_2$ and has class $p$.

Improve this answer
$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .