2
$\begingroup$

Let $G$ be a finite non-abelian $p$-group such that $G$ contains at least an element of order $p^2$ and for every nontrivial normal subgroup $N$, $G/N$ has not any elements of order p^2 and G/Z(G) is non-abelian group and G/G' is elementary abelian group. 1-Is there a group with these properties? 2-Are there up to $p$ non-trivial proper subgroups such that intersection of every non-trivial proper subgroup of $G$ with at least one subgroup of these $p$ subgroups is non-trivial?

$\endgroup$
3
  • $\begingroup$ What do you mean by "for every $N$"? Because it cannot include the trivial subgroup. If you mean for every $N\neq 1$, then it follows that $G$ is extraspecial. $\endgroup$
    – YCor
    Oct 28, 2016 at 16:51
  • $\begingroup$ @YCor: Must it? Take the extraspecial group of order $p^3$ and exponent $p$, and adjoin a central $p$th root to its nontrivial central element, $G=\langle x,y,z\mid x^p=y^p=1, [z,x]=[z,y]=1, [x,y]=z^p\rangle$. The center is generated by $z$, and every nontrivial normal subgroup contains $z^p$; the quotient $G/\langle z^p\rangle$ is elementary abelian of rank $3$. But $G$ is not extra-special. $\endgroup$ Oct 28, 2016 at 20:03
  • $\begingroup$ @ArturoMagidin Yes, sorry I meant a $p$-group whose derived subgroup has order $p$ (the center might be larger). Anyway the only other possibility (in this context) is that the center is cyclic of order $p^2$, and it follows that the only non-extraspecial examples are the central product of an extraspecial group with a cyclic group of order $p^2$. $\endgroup$
    – YCor
    Oct 28, 2016 at 20:55

1 Answer 1

3
$\begingroup$

This is a strange question but I think the answer is yes, at least for $p>2$. It follows from the second assumption that every nontrivial normal subgroup contains the commutator subgroup, so in particular every order p subgroup of the center ontains the commutator subgroup. Because there always is one, the commutator is order p and central. The quotient by this subgroup is an elementary abelian $p$-group.

Let $a$ and $b$ be two elements of order $p$. Then as $a$ and $b$ commute with $[a,b]$, we have $(ab)^p =a^p b^p [a,b]^{p(p-1)/2}=[a,b]^{p(p-1)/2}$. Because $p>2$ and $[a,b]^p =1$, we have $(ab)^p=1$. So the elements of order $p$ actually form a subgroup. By the first assumption this is proper. Of course every non-trivial subgroup intersects it nontrivially.

$\endgroup$
1
  • $\begingroup$ A similar argument works when $P$ is a $2$-group containing an element of order $8$ with the property that $P/N$ is Abelian for each non-identity normal subgroup $N$ of $P$. For then $P$ is nilpotent of class at most two, and the elements of order dividing $4$ form a subgroup. $\endgroup$ Oct 30, 2016 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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