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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?

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  • $\begingroup$ 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 '16 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. $\endgroup$ – Arturo Magidin Oct 28 '16 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 '16 at 20:55
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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.

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  • $\begingroup$ 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. $\endgroup$ – Geoff Robinson Oct 30 '16 at 13:54

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