Extraspecial and normal rank one implies quaternion group

Statement

Let $p$ be a prime number and $P$ be a finite nontrivial $p$-group. Suppose $P$ is an Extraspecial group (?): $P'=\Phi (P)=Z(P)$ and this subgroup has order $p$. Further, suppose that the normal rank of $P$ is one: every Abelian normal subgroup of $P$ is cyclic. Then, $p=2$ and $P$ is the quaternion group.

Related facts

• Classification of finite p-groups of characteristic rank one
• Classification of finite p-groups of normal rank one
• Classification of finite p-groups of rank one

Facts used

1. Maximal among Abelian normal implies self-centralizing in nilpotent
2. Classification of finite p-groups with cyclic normal self-centralizing subgroup

Proof

Proof outline

• We first show that $Z(P)=\Omega _{1}(P)$: Every element of order $p$ is in the center.
• We next show that if $g\in P\setminus Z(P)$, then the cyclic subgroup generated by $g$ has order $p^{2}$, contains $Z(P)$, and is a self-centralizing normal subgroup of $P$.
• We finally use fact (2) to reduce to three possibilities for $P$, and we eliminate two of them by inspection.

Proof details

Given: An extraspecial $p$-group $P$ of normal rank one.

To prove: $p=2$ and $P$ is isomorphic to the quaternion group.

Proof:

We first show that $Z(P)=\Omega _{1}(P)$: Suppose $x\in P\setminus Z(P)$ such that $x$ has order $p$. Then, $x$ commutes with $Z(P)$, so the subgroup generated by $x$ and $Z(P)$ is elementary Abelian of order $p^{2}$. Since it contains the commutator subgroup $Z(P)=P'$, it is also normal. Hence, we have an Abelian normal subgroup of $P$ that is not cyclic -- a contradiction. Thus, $\Omega _{1}(P)\leq Z(P)$. On the other hand, clearly $Z(P)\leq \Omega _{1}(P)$ so $\Omega _{1}(P)=Z(P)$.

Now, since $P$ is extraspecial, $P/Z(P)$ is elementary Abelian, so given any $g\in P\setminus Z(P)$, $g^{p}\in Z(P)$. By the observations just made, $g$ has order $p^{2}$. Let $H$ be the cyclic subgroup generated by $g$. We claim that $H$ is a self-centralizing subgroup of $P$. Clearly, $H$ is maximal among cyclic normal subgroups, since every element has order at most $p^{2}$, and hence, by normal rank one, $H$ is maximal among Abelian normal subgroups. Thus, by fact (1), $H$ is a self-centralizing subgroup of $P$.

Thus, $P$ has a self-centralizing cyclic normal subgroup of order $p^{2}$. Fact (2) reduces us to three possibilities for $p$. We easily see that two of these (the dihedral group of order eight and the non-Abelian group corresponding to an odd prime) do not satisfy the condition of normal rank one. This forces $P$ to be the quaternion group and $p=2$.