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 fact about::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) uses::Maximal among Abelian normal implies self-centralizing in nilpotent
 * 2) uses::Classification of finite p-groups with cyclic normal self-centralizing subgroup

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, $$\Omega_1(P) \le Z(P)$$. On the other hand, clearly $$Z(P) \le \Omega_1(P)$$ 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 $$p = 2$$.