Extraspecial and normal rank one implies quaternion group

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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

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 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.


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.