Classification of finite p-groups of normal rank one

Statement
Suppose $$P$$ is a finite $$p$$-group whose normal rank is one: every Abelian normal subgroup of $$P$$ is cyclic. Then:


 * If $$p$$ is odd, $$P$$ is itself cyclic.
 * If $$p = 2$$, $$P$$ is either cyclic, or it has a cyclic maximal subgroup, with the quotient acting by multiplication by either $$-1$$ (dihedral group and generalized quaternion group) or $$2^{r-2} - 1$$, where $$|P| = 2^r$$.

Related facts

 * Classification of finite p-groups of characteristic rank one
 * Classification of finite p-groups of rank one

Facts used

 * 1) uses::Classification of finite p-groups of characteristic rank one: If $$P$$ is a finite $$p$$-group of characteristic rank one, $$P$$ is a central product of subgroups $$E,R$$ where $$E$$ is trivial or extraspecial, and $$R$$ is cyclic for odd $$p$$, and for $$p=2$$, $$R$$ is either cyclic or it has a cyclic maximal subgroup, with the quotient acting by multiplication by either $$-1$$ (dihedral group and generalized quaternion group) or $$2^{r-2} - 1$$, where $$|R| = 2^r$$.
 * 2) uses::Central factor implies transitively normal
 * 3) uses::Extraspecial and normal rank one implies quaternion group

Proof
Given: A finite $$p$$-group $$P$$ of normal rank one.

To prove: For $$p$$ odd, $$P$$ is cyclic. For $$p = 2$$, $$P$$ is a dihedral group, generalized quaternion group, or is obtained as a semidirect product of a cyclic maximal subgroup with a subgroup acting on it via multiplication by $$2^{r-2} - 1$$ where the order of $$P$$ is $$2^r$$.

Proof': Since $$P$$ has normal rank one, it also has characteristic rank one, and thus $$P$$ is a finite $$p$$-group of characteristic rank one. By fact (1), we can write $$P = ER$$ with the conditions specified in fact (1). Note also that $$E \cap R$$ cannot be trivial, otherwise we would have Abelian normal non-cyclic subgroups such as the product of nontrivial cyclic normal subgroups of both.

Now, since $$E$$ is a central factor of $$P$$, fact (2) states that any normal subgroup of $$E$$ is normal in $$P$$. Thus, any Abelian normal subgroup of $$E$$ is an Abelian normal subgroup of $$P$$, and is thus cyclic. Thus, $$E$$ itself has normal rank one. Applying fact (3) tells us that either $$E$$ is trivial or $$p = 2$$ and $$E$$ is the quaternion group.

For $$p$$ odd, $$E$$ is trivial, and $$P = R$$. Since $$R$$ is cyclic, $$P$$ is cyclic and we are done.

For $$p = 2$$, we make two cases:


 * $$E$$ is trivial: In this case, $$P = R$$, and we observe that the possibilities for $$R$$ are precisely the possibilities we need to prove are permisslbe for $$P$$.
 * $$E$$ is the quaternion group: If $$R$$ has order two, $$R \cap E$$ being nontrivial, must equal $$R$$. In this case, $$G = ER = E$$ is the quaternion group. In case $$R$$ has order bigger than two, the structural possibilities for $$R$$ show that there must exist a cyclic normal subgroup $$L$$ of $$R$$ of order four. Let $$H$$ be a cyclic normal subgroup of order four in $$E$$. Since $$E$$ and $$R$$ commute element-wise, so do $$H$$ and $$L$$ so $$HL$$ is an Abelian normal subgroup of $$P$$ of exponent four. Further, since $$E \cap R$$ is nontrivial and central, $$|E \cap R| = 2$$, so $$|H \cap L| = 2$$ and $$|HL| = 8$$. Thus, $$HL$$ is noncyclic, leading to a contradiction. Thus, we conclude that the case of $$E$$ a quaternion group cannot hold here.

Textbook references

 * , Page 199, Theorem 4.10(i), Section 5.4 ($$p$$-groups of small depth)