# Classification of finite p-groups of normal rank one

This article gives a classification statement for certain kinds of groups of prime power order, subject to additional constraints.
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## 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 $|P| = 2^r$.

## Facts used

1. 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. Central factor implies transitively normal
3. 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.

## References

### Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, More info, Page 199, Theorem 4.10(i), Section 5.4 ($p$-groups of small depth)