Classification of finite p-groups of normal rank one

From Groupprops
Jump to: navigation, search
This article gives a classification statement for certain kinds of groups of prime power order, subject to additional constraints.
View other such statements


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

Related facts

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


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