# Subgroup structure of groups of prime-cube order

Jump to: navigation, search
This article gives specific information, namely, subgroup structure, about a family of groups, namely: groups of prime-cube order.
View subgroup structure of group families | View other specific information about groups of prime-cube order

Note that the tables here work correctly only for groups of order $p^3$ where $p$ is an odd prime. The case $p = 2$ behaves differently. For that case, see subgroup structure of groups of order 8.

## Numerical information on counts of subgroups by isomorphism type

FACTS TO CHECK AGAINST FOR SUBGROUP STRUCTURE: (group of prime power order)
Lagrange's theorem (order of subgroup times index of subgroup equals order of whole group, so all subgroups have prime power orders)|order of quotient group divides order of group (and equals index of corresponding normal subgroup, so all quotients have prime power orders)
prime power order implies not centerless | prime power order implies nilpotent | prime power order implies center is normality-large
size of conjugacy class of subgroups divides index of center
congruence condition on number of subgroups of given prime power order: The total number of subgroups of any fixed prime power order is congruent to 1 mod the prime.

### Number of subgroups per isomorphism type

The number in each column is the number of subgroups in the given group (by row) of the isomorphism type of the (column) group:

Group Second part of GAP ID Nilpotency class Minimum size of generating set group of prime order cyclic group of prime-square order elementary abelian group of prime-square order Total (row sum + 2, for trivial group and whole group)
cyclic group of prime-cube order 1 1 1 1 1 0 4
direct product of ... 2 1 2 $p + 1$ $p$ 1 $2p + 4$
prime-cube order group:U(3,p) 3 2 2 $p^2 + p + 1$ 0 $p + 1$ $p^2 + 2p + 4$
semidirect product of ... 4 2 2 $p + 1$ $p$ 1 $2p + 4$
elementary abelian group of prime-cube order 5 1 3 $p^2 + p + 1$ 0 $p^2 + p + 1$ $2p^2 + 2p + 4$

### Number of conjugacy classes of subgroups per isomorphism type

The number in each column is the number of conjugacy classes of subgroups in the given group (by row) of the isomorphism type of the (column) group:

Group Second part of GAP ID nilpotency class minimum size of generating set group of prime order cyclic group of prime-square order elementary abelian group of prime-square order Total (row sum + 2, for trivial group and whole group)
cyclic group of prime-cube order 1 1 1 1 1 0 4
direct product of ... 2 1 2 $p + 1$ $p$ 1 $2p + 4$
prime-cube order group:U(3,p) 3 2 2 $p + 2$ 0 $p + 1$ $2p + 5$
semidirect product of ... 4 2 2 2 $p$ 1 $p + 5$
elementary abelian group of prime-cube order 5 1 3 $p^2 + p + 1$ 0 $p^2 + p + 1$ $2p^2 + 2p + 4$

### Number of normal subgroups per isomorphism type

Group Second part of GAP ID nilpotency class minimum size of generating set group of prime order cyclic group of prime-square order elementary abelian group of prime-square order Total (row sum + 2, for trivial group and whole group)
cyclic group of prime-cube order 1 1 1 1 1 0 4
direct product of ... 2 1 2 $p + 1$ $p$ 1 $2p + 4$
prime-cube order group:U(3,p) 3 2 2 1 0 $p + 1$ $p + 4$
semidirect product of ... 4 2 2 1 $p$ 1 $p + 4$
elementary abelian group of prime-cube order 5 1 3 $p^2 + p + 1$ 0 $p^2 + p + 1$ $2p^2 + 2p + 4$

### Number of automorphism classes of subgroups per isomorphism type

The number in each column is the number of automorphism classes of subgroups in the given group (by row) of the isomorphism type of the (column) group:

Group Second part of GAP ID nilpotency class minimum size of generating set group of prime order cyclic group of prime-square order elementary abelian group of prime-square order Total (row sum + 2, for trivial group and whole group)
cyclic group of prime-cube order 1 1 1 1 1 0 4
direct product of ... 2 1 2 2 1 1 6
prime-cube order group:U(3,p) 3 2 2 2 0 1 5
semidirect product of ... 4 2 2 2 1 1 6
elementary abelian group of prime-cube order 5 1 3 1 0 1 4

### Number of characteristic subgroups per isomorphism type

Group Second part of GAP ID nilpotency class minimum size of generating set group of prime order cyclic group of prime-square order elementary abelian group of prime-square order Total (row sum + 2, for trivial group and whole group)
cyclic group of prime-cube order 1 1 1 1 1 0 4
direct product of ... 2 1 2 1 0 1 4
prime-cube order group:U(3,p) 3 2 1 0 0 3
semidirect product of ... 4 2 2 1 0 1 4
elementary abelian group of prime-cube order 5 1 3 0 0 0 2

## Numerical information on counts of subgroups by order

### Number of subgroups per order

We have the following:

• Congruence condition on number of subgroups of given prime power order tells us that the number of subgroups of order $p$ is congruent to 1 mod $p$, and so is the number of subgroups of order $p^2$. Since the non-normal subgroups form conjugacy classes of size equal to a nontrivial power of $p$, the number of normal subgroups of each order is also congruent to 1 mod $p$.
• All the subgroups of order $p^2$ are normal, because prime power order implies nilpotent and nilpotent implies every maximal subgroup is normal. In particular, we obtain that (number of subgroups of order $p^2$) = (number of normal subgroups of order $p^2$).
• The subgroups of order $p^2$ all contain the Frattini subgroup, and thus correspond, via the fourth isomorphism theorem, to the maximal subgroups of the Frattini quotient, which is an elementary abelian group of order $p^r, 1 \le r \le 3$, where $r$ is the minimum size of generating set. The number of subgroups is $(p^r - 1)/(p - 1)$, which could be one of the numbers $1, p + 1, p^2 + p + 1$.
• The normal subgroups of order $p$ are precisely the subgroups of order $p$ inside the socle, which is the first omega subgroup of the center. In particular, it is an elementary abelian group of order $p^s, 1 \le s \le 3$, and the number of subgroups of order $p$ in it is $(p^s - 1)/(p - 1) = p^{s-1} + p^{s-2} + \dots + p + 1$. Thus, the number of normal subgroups of order $p$ is $1, p + 1, p^2 + p + 1$. For a non-abelian group, it must be 1.
• For abelian groups, every subgroup is normal. Moreover, we have subgroup lattice and quotient lattice of finite abelian group are isomorphic, so we get (number of subgroups of order $p$) = (number of normal subgroups of order $p$) = (number of subgroups of order $p^2$) = (number of normal subgroups of order $p^2$).
Group Second part of GAP ID number of subgroups of order $p$ number of normal subgroups of order $p$ number of subgroups of order $p^2$ number of normal subgroups of order $p^2$
cyclic group of prime-cube order 1 1 1 1 1
direct product of ... 2 $p + 1$ $p + 1$ $p + 1$ $p + 1$
prime-cube order group:U(3,p) 3 $p^2 + p + 1$ 1 $p + 1$ $p + 1$
semidirect product of ... 4 $p + 1$ 1 $p + 1$ $p + 1$
elementary abelian group of prime-cube order 5 $p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$