# Subgroup structure of groups of prime-fourth order

## Contents

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

Note that the information here applies fully only for groups of order $p^4$ where $p$ is a prime at least equal to 5. It does not apply to the cases $p = 2,3$, which are anomalous. For more on the anomalous primes, see:

## Numerical information on counts of subgroups by order

### Number of subgroups per order

Note the following:

• Since prime power order implies nilpotent and nilpotent implies every maximal subgroup is normal, the subgroups of order $p^3$ (and hence index $p$) are all normal subgroups. Thus, we have (number of subgroups of order $p^3$) = (number of normal subgroups of order $p^3$).
• The subgroups of order $p^3$, i.e., the maximal subgroups, all contain the Frattini subgroup, and hence, by the fourth isomorphism theorem, correspond to the maximal subgroups of the Frattini quotient. This is an elementary abelian group of order $p^r, 1 \le r \le 4$, where $r$ is the minimum size of generating set for the group. The number of such subgroups is $(p^r - 1)/(p - 1)$, which is thus one of the numbers $1, p + 1, p^2 + p + 1, p^3 + p^2 + p + 1$.
• The normal subgroups of order $p$ are precisely the subgroups of order $p$ in the socle, which is the first omega subgroup of the center, and is an elementary abelian group of order $p^s, 1 \le s \le 4$, where $s$ is the rank of the center. The number of normal subgroups of order $p$ is thus $(p^s - 1)/(p - 1)$, which is thus one of the numbers $1, p + 1, p^2 + p + 1, p^3 + p^2 + p + 1$. Further, for a non-abelian group, the center has order at most $p^2$ (because cyclic over central implies abelian forces the center to not be a maximal subgroup), hence the only possibilities are 1 and $p + 1$.
• For an abelian group of order $p^4$, all subgroups are normal. Further, since subgroup lattice and quotient lattice of finite abelian group are isomorphic, the number of subgroups of order $p^j$ equals the number of subgroups of order $p^{4 - j}$ for all $j$. In particular, we get (number of subgroups of order $p$) = (number of normal subgroups of order $p$) = (number of subgroups of order $p^3$) = (number of normal subgroups of order $p^3$). Separately, (number of subgroups of order $p^2$) = (number of normal subgroups of order $p^2$.
Group GAP ID 2nd part Nilpotency class Mimimum size of generating set 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 $p^2$ Number of subgroups of order $p^3$ Number of normal subgroups of order $p^3$
cyclic group of prime-fourth order 1 1 1 1 1 1 1 1 1
direct product of cyclic p^2 and cyclic p^2 2 1 2 $p + 1$ $p + 1$ $p^2 + p + 1$ $p^2 + p + 1$ $p + 1$ $p + 1$
SmallGroup(p^4,3) 3 2 2 $p^2 + p + 1$ $p + 1$ $2p^2 + p + 1$ $p + 1$ $p + 1$ $p + 1$
SmallGroup(p^4,4) 4 2 2 $p + 1$ $p + 1$ $p^2 + p + 1$ $p + 1$ $p + 1$ $p + 1$
direct product of cyclic p^3 and cyclic p 5 1 2 $p + 1$ $p + 1$ $p + 1$ $p + 1$ $p + 1$ $p + 1$
SmallGroup(p^4,6) 6 2 2 $p + 1$ 1 $p + 1$ $p + 1$ $p + 1$ $p + 1$
SmallGroup(p^4,7) 7 3 2 $p^3 + p^2 + p + 1$ 1 $2p^2 + p + 1$ 1 $p + 1$ $p + 1$
SmallGroup(p^4,8) 8 3 2 $p^2 + p + 1$ 1 $2p^2 + p + 1$ 1 $p + 1$ $p + 1$
SmallGroup(p^4,9) 9 3 2 $p^2 + p + 1$ 1 $p^2 + p + 1$ 1 $p + 1$ $p + 1$
SmallGroup(p^4,10) 10 3 2 $p^2 + p + 1$ 1 $p^2 + p + 1$ 1 $p + 1$ $p + 1$
direct product of cyclic p^2 and elementary abelian p^2 11 1 3 $p^2 + p + 1$ $p^2 + p + 1$ $2p^2 + p + 1$ $2p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$
SmallGroup(p^4,12) 12 2 3 $p^3 + p^2 + p + 1$ $p + 1$ $p^3 + 2p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$
SmallGroup(p^4,13) 13 2 3 $p^2 + p + 1$ $p + 1$ $2p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$
SmallGroup(p^4,14) 14 2 3 $p^2 + p + 1$ 1 $p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$ $p^2 + p + 1$
elementary abelian group of prime-fourth order 15 1 4 $p^3 + p^2 + p + 1$ $p^3 + p^2 + p + 1$ $p^4 + p^3 + 2p^2 + p + 1$ $p^4 + p^3 + 2p^2 + p + 1$ $p^3 + p^2 + p + 1$ $p^3 + p^2 + p + 1$

Here's how the anomalous primes differ:

• For abelian groups, the formulas work for $p = 2$ and $p = 3$ if we consider the corresponding groups (Note: The GAP IDs don't correspond exactly for $p = 2$, because of there being only 14 groups).
• For groups of class two, the formulas work for $p = 3$, if we consider the corresponding groups. The GAP IDs work.
• In addition, somewhat unexpectedly, there is a match for $p = 2$ on IDs 3 (with 3), 4 (with 4), 6 (with 6), and 13 (with 14).