Subgroup structure of groups of order 3.2^n

This article gives specific information, namely, subgroup structure, about a family of groups, namely: groups of order 3.2^n.
View subgroup structure of group families | View other specific information about groups of order 3.2^n

Particular cases $n$ $2^n$ $3 \cdot 2^n$ Information on groups of order $3 \cdot 2^n$ Information on subgroup structure of groups of order $3 \cdot 2^n$
0 1 3 only cyclic group:Z3 subgroup structure of cyclic group:Z3
1 2 6 groups of order 6 subgroup structure of groups of order 6 (see specifically subgroup structure of symmetric group:S3 and subgroup structure of cyclic group:Z6)
2 4 12 groups of order 12 subgroup structure of groups of order 12
3 8 24 groups of order 24 subgroup structure of groups of order 24
4 16 48 groups of order 48 subgroup structure of groups of order 48
5 32 96 groups of order 96 subgroup structure of groups of order 96
6 64 192 groups of order 192 subgroup structure of groups of order 192
7 128 384 groups of order 384 subgroup structure of groups of order 384
8 256 768 groups of order 768 subgroup structure of groups of order 768
9 512 1536 groups of order 1536 subgroup structure of groups of order 1536

Numerical information on counts of subgroups by order

FACTS TO CHECK AGAINST FOR SUBGROUP STRUCTURE: (finite solvable group)
Lagrange's theorem (order of subgroup times index of subgroup equals order of whole group, so both divide it), |order of quotient group divides order of group (and equals index of corresponding normal subgroup)
Sylow subgroups exist, Sylow implies order-dominating, congruence condition on Sylow numbers|congruence condition on number of subgroups of given prime power order
Hall subgroups exist in finite solvable|Hall implies order-dominating in finite solvable| normal Hall implies permutably complemented, Hall retract implies order-conjugate
MINIMAL, MAXIMAL: minimal normal implies elementary abelian in finite solvable | maximal subgroup has prime power index in finite solvable group

There are only two prime factors of this number. Order has only two prime factors implies solvable (by Burnside's $p^aq^b$-theorem) and hence all groups of this order are solvable groups (specifically, finite solvable groups). Another way of putting this is that the order is a solvability-forcing number. In particular, there is no simple non-abelian group of this order.

By Lagrange's theorem, the only possible orders of subgroups are of the form $2^r, 0 \le r \le n$, and $3 \cdot 2^r, 0 \le r \le n$.

Here is some information on the existence and counts of subgroups of various orders:

1. Congruence condition on number of subgroups of given prime power order: If $p$ is a prime and $p^r$ divides the order of the group, the number of subgroups of order $p^r$ is congruent to 1 mod $p$.
• Case $p = 2$: The number of subgroups of order 2 is congruent to 1 mod 2, i.e., it is odd. Similarly, the number of subgroups of order $2^r$ is odd for each fixed $r$ satisfying $0 \le r \le n$.
• Case $p = 3$: The number of subgroups of order 3 is congruent to 1 mod 3.
2. By the fact that Sylow implies order-conjugate, we obtain that Sylow number equals index of Sylow normalizer, and in particular, divides the index of the Sylow subgroup.
• Case $p = 2$: The number of 2-Sylow subgroups (in this case, subgroups of order $2^n$ forming a single conjugacy class) divides 3, hence it is either 1 or 3. If it is 1, the 2-Sylow subgroup is a normal Sylow subgroup and the whole group is an internal semidirect product of that by the action of the 3-Sylow subgroup. If the action is trivial, we get a finite nilpotent group, otherwise we get a non-nilpotent group.
• Case $p = 3$: The number of 3-Sylow subgroups (in this case, all isomorphic to cyclic group:Z3 and all forming a single conjugacy class) is one of the numbers $1,2,4,8,\dots,2^n$. Combining with the congruence condition, we can eliminate odd powers of 2, so we get $1,4,16,\dots,2^l$ where $l$ is $n$ or $n - 1$ depending on whether $n$ is even or odd.
3. In the case of a finite nilpotent group, the number of subgroups of a given order is the product of the number of subgroups of order equal to each of its maximal prime power divisors, in the corresponding Sylow subgroup. In particular, we get (number of subgroups of order 3) = (number of subgroups of order $2^n$) = 1. Also, we get that, for each $1 \le r \le n - 1$, (number of subgroups of order $2^r$) = (number of subgroups of order $3 \cdot 2^r$).
4. In the special case of a finite abelian group, we have subgroup lattice and quotient lattice of finite abelian group are isomorphic. Thus, we get that (number of subgroups of order $2^r$) = (number of subgroups of order $2^{n-r}$) = (number of subgroups of order $3 \cdot 2^r$) = (number of subgroups of order $3 \cdot 2^{n-r}$), for each $1 \le r \le n - 1$. Also, we get (number of subgroups of order 3) = (number of subgroups of order $2^n$) = 1.
5. Finite supersolvable implies subgroups of all orders dividing the group order: For any finite supersolvable group, there are subgroups of every possible order, i.e., there are subgroups of order $2^r$ and $3 \cdot 2^r$ for $0 \le r \le n$.

Constraints on number of subgroups of order a power of 2

By (1), the number must be odd. Also, the number for the largest power of 2 (i.e., $2^n$) must be either 1 or 3.

Number of subgroups of order 2

Value of $n$ $2^n$ $3 \cdot 2^n$ Subgroup structure page Theoretical constraint on number of subgroups, with explanation Actual possibility set of number of subgroups
1 2 6 subgroup structure of groups of order 6 1 or 3, based on either of (1) or (2) 1, 3 (full possibility set)
2 4 12 subgroup structure of groups of order 12 odd based on (1). Also, at most 12 by size constraints 1, 3, 7
3 8 24 subgroup structure of groups of order 24 odd based on (1). Also, at most 24 by size constraints 1, 3, 5, 7, 9, 13, 15
4 16 48 subgroup structure of groups of order 48 odd based on (1). Also, at most 48 by size constraints 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 27, 31
5 32 96 subgroup structure of groups of order 96 odd based on (1). Also, at most 96 by size constraints 1, 3, 7, 9, 11, 15, 17, 19, 23, 25, 27, 31, 33, 35, 39, 43, 47, 49, 51, 55, 63

The following additionally turn out to be true:

• The maximum number of subgroups of order 2 is $2^{n+1} - 1$, and this value is attained for the direct product of symmetric group:S3 and the elementary abelian group of order $2^{n-1}$.
• For $1 \le m \le n + 1$, we can find groups of order $3 \cdot 2^n$ that contain $2^m - 1$ subgroups of order 2. There are many straightforward constructions of this.
• For every odd integer $u$, there exists $n_0$ such that for $n \ge n_0$, there exists a group of order $3 \cdot 2^n$ containing $u$ subgroups of order 2.