Groups of order 3.2^n

This article discusses the groups of order $3 \cdot 2^n$, where $n$ varies over nonnegative integers. Note that any such group has a 3-Sylow subgroup which is cyclic group:Z3, and a 2-Sylow subgroup, which is of order $2^n$. Further, because order has only two prime factors implies solvable, any such group is a solvable group.

Number of groups of small orders

Exponent $n$	Value $2^n$	Value $3 \cdot 2^n$	Number of groups of order $3 \cdot 2^n$	Reason/explanation/list
0	1	3	1	only cyclic group:Z3, see equivalence of definitions of group of prime order
1	2	6	2	cyclic group:Z6 and symmetric group:S3; see classification of groups of order a product of two distinct primes
2	4	12	5	See groups of order 12
3	8	24	15	See groups of order 24
4	16	48	52	See groups of order 48
5	32	96	231	See groups of order 96
6	64	192	1543	See groups of order 192
7	128	384	20169	See groups of order 384
8	256	768	1090235	See groups of order 768
9	512	1536	408641062	See groups of order 1536

Arithmetic functions

Derived length

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.

$n$	$3 \cdot 2^n$	total number of groups	length 1 (abelian groups)	length 2	length 3	length 4	length 5
0	3	1	1
1	6	2	1	1
2	12	5	2	3
3	24	15	3	10	2
4	48	52	5	41	4	2
5	96	231	7	198	18	8
6	192	1543	11	1405	77	50
7	384	20169	15	19338	459	357

Composition length

As noted above, all groups of order $3 \cdot 2^n$ are solvable groups. Thus, the composition factors are all simple abelian groups, and hence cyclic groups of prime order. In particular, there are $n$ composition factors isomorphic to cyclic group:Z2 and one composition factor isomorphic to cyclic group:Z3. The composition length is thus $n + 1$.

Frattini length

$n$	$3 \cdot 2^n$	total number of groups	length 1	length 2	length 3	length 4	length 5
0	3	1	1
1	6	2	2
2	12	5	3	2
3	24	15	4	9	2
4	48	52	5	29	16	2
5	96	231	6	122	85	16	2
6	192	1543	7	653	757	108	16	2
7	384	20169	8	9190	9796	1049	108	16	2

Fitting length

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.

Since all groups of this order are solvable, the Fitting lengths are well defined and meaningful. Also note that the Fitting length is bounded from above by the derived length.

Note that Fitting length 1 means that the group is a nilpotent group (specifically, a finite nilpotent group) which in this case means it is a direct product of its 2-Sylow subgroup and 3-Sylow subgroup, the latter being cyclic group:Z3. The number of such groups equals the number of groups of order $2^n$. See number of nilpotent groups equals product of number of groups of order each maximal prime power divisor, which in turn follows from equivalence of definitions of finite nilpotent group.

$n$	$3 \cdot 2^n$	total number of groups	length 1 (nilpotent groups)	length 2	length 3	length 4	length 5
0	3	1	1
1	6	2	1	1
2	12	5	2	3
3	24	15	5	9	1
4	48	52	14	34	4
5	96	231	51	163	17
6	192	1543	267	1190	86
7	384	20169	2328	17305	536