# Groups of order 3.2^n

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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