Abelian subgroup structure of groups of order 2^n

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This article gives specific information, namely, abelian subgroup structure, about a family of groups, namely: groups of order 2^n.
View abelian subgroup structure of group families | View other specific information about groups of order 2^n

Abelian subgroup structure by order of groups

n 2^n Number of groups of order 2^n Information on groups Information on abelian subgroup structure
0 1 1 trivial group --
1 2 1 cyclic group:Z2 subgroup structure of cyclic group:Z2
2 4 2 groups of order 4 subgroup structure of groups of order 4
3 8 5 groups of order 8 subgroup structure of groups of order 8
4 16 14 groups of order 16 abelian subgroup structure of groups of order 16
5 32 51 groups of order 32 abelian subgroup structure of groups of order 32
6 64 267 groups of order 64 abelian subgroup structure of groups of order 64
7 128 2328 groups of order 128 abelian subgroup structure of groups of order 128
8 256 56092 groups of order 256 abelian subgroup structure of groups of order 256
9 512 10494213 groups of order 512 abelian subgroup structure of groups of order 512

Summary on existence, congruence, and replacement

Summary on existence

Note that if there exists an abelian normal subgroup of a particular order, there exist abelian normal subgroups of all smaller orders, because any normal subgroup contains normal subgroups of all orders dividing its order.

n 2^n Largest k for which every group of order 2^n contains an abelian subgroup of order 2^k Corresponding value 2^k Largest k for which every group of order 2^n contains an abelian normal subgroup of order 2^k Corresponding value 2^k
0 1 0 1 0 1
1 2 1 2 1 2
2 4 2 4 2 4
3 8 2 4 2 4
4 16 3 8 3 8
5 32 3 8 3 8
6 64 4 16 4 16
7 128 4 16 4 16

Summary on congruence condition

Below are values of n and k, the n values in the rows and the k values in the columns. A "Yes" indicates that in any group of order 2^n, the number of abelian subgroups of order 2^k is either 0 or odd. A "No" indicates that there exists a group of order 2^n where the number of abelian subgroups of order 2^k is a nonzero even number.

n 2^n k = 0, 2^k = 1 k = 1, 2^k = 2 k = 2, 2^k = 4 k = 3, 2^k = 8 k = 4, 2^k = 16 k = 5, 2^k = 32 k = 6, 2^k = 64
0 1 Yes
1 2 Yes Yes
2 4 Yes Yes Yes
3 8 Yes Yes Yes Yes
4 16 Yes Yes Yes Yes Yes
5 32 Yes Yes Yes Yes Yes Yes
6 64 Yes Yes Yes Yes Yes Yes Yes
7 128 Yes Yes Yes Yes Yes Yes (?) Yes Yes
8 256 Yes Yes Yes Yes Yes Yes (?) No Yes Yes