Abelian subgroup structure of groups of order 2^n

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