Regular abelian groups of order $2^m$ and type $(2,2,\ldots,2)$ I was wondering if there is a systematic way to construct the regular Abelian groups of order $2^m$ and type $(2,2,\ldots,2)$.  Since the permutation group needs to be regular, it should act transitively on a set of size $2^m$.  For example, if $m=2$, the unique regular permutation group of order 4 and type $C_2 \times C_2$ is $\{1, (12)(34), (13)(24), (14)(23)\}$.  
How about a permutation group $G \le S_8$ of order 8, acting regularly on $\{1,\ldots,8\}$, isomorphic to $C_2^3$?  Since $G$ is semiregular, every one of the 7 nonidentity permutations has no fixed points.  Since each of these permutations must have order 2, each is a product of exactly 4 transpositions.  Thus, the ${8 \choose 2}=28$ transpositions need to be partitioned into 7 subsets, each containing 4 transpositions, such that we have an Abelian group structure.  Is there a systematic way to do this?  
Actually, the context of my question is as follows.  [Wielandt, Finite Permutation Groups, Section 13] mentions in his text the following assertion:  Let $G$ be primitive on $\Omega$ and $G_\Delta$ transitive on $\Omega-\Delta=\Gamma$, $1 < |\Gamma| \le \frac{1}{2}|\Omega|$; then, $|\Gamma| < \frac{1}{2}|\Omega|$ implies $G$ is alternating or symmetric.  
The text then says that the hypothesis $|\Gamma| < \frac{1}{2}|\Omega|$ may be omitted in the assertion unless $|\Omega|=2^m, m \ge 3$, and that in this exceptional case a counterexample is provided by the normalizer, in $S^\Omega$, of the regular Abelian group of order $2^m$ and type $(2,2,\ldots,2)$.  So, I was wondering what these regular Abelian groups look like, what their normalizers would be, and why they provide a counterexample.  
 A: I haven't checked the details but there are some suggestive patterns
$(12)(34)(56)(78)$
$(13)(24)(57)(68)$
$(14)(23)(58)(67)$
$(15)(26)(37)(48)$
$(16)(25)(38)(47)$
$(17)(28)(35)(46)$
$(18)(27)(36)(45)$
$1$ pairs with $2,3,4,5,6,7,8$ which is the organising scheme, together with including the four-group at the beginning
$2$ pairs with $1,4,3,6,5,8,7$
$3$ pairs with $4,1,2,7,8,5,6$
$4$ pairs with $3,2,1,8,7,6,5$
$5$ pairs with $6,7,8,1,2,3,4$
$6$ pairs with $5,8,7,2,1,4,3$
$7$ pairs with $8,5,6,3,4,1,2$
$8$ pairs with $7,6,5,4,3,2,1$
The patterns for $1,2,4,8$ could be continued in the next iteration for $16$, and the pattern for $5$ looks as though it could be carried over to $9$ in the next iteration. I am sure there is a simple way of describing this - others may have it to hand.

Further to the comment, if the first column is $i$ pairs with $i$ - the identity - and we start with the trivial group, we have the squares
$2^0$
$1$
We replace $n$ by the block
$2n-1, 2n$
$2n, 2n-1$
We get successively - order 2
$1,2$
$2,1$
Then - order 4
$1,2,3,4$
$2,1,4,3$
$3,4,1,2$
$4,3,2,1$
Then order $8$ as above. The squares are symmetric, and we can read the permutations vertically or horizontally.
There are lots of constructions like this for square objects $2^n \times 2^n$ - this must be related to something else.
