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.
