A finite group $B$ is said to be a B-*group* if every primitive permutation group having a regular (transitive) subgroup isomorphic to B is $2$-transitive. 


Schur proved that a cyclic group of composite order is a B-group. Wielandt showed that no group of the form $B\_1 \times \cdots \times B\_d$ with $|B\_1| = \ldots = |B\_d| \ge 3$ and $d \ge 2$ is a B-group. (See Theorems 25.3 and 25.7 in Wielandt, *Finite permutation groups*.) 


Exercise 3.5.6 in Dixon and Mortimer *Permutation groups* asks for a proof that no elementary abelian $p$-group is a B-group. However this is false: the permutation groups of degree $4$ containing $C\_2 \times C\_2$ as a regular subgroup are $\langle (12)(34), (13)(24)\rangle$ and $\mathrm{Dih}(4)$ (both imprimitive) and $A\_4$, $S\_4$ (both $2$-transitive). My question is:



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> For which $d \in \mathbb{N}$ is the elementary abelian group $C\_2^d$ a B-group?
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Necessary conditions for a primitive permutation groups to have a regular subgroup are given in Corollary 3 of Liebeck, Praeger, Saxl, [*Transitive Subgroups of Primitive Permutation Groups*](http://ac.els-cdn.com/S0021869300985478/1-s2.0-S0021869300985478-main.pdf?_tid=48e4262a-cebc-11e6-9984-00000aab0f26&acdnat=1483122055_00f0d3ffcdb38b4bc79ca9e73ef6f3f3), J. Alg **234**, 291–361 (2000). The main theorem of Li, [The finite primitive permutation groups containing an abelian regular subgroup](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0024611503014266), Proc. London Math. Soc. **87**, 725–747 (2003), gives a sharper result. It seems non-trivial to use either of these papers to answer the question, but I'd welcome a correction.


If $H \le \mathrm{GL}(\mathbb{F}\_2^d)$ then the subgroup $\mathbb{F}\_2^d \rtimes H$ of the affine general linear group $\mathrm{AGL}(\mathbb{F}\_2^d)$ is primitive if and only if $H$ is irreducible. So if there exists an irreducible $H$ acting intransitively on $\mathbb{F}\_2^d \backslash \{0\}$, then $C\_2^d$ is not a B-group. Such groups exist whenever $d$ is composite, but not always when $d$ is prime. For example, the three irreducible subgroups of $\mathrm{GL}(\mathbb{F}\_2^3)$ all contain a Singer element of order $7$, so give $2$-transitive subgroups of $\mathrm{AGL}(\mathbb{F}\_2^3)$. (The only other primitive permutation groups of degree $8$ containing a subgroup isomorphic to $C\_2 \times C\_2 \times C\_2$ are $A\_8$ and $S\_8$, so it follows that $C\_2 \times C\_2 \times C\_2$ is a B-group.) This motivates my second question:



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> For which primes $p$ is there an irreducible subgroup $H$ of $\mathrm{GL}(\mathbb{F}\_2^p)$ such that $H$ acts intransitively on $\mathbb{F}\_2^p \backslash \{0\}$?
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Regular elementary abelian subgroups of primitive permutation groups
This is a bit long for a comment. To answer the question it would be very useful to know what groups can act primitively on a set of order $2^a$. Looking at O'Nan--Scott--Aschbacher, we see that such groups must be either affine or almost simple, or product action. Let's consider these three families one at a time.


**Almost simple**. In this case, the following paper is useful:


*Robert M. Guralnick*, MR 700286 [**Subgroups of prime power index in a simple group**](http://dx.doi.org/10.1016/0021-8693(83)90190-4), *J. Algebra* **81** (1983), no. 2, 304--311.


The main theorem of the paper classifies the objects in the title. A very useful corollary for us is this:



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> **Corollary 2**. Let $G$ be a nonabelian simple group acting transitively
>  on $\Omega$ with $|\Omega|=p^a$ for some prime $p$. Then $G$ acts $2$-transitively on R unless $G \cong {\rm PSU}\_4(2)$
>  and $|\Omega| = 27$ in which case $G$ acts as a rank $3$ primitive permutation group and $G$ has orbits of size $1$, $10$, and $16$. 
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This result means that, for the purposes of the question we can forget about simple groups containing regular elementary abelian subgroups -- they will never stop these subgroups being $B$-groups.


**Affine**. Now, as you point out, this reduces to studying irreducible subgroups $S$ of ${\rm GL}\_a(2)$ that are intransitive on non-zero vectors. As Derek says we can assume that $a$ is prime and that $2^a-1$ is prime (otherwise $E\_{2^a}$ cannot be a $B$-group).


Now, as Derek says, looking at Aschbacher's classification of subgroups of ${\rm GL}\_a(q)$ (or maybe Nori's results would suffice), we can assume that $S$ is almost simple. So we're interested in almost simple groups $S$ that have a representation over $\mathbb{F}\_2$ of degree a prime $p$ with $2^p-1$ prime, and are intransitive on non-zero vectors.


We can deal with that last condition -- intransitivity on non-zero vectors -- by referring to the main theorem in Guralnick's paper cited above. If $S$ is transitive on non-zero vectors, then this is a primitive action of prime degree and so is on Guralnick's list. I think with a little work you can probably rule out (nearly) all of the possibilities on this list, i.e. (with some possible exceptions) any almost simple group will be intransitive on vectors.


**Product action**. In this case the group $G\leq U\wr S\_k$ where $U$ is almost simple with an action of degree $2^b$ with $b$ a proper divisor of $a$. In particular $a$ is composite, and we already know that groups $E\_{2^a}$ with $a$ composite cannot be $B$-groups.


**CONCLUSION**. 
The only outstanding case is the affine one which means that we've reduced Question 1 to Question 2. And, given the progress on the affine case described above, based on Derek's remarks we would conclude the following (modulo the exceptions in the final paragraph of the affine case):



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> $E\_{2^a}$ is a $B$-group if and only if $a$ is prime, $2^a-1$ is prime, and there exists no almost simple group (except ${\rm GL}\_a(2)$) with an irreducible representation of degree $a$ over $\mathbb{F}\_2$.
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The problem of ascertaining when almost simple groups have such representations is currently beyond me.


