Are there any abelian 2 groups with Complete Holomorphs Other than $C^2_2$ and $C^4_2$? The title says it. I'm reviewing some group theory concepts I haven't touched in quite a while, since I don't teach abstract algebra in my current position, and could not find the answer to this question.  I've searched on google and found some papers that discuss other types of groups, but not 2-groups.  I know that the holomorph of all non-trivial finite abelian groups of odd order are complete groups, due to a theorem by Miller(1908), where complete means trivial center and all automorphisms are inner.  Also, since the holomorph of a finite abelian group is the direct product of the holomorphs of its Sylow subgroups, again Miller(1903?), does that imply the following:
Let G be an even ordered abelian group.  If the holomoprh of the Sylow 2-subgroup of G is complete then the holomorph of G is complete.  
As an added note, the automorphism group of $C^n_2$ is isomorphic to $PSL(n,2)$ since $C^n_2$ can be thought of as an $n$ dimensional vector space over the finite field $Z_{2}$.  In the case where $n$ = 4, $PSL(4,2)$ is isomorphic to $A_{8}$.  Is this connected to the holomorph being complete?  Is the answer known for $C^n_2$, if not for all 2-groups?
Thanks
 A: The completeness of ${\rm AGL}(n,2)$ for $n \ne 3$ follows easily from the following two facts:


*

*${\rm GL}(n,2)$ is complete for all $n \ge 1$.


CORRECTION: Sorry - that was very careless of me! As Greg and Jack have pointed out, ${\rm GL}(n,2)$ is NOT complete for $n \ge 3$. The inverse-transpose automorphism which, in terms of groups of Lie type is the graph automorphism of groups of Lie-type $A_n$, is an outer automorphism of ${\rm GL}(n,2)$. However, as Jack pointed out, this does not induce an automorphism of ${\rm AGL}(n,2)$, because it interchanges subspaces of $V$ of dimension $r$ with subspaces of dimension $n-r$, and so does not act on $V$ itself.


*

*The cohomology group $H^1({\rm GL}(n,2), V)$ (with $V$ the natural module) is trivial for $n \ne 3$. This is equivalent to saying that for $n \ne 3$ all complements of the elementary abelian normal subgroup of order $2^n$ in ${\rm AGL}(n,2)$ are conjugate in ${\rm AGL}(n,2)$.


I believe that the second of these was first proved in:
D.G. Higman, Flag-transitive collineation groups of finite projective spaces. Illinois J. Math. 6 (1962), 434-466.
I don't know when the first statement was first proved, but it follows from then general theory of automorphism groups of finite groups of Lie type.
Note that, for $n=3$, there are two conjugacy classes of complements of the $2^3$ normal subgroup, and they are interchanged by an outer automorphism of ${\rm AGL}(3,2)$.
A: I believe the holomorph of the elementary abelian group of order 2n is complete unless n=1 or 3.  It is called AGL(n,2), the affine general linear group of dimension n over the field of size 2.
In other words, n=4 isn't the exception, but rather the rule.  GL(3,2) is all sorts of weird, and I guess GL(1,2) is just too small.
The holomorph of 4×4×2×2 is also complete.  There are no other (non-elementary) abelian groups of order dividing 32 whose holomorph is complete.
