## Are there any abelian 2 groups with Complete Holomorphs Other than $C^2_2$ and $C^4_2$?

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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

-

## 2 Answers

The completeness of ${\rm AGL}(n,2)$ for $n \ne 3$ follows easily from the following two facts:

1. ${\rm GL}(n,2)$ 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.

1. 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 ${\rm AGL}(3,2)$.

-
Thanks for the reference! This might help with my older question: math.stackexchange.com/questions/5226/… ? At the very least, this gives a very clear sufficient condition. – Jack Schmidt Dec 6 2010 at 14:36
@Derek Holt Sorry, I'm a little confused. GL(4,2) is isomorphic to $A_8$ which a non-abelian finite simple group, but is not complete. – Greg Gibson Dec 6 2010 at 17:54
@Greg: Good catch! The general theory shows that Out(GL(n,2)) = 2 for n ≥ 3. The extra automorphism takes a matrix to its inverse-transpose (conjugates by a transposition in the A8 form). I guess this automorphism is induced by conjugation by some element of V. Derek's argument fits really nicely into a decomposition of Aut(G|xV) for any semi-direct product, not just holomorphs, but it does need G complete to clearly work. – Jack Schmidt Dec 6 2010 at 19:48

3
Oh, I see: the inverse transpose automorphism of GL(n,2) does not survive to AGL(n,2), because it takes V to a non-isomorphic module, V*. It is an automorphism of GL, but it does not commute with the action on V (the C2^n). Let me know if you want me to write up the semi-direct product stuff. Derek probably has a clearer description. – Jack Schmidt Dec 6 2010 at 19:52
@Jack Schmidt Thanks, I'd appreciate some more insight. – Greg Gibson Dec 7 2010 at 2:22
show 1 more comment

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9385370016098022, "perplexity_flag": "head"}