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Let $AGL(2d,2)$ be the affine general linear group acting natrually on a $2d$-dimensional vector space over $GF(2)$. Is there a regular subgroup of $AGL(2d,2)$ isomorphic to $Z_{2^d}:Z_{2^d}$ for $d=3$ and $4$ respectively?

I tried to compute all regular subgroups of a Sylow $2$-subgroup of $AGL(2d,2)$ using the magma command "RegularSubgroups" but failed since it's too memory-consuming.

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As a computational problem, I think this is very hard. I have some interest in this, because I (and a few other others) are trying to use Magma to compute the regular subgroups of ${\rm ASp}(8,2)$. We expect to succeed eventually, but it will involve a huge amount of computer time and memory. My guess is that this computation would be possible, but difficult, for ${\rm AGL}(6,2)$ and impossible for ${\rm AGL}(8,2)$. –  Derek Holt Nov 22 '12 at 10:50
    
@DerekHolt Yes, I hope to compute the problem for AGL(8,2) and I think the only possibility to succed on my laptop is utilizing the specified structure $Z_8{:}Z_8$ rather than constructing all the regular subgroups of the Sylow $2$-subgroup $P$ directly. I am able to construct all the semiregular cyclic subgroups of order $8$ in $P$, but failed in searching regular subgroups $Z_8{:}Z_8$ in their normalisers because of lack of memory. –  Binzhou Xia Nov 22 '12 at 15:43
    
By the way, why were you trying to calculate the regular subgroups of a Sylow 2-subgroup of $G$ rather than of $G$ itself? There will be far fewer conjugacy classes of subgroups in $G$, so it is quicker to do it in $G$. Try it with ${\rm AGL}(5,2)$ with $G$ and with a Sylow 2-subgroup. –  Derek Holt Nov 22 '12 at 17:47
    
@DerekHolt If nothing is wrong, the codes in the next comment for magma prove there does not exist regular subgroup $Z_8{:}Z_8$ in $AGL(6,2)$. My laptop runs about 15 minutes. But if I substitue P for G in the definition of S, then it will run out of memory. (The available memory of my laptop is no more than 2G.) –  Binzhou Xia Nov 23 '12 at 2:05
    
G:=AGL(6,2); G; P:=Sylow(G,2); S:=[a:a in CyclicSubgroups(P:OrderEqual:=8)|IsSemiregular(a`subgroup) eq true]; #S; num:=0; for x in S do N:=Normaliser(P,x`subgroup); for y in [a:a in CyclicSubgroups(N:OrderEqual:=8)|IsSemiregular(a`subgroup) eq true] do if #(x`subgroup meet y`subgroup) eq 1 then H:=sub<N|x`subgroup,y`subgroup>; if IsSemiregular(H) eq true then H; end if; end if; end for; num:=num+1; num; end for; –  Binzhou Xia Nov 23 '12 at 2:05

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