Let $G_n$ be the extraspecial group of order $2^{1+2n}$. Its outer automorphism group is known to be isomorphic to the general orthogonal group $GO(2n)$. I'd like to get an explicit map of this isomorphism in gap. Here's what I tried so far:

Print("|out| = ",Size(aut)/Size(inn),"\n");
Print("|ort| = ",Size(ort),"\n");
Print("generators of aut = ",Length(gen_aut),"\n");
Print("generators of ort = ",Length(gen_ort),"\n");

|out| = 348364800
|ort| = 348364800
generators of aut = 10
generators of ort = 3

The size of the two groups (out and ort) matches as expected, but I'm not sure how to proceed from here. How would you define the outer automorphism group in GAP and how would you align its generators to those of the orthogonal group and find the isomorphism.


1 Answer 1


The isomorphism comes from the induced action of the automorphism group on the centre factor. Lets build this, by writing down matrices that describe the action of the automorphisms on a basis of $G/Z$:

gap> z:=Centre(grp);
Group([ f9 ])
gap> pcgs:=ModuloPcgs(grp,z); # like a basis for G/Z
[ f1, f2, f3, f4, f5, f6, f7, f8 ]
gap> mats:=List(GeneratorsOfGroup(aut),x->List(pcgs,
> y->ExponentsOfPcElement(pcgs,Image(x,y))*One(GF(2))));;
gap> mg:=Group(mats);
<matrix group with 10 generators>
gap> Size(mg);

So now mats are the matrix generators. Of course it is possible that we have chosen a basis that does not fit with the correct form, but we're lucky:

gap> mg=ort;

Note that this luck is not promised by the manual. It is possible that if you chose another prime/dimension you get a discrepancy and need to conjugate to get the right form. You can build the homomorphism as:

gap> hom:=GroupHomomorphismByImages(aut,mg,GeneratorsOfGroup(aut),mats);;
gap> Kernel(hom)=inn;

Note that this takes a momentm, as GAP verifies that it is a homomorphism. Use GroupHomomorphismByImagesNC to skip the test.

  • $\begingroup$ thanks for another excellent answer. How would this be adapted to work in the reverse direction? that is if I pick a random element of ort, how would I find its image in aut? $\endgroup$
    – unknown
    Commented Jun 11, 2021 at 19:53
  • 1
    $\begingroup$ You would take the PreImagesRepresentative of a matrix under the homomorphism hom. $\endgroup$
    – ahulpke
    Commented Jun 11, 2021 at 21:05

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