In GAP, for special group $M$ of order 512 how we find one answer of equality Inn(X)\cong M,?

Question

Let $M$ be the under group,

F:=FreeGroup("a","b");;
R:=ParseRelators(F,"a^8=b^8=[a,b]^8=1, [[a,b],b]=[[a,b],a]=1");;
M:=F/R;;

Now we want to find a group $X$ such that $Inn(X)\cong M$. I wrote the under program:

n:=9;;
i:=0;;
T:=SmallGroup(2^n, i+1);;
C:=Center(T);;
G:=T/C;;
while 
    IsomorphismGroups(G, M)=fail do
    i:=(i+1) mod NumberSmallGroups(2^n);
    n:=n+Maximum(1-i, 0);
    T:=SmallGroup(2^n, i+1);
    C:=Center(T);
    G:=T/C;
od;

But my computer can not load it. Do there exists a easy or better program for this question?

thank you

Answer 1

I would look at a Schur Cover as being a good candidate (there is no guarantee that it will satisfy your condition, but you can hope to be lucky):

gap> hom:=EpimorphismSchurCover(M);
[ f1, f2 ] -> [ a, b ]
gap> C:=Source(hom);
<fp group on the generators [ f1, f2 ]>

Note that the source group (in the current version) always is an Fp group, this somehow limits the applicability if the group is larger.

gap> Size(C);
16384
gap> Index(C,Centre(C));
512
gap> IsomorphismGroups(C/Centre(C),M);
[ (1,483,79,227,361,212,70,489)(2   [...]

So we were lucky and found a group $X$ as you wanted.