Dear All,
A student Fawad asked me the following question.
Let G:=Z_2XZ_3= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}be a cyclic group, and 'd'
is the divisor of the order of a group G i.e ( d/IGI =>d=1,2,3,6)how
to list all the automorphisms of a group Aut(G), and how to find
explicitly all those automorphisms fixing 'd' elements, where fix means
f(x)=x ; for some x belongs to G & for f belongs to Aut(G).f(e)=e : by
homomorphism property,
i.e For d=1,A={set of all those automorphisms of Aut(G) fixing d=1 element
only ( Identity element)}For d=2,B={set of all those automorphisms of Aut(G)
fixing d=2 element}
For d=3C={set of all those automorphisms of Aut(G) fixing d=3 element}For
d=6D={ I : because identity is the only auto fixing all the elements of a group
G }
I wrote a function Fawad () which I think is just a step away from what he
wants.
I
would invite your comments on it that whether it is correct or not and
what will be a more correct and efficient way in this case.
Here is Fawad.
Fawad:=function(G)
local A,B,d,f,i,Aut;
A:=[];
B:=[];
Aut:=AutomorphismGroup(G);
for d in DivisorsInt(Size(G)) do
for f in Aut do
if Number([1..Size(G)],i -> Image(f,AsList(G)[i])=AsList(G)[i])=d
then Add(A,f);
Add(B,d);
fi;
od;
od;
return [B,A];
end;
All the best,
Muhammad Shah
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