4
$\begingroup$

Question.
Is there any way to get and use a tuple representation of a DirectProduct?

Explaination.
Let's assume we have groups $G, H$ which are cyclic group and symmetric group respectively. Let's take $L$ as a direct product of $G$ and $H$. I am aware of Projection and Embedding functions as the example below presents.

gap> G := CyclicGroup(6);
<pc group of size 6 with 2 generators>
gap> H := SymmetricGroup(6);
Sym( [ 1 .. 6 ] )
gap> L := DirectProduct(G,H);
<group of size 4320 with 4 generators>
gap> Embedding(L,1);
MappingByFunction( <pc group of size 6 with 
2 generators>, <group of size 4320 with 4 generators>, function( elm ) ... end )
gap> Image(Embedding(L,1));
<group of size 6 with 2 generators>
gap> Image(Embedding(L,2));
<group of size 720 with 2 generators>
gap>

My goal is to take an exact element of $g \in G$ and $h \in H$ and get an element $(g,h) \in L$. I have noticed there is an object called DirectProductElement, but there is no documentation about it.

$\endgroup$
3
  • $\begingroup$ @DerekHolt, Don't you mean? Image(Embedding(L, 1), GeneratorsOfGroup(G)[1]);? $\endgroup$
    – limakzi
    May 19 at 14:12
  • $\begingroup$ So, either I do not understand or your code does not compile. $\endgroup$
    – limakzi
    May 19 at 14:51
  • $\begingroup$ OK, I'll write it as an answer, using your example. $\endgroup$
    – Derek Holt
    May 19 at 14:57

1 Answer 1

Reset to default
7
$\begingroup$ 
gap> G := CyclicGroup(6);;                                                     
gap> H := SymmetricGroup(6);;
gap> L := DirectProduct(G,H);;
gap> g:=Random(G);;
gap> h:=Random(H);;
gap> m1 := Embedding(L,1);;
gap> m2 := Embedding(L,2);;
gap> Image(m1,g) * Image(m2,h);
DirectProductElement( [ f1, (3,5,4,6) ] )
$\endgroup$
2
  • $\begingroup$ This is the answer that works and it gives what I wanted. Actually, getting back to your comment. In your comment m1 was not Embedding, but an Image. That's the reason, I doubt your code in your comment will work. Thank you Derek! $\endgroup$
    – limakzi
    May 19 at 15:01
  • $\begingroup$ OK, I,'ll delete the comment! $\endgroup$
    – Derek Holt
    May 19 at 15:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.