Let K={(1),(12)(34),(13)(24),(14)(23)} show that K is a normal subgroup of the alternating group of degree A4 and find all the members of A4/K and write down the group table for A4/K.
i know that K is the Klein four group and i have already proven it is a normal subgroup but i need a start on approaching the A4/K part of this question no direct answers are necessary,just ideas thank you! :)
You already showed that the group $K$ is a subgroup of $A_4$ and indeed it is Klein $4$-group which is the only proper normal subgroup of $A_4$. For doing your question I employ GAP and this link as follows:
gap> a4:=AlternatingGroup(4);;
k:=Group((),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3));;
s:=FactorGroup(a4,k);
gap> Cayley:=function(G)
local s,i,l,m,j,k,max;
l:=Elements(G);
max:=1;
for i in [1..Length(l)] do
for j in [1..Length(l)] do
m:=l[i]*l[j];
s:=String(m);
if max<Length(String(s)) then
max:=Length(String(s));
fi;
od;
od;
s:=String(" ",max);
Print(s," ");
for i in [1..Length(l)] do
s:=String(l[i],max);
Print(s," ");
od;
Print("\n");
s:=String(" ",max);
Print(s,"_");
for i in [1..Length(l)*max+2*(Length(l)-1)] do
Print("_");
od;
Print("\n");
for i in [1..Length(l)] do
s:=String(l[i],max);
Print(s,"|");
for j in [1..Length(l)] do
m:=l[i]*l[j];
s:=String(m,max);
Print(s," ");
od;
Print("\n");
od;
end;
gap> Cayley(s);
<identity> of ... f1 f1^2
________________________________________________________
<identity> of ...|<identity> of ... f1 f1^2
f1| f1 f1^2 <identity> of ...
f1^2| f1^2 <identity> of ... f1
However $A_4/K\cong\mathbb Z_3$ and you maybe construct its table by hand.
Edit
Remarks by A.K. (adding them here, since I can't fit the GAP session in the comment):
First, the line
k:=Group((),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3));;
contains the identity permutation () but only to show how straightforwardly the question may be asked in GAP. Of course, () may be removed from the list of generators.
Second, in the real life (that is in the interactive GAP session) it's also possible to see the pattern in the Cayley table without caring too much about the pretty-printing and writing special function for that. The simplest way may look like this:
gap> l:=AsList(s);
[ <identity> of ..., f1, f1^2 ]
gap> t:=MultiplicationTable(l);
[ [ 1, 2, 3 ], [ 2, 3, 1 ], [ 3, 1, 2 ] ]
gap> Display(t);
[ [ 1, 2, 3 ],
[ 2, 3, 1 ],
[ 3, 1, 2 ] ]
or, if you wish to see actual elements of the group instead of numbers, like this:
gap> t1:=List(t,x -> List(x,i->l[i]));
[ [ <identity> of ..., f1, f1^2 ], [ f1, f1^2, <identity> of ... ],
[ f1^2, <identity> of ..., f1 ] ]
gap> PrintArray(t1);
[ [ <identity> of ..., f1, f1^2 ],
[ f1, f1^2, <identity> of ... ],
[ f1^2, <identity> of ..., f1 ] ]
And, of course, instead of looking at the table one could use StructureDescription to see that the group is cyclic, or just conclude that from its order:
gap> StructureDescription(s);
"C3"
gap> Size(s);
3
etc.
A long hint : In principle, all the elements of $A_4/K$ are of the form $$ \sigma K $$ for some $\sigma \in A_4$.
However, you know that $\sigma K = \tau K$ iff $\sigma\tau^{-1} \in K$. So you need to check when this "collapsing" happens.
For instance, you know that all non-trivial elements of $A_4/K$ have order 2, so you know that $\sigma^2 \in K$ for all $\sigma \in A_4$. You can thus list down $\{\sigma^2 : \sigma \in A_4\}$, which will eliminate many elements, if not all.
