Operation table for the quotient group $S_4/V$ So I need to write out an operation table for the quotient group $S_4/V$, where $$V = \{(e),(12)(34),(13)(24),(14)(23)\}$$
thus $|V|$ = 4 and $|S_4|$ = 24. 
My question is:
Do I need to really write out all the elements of $S_4\times V$, which is $4\times 24= 96$ elements total!?
Say it isn't so........
 A: We already know that one of two proper normal subgroups of $S_4$ is the Klein $4-$group $$V={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$$ with quotient $S_3=S_4/V$ of order $6$. Knowing the following presentation of $S_3$, we can make the Cayley table of it easily:
$$S_3=\langle a,b\mid a^2=b^3=(ab)^2=1\rangle$$
Since $(ab)^2=1$ so we get $abab=1$ and so $$aba=b^2,~~bab=a$$ Now can make th Operation table of $S_3$ and simplify some relations in it:

I left some boxex for you to fill. Also, considering a computational way, by using GAP, we have:
   gap> s4:=SymmetricGroup(4);;
        V:=Group((1,2)(3,4),(1,3)(2,4),(1,4)(2,3));;
        s:=FactorGroup(s4,k);;
        f:=Image(IsomorphismPermGroup(s));


 Group([ (1,4)(2,6)(3,5), (1,2,3)(4,5,6) ])


   gap> StructureDescription(f);


 "S3"


   l:=AsList(f);
   t:=MultiplicationTable(f);

[ (), (1,2,3)(4,5,6), (1,3,2)(4,6,5), (1,4)(2,6)(3,5), (1,5)(2,4)(3,6), 
 (1,6)(2,5)(3,4) ]

[ [ 1, 2, 3, 4, 5, 6 ], [ 2, 3, 1, 6, 4, 5 ], [ 3, 1, 2, 5, 6, 4 ], 
[ 4, 5, 6, 1, 2, 3 ], [ 5, 6, 4, 3, 1, 2 ], [ 6, 4, 5, 2, 3, 1 ] ]

