Operation table for the quotient group $S_4/V$

Question

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........

This is confusing: do you need $;S_4/V;$ or $;S_4\times V;$ ?? — DonAntonio, Nov 21 '13 at 04:45
Sorry, it just asked for the operation table for the group of $S_4$ / $V$ — Alice, Nov 21 '13 at 04:46
Well, then as said that group only has $;\frac{|S_4|}{|V|}=\frac{24}4=6;$ elements... — DonAntonio, Nov 21 '13 at 05:10

score 4 · Answer 1 · edited Apr 13 '17 at 12:21

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:

enter image description here

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 ] ]

How'd you make the table? (+1) – amWhy Dec 23 '13 at 12:33 — amWhy, Dec 23 '13 at 12:33

Operation table for the quotient group $S_4/V$

1 Answers1