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Show that $V = \{\mathrm{id}, (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $\operatorname{Sym}(4)$, and that $G/V$ is isomorphic to $\operatorname{Sym}(3)$. (The group $V$ is the Klein $4$-group.)

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2 Answers

up vote 3 down vote accepted

Definitions are your friend! Review: "Normal subgroup", "quotient group", "isomorphic"...

First, confirm that $V$ is indeed a subgroup of $S_4$, which is easy enough to do.

Then: Can you show that $gVg^{-1} = V$ for all $g \in G = S_4$? That's one way to show that $V$ is normal in $S_4$.

Once you've shown that $V$ is a normal subgroup of $S_4$, then determine $[S_4 : V] = |S_4|/V|$, and show that the quotient group $S_4/V$ must be be isomorphic to $S_3$.

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You also need to prove that it is a subgroup! –  user1729 Feb 12 '13 at 16:20
    
How would I prove that it is a subgroup? I know that sym(4) has 24 permutations, of which 4 belong to the Klein four-group so it is a subgroup but I don't think that's a very good proof and I wouldn't want to list all 24 permutations. –  user61854 Feb 13 '13 at 16:19
    
You can show that it is a subgroup of Sym(4) by showing it obviously contains the identity of Sym(4), every non-identity element is its own inverse (so closed under inversion), and that permuting any two elements in V gives an element in V, and hence is closed. There you go, V is therefore a subgroup of Sym(4). –  amWhy Feb 13 '13 at 16:23
    
For showing it's a normal subgroup, I am instructed to use Hg=gH for all g in G. This requires the use of cosets I believe but I don't know what each g will be. How do I find this right g and this left g? Are they permutations? Thanks for all your help, I'm sorry I must seem so awful at algebra! –  user61854 Feb 13 '13 at 22:44
    
Yes, they are permutations: Sym(4) are permutations of four numbers $\{1, 2, 3, 4\}$. For showing its a normal subgroup, pick an arbitrary $g$ in Sym(4): and use h in V: you can show that all g in Sym(4) belong to one of 6 cosets of V: and the left cosets of V = right cosets of V. –  amWhy Feb 13 '13 at 22:51
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Let's to examine the concrete approach which @amWhy pointed by GAP 4.7.2 as follows:

gap> S4:=SymmetricGroup(IsPermGroup,4);;
     V:= Group((),(1,2)*(3,4),(1,3)*(2,4),(1,4)*(2,3));;
     e:=Elements(S4);;
     f:=Elements(V);;
gap> for i in [1..Size(g)] do 
       for j in [1..4] do 
     Print(e[i]^(-1)*f[j]*e[i] in f, "\n");
       od;
     od;

 true
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Well done by our expert gap ;-)+1 –  Sami Ben Romdhane Dec 7 '13 at 19:27
    
@SamiBenRomdhane: Hello Sami. I am going to bed. :-) –  B. S. Dec 7 '13 at 19:28
    
Good night and tomorrow ان شاء الله:-) –  Sami Ben Romdhane Dec 7 '13 at 19:35
1  
$\ddot\smile +1$ –  amWhy Dec 7 '13 at 22:46
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