# Quotient groups of permutations

1. May 4, 2012

### Jesssa

hey guys,

I just want grasp the whole concept of quotient groups,

I understand say, D8/K where K={1,a2}

I can see the quotient group pretty clearly without much trouble however I start to get stuck when working with larger groups, say S4

For instance S4/L where L is the set of (xx)(xx) elements of S4 including e,

Without going through and doing all the calculations I know that the order of the quotient group will be 6, however I can't see what the group will be,

L is a normal subgroup of S4 since each element in L contains its conjugate so left and right cosets equal.

I first thought the quotient group would be something like

{1,(xx)L,(xx)(xx)L,(xxx)L,(xxxx)L,?} (different length cycles x L) but there are only 5 different length cycles.

Is there a way to find these quotient groups without having to multiply L but every element of S4?

2. May 4, 2012

### DonAntonio

We can try the following approach: in the quotient group $\,\,S_4/L\,\,$ , with $L:=\{(1),\, (12)(34),\, (13)(24),\, (14)(23)\}\,\,$ , we have

that $\,\,(12)L\neq (123)L\,\,$ , since $\,\,(12)(123)^{-1}=(12)(132)=(13)\notin L$ .

Thus, by Lagrange's theorem the group $\,\,S_4/L\,\,$ has order 6 and by the above it isn't abelian, thus it is isomorphic with...

DonAntonio

3. May 4, 2012

### Jesssa

Sorry I don't follow,

Doesn't having (12)L ≠ (123)L just mean there two elements of S4/L and the rest shows (13)L≠L, but (13)L could equal (some other permuation in S4)L,

Is there no way to find all the elements of S4/L without doing gL for all gεS4 and finding which ones are equal, that would give 24 cosets and 18 of the cosets would have to be duplicates of the remaining 6 to give an order 6 group right?

I could give a guess, is it isomorphic to D6?

I was just kind of hoping there was a fast way to find quotient groups, without, in this case, having to multiply L by 24 permutations and find which of these cosets are equal

edit

oh i think i worked it out,

then you get {(12),(34),( 1 4 2 3 ),( 1 3 2 4 )}

(12)(13)(24) = (1423), so (13)(24)=(12)(1423) then (13)(24)(1423)=(12)

so (13)(24)(1324) = (12)

and its a normal subgroup so left and right cosets are equal so (12)L = (1324)L

similar argument you can get (12)L=(1423)L=(34)L

so you find 4 cosets in one, so you'll only have to do 6 calculations to find the quotient group

Last edited: May 4, 2012
4. May 5, 2012

### DonAntonio

Ok, perhaps I wasn't clear enough. What I thought is that you'd check the elements $\,\,(12)(123)L\,\,,\,\,(123)(12)L\,\,$ and reached the

(correct, even) conclusion that they're different, thus making the quotient a non-abelian group without checking all the different 24 products, or even close.

And yes, the quotient is isomorphic with $\,\,D_6\cong S_3\,\,$ , though (the most common notation is the second one), and the

more usual name for the dihedral group of order 2n is $\,\,D_n\,\,,\,\,not\,\,D_{2n}$

DonAntonio
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