Quotient groups of permutations

[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?

 

[DonAntonio]

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?

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

 

[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,

start with (12)L

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

 

[DonAntonio]

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,

start with (12)L

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

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

 

[CellCoree]

Thank you for your question. Quotient groups of permutations can be a bit tricky to grasp, but once you understand the concept, it can be a powerful tool in understanding the structure of groups.

To start, let's review what a quotient group is. A quotient group is formed by taking a group G and a normal subgroup N, and then defining a new group using the cosets of N in G. The elements of the quotient group are the cosets, and the operation is defined as the coset multiplication. This means that the elements of the quotient group are not individual elements of G, but rather sets of elements of G.

Now, let's apply this to your example of S4/L. The first step is to determine the cosets of L in S4. Since L is a normal subgroup, the left and right cosets will be the same. We can use the formula for coset multiplication to find the cosets:

L = {(1), (12)(34), (13)(24), (14)(23)} (since L contains all elements of S4 that are of the form (xx)(xx))

The cosets of L in S4 are then:

(1)L = {(1), (12)(34), (13)(24), (14)(23)}

(12)L = {(12), (1234), (1342), (1423)}

(13)L = {(13), (1324), (1243), (1432)}

(14)L = {(14), (1423), (1234), (1342)}

Now, we can define the elements of the quotient group S4/L as the set of these cosets:

S4/L = {(1)L, (12)L, (13)L, (14)L}

The operation in this quotient group is coset multiplication, which means that (12)L * (13)L = (1234)L. This is because (12)(13) = (1234) and (L)(L) = L.

To find the order of the quotient group, we simply count the number of distinct cosets. In this case, there are 4, so the order of the quotient group S4/L is 4.

In general, it is not necessary to multiply every element of S4 in order to find the quotient group. As you mentioned, L is a normal subgroup and therefore, we can use the formula for coset multiplication to