Expert: Paul Klarreich - 11/16/2008

Question
I have 3 questions:
1) Represent all cosets of H = S3 in G = S4
I'm confused about how to use the lemma (Lagrange's theorem:
a "is in" aH
aH=H if and only if a "is in" H
aH=bH or aH "intersection" bH = "Empty set"
aH=bH if and only if (a^-1)b "is in H"
order(aH) = order(bH)
aH =Ha if and only if H =aHa^-1
aH is a subgroup of G if and only if a "is in" H

I know that the elements in H=S3 are: (1), (12), (13), (23)...  
and the elements of G=S4 are: (1), (12)(34),(13)(24),(14)(23), (123), (243),(142),(234),(132),(143),(234), (124)...   
but how can I represent all the cosets of S3 in S4?
I tried:
Let a "be in" S3 and b "be in" S4 where a= (1), (12), (13), (23)...  
and b=(1), (12)(34),(13)(24),(14)(23), (123),(243),(142),(234),(132), (143),(234), (124)...   
then the cosets are a*b, a2*b2, a3*b3... where there are no repetitions of elements among the cosets   
but that is not correct.
How can I represent the cosets? ***************************************************
2) what are all the ways to write G=Z24+Z18 (+ means direct product) fo Zn groups. Arrange them in descending order according to size   

I find most difficult to write G=Z24+Z18 as a direct product of Zn
are the elemetns going to look like (a,b,c) or just (a,b)? I don't get what Zn does to the Z24+Z18   
I know that Z24+Z18 can be split into Z with subsets that are relatively prime: Z24 + Z18 split into Z3+Z8 + Z9+Z2 Am I doing this right?
we don't know what to do next  or even if we are right *********************************************************

3) Prove that G is Isomorphic to G G=R+R(means direct product)and G(this is the Range)={ax+b,where a,b are in R} under polynomial addition.   
What do the elements looke like? (1,1) for the first Group (domain) and will the elements of G (range) look like 1*1+3 ?     

this is my incorrect proof:  
fi: (Greek letter) G->G is an Isomorphism   
Onto:  Let (x,y) be in G (this is the 2nd group, range)  
      y is in R+R("+" means direct product)
      Define a=x in G={ax+b} then fi(a)=( (c,d)+(ax+b) ) =(x,y)    

One to one:
          let (v,w),(x,y) be in G (range) such that fi(v,w)=fi(x,y) so((v,w),(ax+b)) = ((x,y)(cx+d)) making (v,w)=(x,y)   

Operation Preserved: let x,y be in g, then x=(a,b) y=(c,d) where a,b,c,d are Rationals  
then x*y =(a,b)*(c,d)=(a+c, b+d) since a+c, b+d are Rationals and fi(x)*fi(y)=(ax+b)(cx+d)=(adx+bd+bcx+acxx)=(bc+acx)x+(adx+bd)=fi(xy)  
therefore fi is Operation preserved.
As a result G is Isomorphic to G.   

I can't think of how the elements would look like. That is my problem I don't know how to write the elements

Answer
Questioner:   Kate
Category:  Calculus
Private:  Yes
 
Subject:  Lagrange's theorem
Question:  I have 3 questions:
1) Represent all cosets of H = S3 in G = S4
I'm confused about how to use the lemma (Lagrange's theorem:
a "is in" aH
aH=H if and only if a "is in" H
aH=bH or aH "intersection" bH = "Empty set"
aH=bH if and only if (a^-1)b "is in H"
order(aH) = order(bH)
aH =Ha if and only if H =aHa^-1
aH is a subgroup of G if and only if a "is in" H

I know that the elements in H=S3 are: (1), (12), (13), (23)...  
and the elements of G=S4 are: (1), (12)(34),(13)(24),(14)(23), (123), (243),(142),(234),(132),(143),(234), (124)...   
but how can I represent all the cosets of S3 in S4?
I tried:
Let a "be in" S3 and b "be in" S4 where a= (1), (12), (13), (23)...  
and b=(1), (12)(34),(13)(24),(14)(23), (123),(243),(142),(234),(132), (143),(234), (124)...   
then the cosets are a*b, a2*b2, a3*b3... where there are no repetitions of elements among the cosets   
but that is not correct.
How can I represent the cosets? ***************************************************
2) what are all the ways to write G=Z24+Z18 (+ means direct product) fo Zn groups. Arrange them in descending order according to size   

I find most difficult to write G=Z24+Z18 as a direct product of Zn
are the elemetns going to look like (a,b,c) or just (a,b)? I don't get what Zn does to the Z24+Z18   
I know that Z24+Z18 can be split into Z with subsets that are relatively prime: Z24 + Z18 split into Z3+Z8 + Z9+Z2 Am I doing this right?
we don't know what to do next  or even if we are right *********************************************************

3) Prove that G is Isomorphic to G G=R+R(means direct product)and G(this is the Range)={ax+b,where a,b are in R} under polynomial addition.   
What do the elements looke like? (1,1) for the first Group (domain) and will the elements of G (range) look like 1*1+3 ?     

this is my incorrect proof:  
fi: (Greek letter) G->G is an Isomorphism   
Onto:  Let (x,y) be in G (this is the 2nd group, range)  
     y is in R+R("+" means direct product)
     Define a=x in G={ax+b} then fi(a)=( (c,d)+(ax+b) ) =(x,y)    

One to one:
         let (v,w),(x,y) be in G (range) such that fi(v,w)=fi(x,y) so((v,w),(ax+b)) = ((x,y)(cx+d)) making (v,w)=(x,y)   

Operation Preserved: let x,y be in g, then x=(a,b) y=(c,d) where a,b,c,d are Rationals  
then x*y =(a,b)*(c,d)=(a+c, b+d) since a+c, b+d are Rationals and fi(x)*fi(y)=(ax+b)(cx+d)=(adx+bd+bcx+acxx)=(bc+acx)x+(adx+bd)=fi(xy)  
therefore fi is Operation preserved.
As a result G is Isomorphic to G.   

I can't think of how the elements would look like. That is my problem I don't know how to write the elements
..........................................
Hi, Kate,

This is a lot of stuff, and I can't do it all at once.  So here is a little bit (maybe part 1) -- I'll send updates if I can manage them.

The elements in H=S3 are: (1), (12), (13), (23), (123), (132)  < you didn't write them all out, so I did.

and the elements of G=S4 are: (1), (12)(34),(13)(24),(14)(23), (123), (243),(142),(234),(132),(143),(234), (124)...   
................
Of course there are  24 elements of S4 including those,

Since Ord(G) = 24, Ord(H) = 6, there should be four cosets (including H = S3, of course.)

Pick an element of S4, but not in H, and call it 'a'.  Obviously, pick one involving (4), such as:

a = (34) -- leave 1,2 alone, switch 3,4.

Now find the product of aH, meaning each element of H is composed with a.  Compute the products: (Wish me luck -- I don't do this a lot.  In fact, it's the first time.)

(34)*(1), = (34)
(34)*(12),= (12)(34)
(34)*(13) = abdc-abdc-dbac = (241)
(34)*(23),= abcd-abdc-adbc = (342)
(34)*(123)= abcd-abdc-dabc =  (2341)
(34)*(132)= abcd-abdc-bdac = (3142)

So our first coset is (34)H =
{ (34),(12)(34),(241),(342),(2341),(3142) }

Now pick another 'a' in G, but not already listed.  Use it to produce a coset.

Finally, the last six will be the fourth coset.  

-------------------------------------
Note: Don't send Private questions -- I change it to public anyway.  If it really has to be private, don't send it to me.
-------------------------------------

2) What are all the ways to write G = Z24+Z18 (+ means direct product) for Zn

groups. Arrange them in descending order according to size   

I really hate it when questioners assume I am familiar with their terminology,

as if I just finished teaching the course with their textbook.

I assume that Z24 means the integers modulo 24, so

Z24 = {0,1..23} and + means addition mod 24.
Z18 = {0,1..23} and + means addition mod 18.

Now the direct product is all the 24*18 = 432 ordered pairs of these, using the

operations above, so, for example,

(17,11) + (9,10) = (26,21) --> (2,3)

Now, exactly what are you asking?  What is your question?

Are you looking for subgroups?  What?

[I suspect that if you are forced to define all your terms and your terminology,

not only will I understand your question, but so will you.]
....................
I am guessing here that you could form a subgroup in this way:

The overall group (direct product) looks like this:

(0,0),(1,0),...(23,0)
(0,1),(1,1),...(23,1)
(0,2),(1,2),...(23,2)
...
(0,17),(1,17),...(23,17)
[24 by 18]

But suppose we take only every multiple of 3 from the Z24's:

(0,0),(3,0),...(21,0)
(0,1),(3,1),...(21,1)
(0,2),(3,2),...(21,2)
...
(0,17),(3,17),...(21,17)
This is 8 across, 18 down, [8 by 18] = 144

And we could take Z24[3] as above, and Z18[2], perhaps.

(0,0),(3,0),...(21,0)
(0,2),(3,2),...(21,2)
(0,4),(3,4),...(21,4)
...
(0,16),(3,16),...(21,16)
[8 by 9] = 72.

Now you could work on this, I think.  Maybe you will find that ACROSS, you can

have  1,2,3,4,6,8,12,24 elements, by DOWN, 1,2,3,6,9,18 elements.

Figure all those products and arrange in size order from 1 X 1 up to 24 X 18.

Make any sense?