Finding Distinct Elements of G/H in Symmetric Group S4

[Shackleford]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_134032.jpg?t=1312484230

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_134038.jpg?t=1312484242

I found that the order of G/H is 6.

According to the Lagrange's Thereom,

order of G = order of H * index of H in G = order of H * order of G/H

The index of H in G is the number of distinct left cosets. The quotient group G/H is the cosets of H in G. Since H is normal, the left and right cosets are equal. Thus, the distinct left cosets equal to cosets of H in G.

Find the distinct elements of G/H? G = S₄ has 4! = 24 elements. Do I need to operate each of the 24 elements of G on the 6 elements of H? Or, since it is a symmetric group, is there a faster way to find the distinct elements?

 

[micromass]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_134032.jpg?t=1312484230

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_134038.jpg?t=1312484242

I found that the order of G/H is 6.

According to the Lagrange's Thereom,

order of G = order of H * index of H in G = order of H * order of G/H

The index of H in G is the number of distinct left cosets. The quotient group G/H is the cosets of H in G. Since H is normal, the left and right cosets are equal. Thus, the distinct left cosets equal to cosets of H in G.

Find the distinct elements of G/H? G = S₄ has 4! = 24 elements. Do I need to operate each of the 24 elements of G on the 6 elements of H? Or, since it is a symmetric group, is there a faster way to find the distinct elements?

Yes, you will need to operate each of the 24 elements on the 4 elements on H, and you'll need to see what pops out.

However, you don't need to operate every single on of the 24 elements. If you're smart, you'll only need to operate 6 elements!

For example, if I operate (1) on H, then I of course get {(1),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}.
But if I operate (1 2)(3 4) on H, then I already know that I get the same coset. So I don't need to operate (1 2)(3 4) on H, since I already know the result. Using such a logic, I claim that you can find all 6 cosets by operating only 6 elements on H.

 

[Shackleford]

Yes, you will need to operate each of the 24 elements on the 4 elements on H, and you'll need to see what pops out.

However, you don't need to operate every single on of the 24 elements. If you're smart, you'll only need to operate 6 elements!

For example, if I operate (1) on H, then I of course get {(1),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}.
But if I operate (1 2)(3 4) on H, then I already know that I get the same coset. So I don't need to operate (1 2)(3 4) on H, since I already know the result. Using such a logic, I claim that you can find all 6 cosets by operating only 6 elements on H.

How do you know you get the same coset? Is it because (1 2)(3 4) is in (1)H?

Also, I think I made some progress on the other problem. I'll scan my work in a little bit.

 

[micromass]

Indeed, in general: If aH is a coset and if $b\in aH$, then aH=bH.
In this case, (1)H is a coset and (1 2)(3 4) is in H, so (1)H=(1 2)(3 4)H.

 

[Shackleford]

Indeed, in general: If aH is a coset and if $b\in aH$, then aH=bH.
In this case, (1)H is a coset and (1 2)(3 4) is in H, so (1)H=(1 2)(3 4)H.

Crap. I vaguely remember that. I had a hard time paying attention in class for these sections.

 

[stringy]

Crap. I vaguely remember that. I had a hard time paying attention in class for these sections.

Perhaps it would help to remember that cosets are simply equivalence classes...? Then if $b\in aH$, it's pretty clear why $aH=bH$ should be true. For if two equivalence classes have an element in common, then they are equal.

 

[Shackleford]

Perhaps it would help to remember that cosets are simply equivalence classes...? Then if $b\in aH$, it's pretty clear why $aH=bH$ should be true. For if two equivalence classes have an element in common, then they are equal.

A coset is an equivalence class?

No. That's not clear. I don't see it in the book. Let me check the notes.

 

[stringy]

Suppose G is a group and let H be a subgroup. Define an equivalence relation $\sim$ between two group element as

$$a\sim b$$

if $b=ah$ for some $h \in H$.

Proving transitivity, symmetry, and reflexivity is pretty straitforward. And the equivalence classes of this relation are the cosets of H! Then you can throw everything you know about equivalence classes at cosets: they partition the group, for any two they are either equal or disjoint, etc. etc.

 

[Shackleford]

I still don't understand why. I hope this is in the book.

I figured out the six subgroups, but I'm not getting a closed table.

(1,2)(1,4) = (1,4,2)

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_204746.jpg?t=1312509725

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_205810.jpg?t=1312509740

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_204806.jpg?t=1312509750

 

[micromass]

I still don't understand why. I hope this is in the book.

I figured out the six subgroups, but I'm not getting a closed table.

(1,2)(1,4) = (1,4,2)

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_204746.jpg?t=1312509725

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_205810.jpg?t=1312509740

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_204806.jpg?t=1312509750

Yes, and (1 4 2)H=(1 2 3)H.

So

$$(1~2)H.(1~4)H=(1~2)(1~4)H=(1~4~2)H=(1~2~3)H$$

 

[Shackleford]

Yes, and (1 4 2)H=(1 2 3)H.

So

$$(1~2)H.(1~4)H=(1~2)(1~4)H=(1~4~2)H=(1~2~3)H$$

-_- I could have checked the subgroups.

I'm very burnt out. I'm ready to be finished with this class and thus the summer semester(s). This homework is due in the morning, then one Monday, and the final is Wednesday.

I have another question for you in a minute.

 

[Shackleford]

Okay. I finished this problem, painfully so.

Here is the next one. I'm not sure I'm doing enough to show this. Just by looking at the subgroup H with I and -I, you could reason that H is a normal subgroup.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_185509.jpg?t=1312511637

 

[micromass]

Okay. I finished this problem, painfully so.

Here is the next one. I'm not sure I'm doing enough to show this. Just by looking at the subgroup H with I and -I, you could reason that H is a normal subgroup.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_185509.jpg?t=1312511637

What you wrote down is completely correct!

 

[Shackleford]

What you wrote down is completely correct!

It just seems like a trivial answer. Given some of the other abstract algebra problems, it doesn't feel like I'm doing enough! Haha.