left and right cosets

jmoney90 · #1 $Old$ November 4th, 2009, 05:10 PM

Here is the problem in the text, I have a specific quetsion about it:

WOrk out the left and right cosets of H in G when G=A4 (alternating group that permutates 4 numbers)

H={e,(12)(34),(13)(24),(14)(23)}

and

G=A4 H={e, (123), (132)}

Okay, I know how to find cosets, my question here is whether I need to go through all the work of finding the cosets of each element in G. My idea is that LaGranges theorem tells us that |G|/|H| gives us the number of unique cosets, so I only need to work out the 4 unique cases. Am I correct here?

Drexel28 · #2 $Old$ November 4th, 2009, 05:32 PM

Quote:

Originally Posted by jmoney90 $View Post$

Here is the problem in the text, I have a specific quetsion about it:

WOrk out the left and right cosets of H in G when G=A4 (alternating group that permutates 4 numbers)

H={e,(12)(34),(13)(24),(14)(23)}

and

G=A4 H={e, (123), (132)}

Okay, I know how to find cosets, my question here is whether I need to go through all the work of finding the cosets of each element in G. My idea is that LaGranges theorem tells us that |G|/|H| gives us the number of unique cosets, so I only need to work out the 4 unique cases. Am I correct here?

You are correct. We know that the number of elements in the alternating group $A_n$ is $\left|A_n\right|=\frac{n!}{2}$ and Lagrange's theorem tells us $|G|=|H|\left[G:H\right]$ . Therefore for $G=A_4$ and $|H|=4$ this shows that $\left[A_4:H\right]=\frac{24}{4\cdot2}=3$

For the second we have $|H|=3$ so $\left[A_4:H\right]=\frac{24}{3\cdot2}=4$

You have no trouble finding the cosets?

jmoney90 · #3 $Old$ November 4th, 2009, 05:42 PM

Quote:

Originally Posted by Drexel28 $View Post$

You are correct. We know that the number of elements in the alternating group $A_n$ is $\left|A_n\right|=\frac{n!}{2}$ and Lagrange's theorem tells us $|G|=|H|\left[G:H\right]$ . Therefore for $G=A_4$ and $|H|=4$ this shows that $\left[A_4:H\right]=\frac{24}{4\cdot2}=3$

For the second we have $|H|=3$ so $\left[A_4:H\right]=\frac{24}{3\cdot2}=4$

You have no trouble finding the cosets?

No, the cosets are the easy part lol, I just wanted to make sure I was correct in my assumption that I only need to work out 2 cases (3 possible variations, and 1 of them is my subgroup H). I was just validating that I can be lazy rather than work out all 12 possible cosets

Drexel28 · #4 $Old$ November 4th, 2009, 05:57 PM

Quote:

Originally Posted by jmoney90 $View Post$

No, the cosets are the easy part lol, I just wanted to make sure I was correct in my assumption that I only need to work out 2 cases (3 possible variations, and 1 of them is my subgroup H). I was just validating that I can be lazy rather than work out all 12 possible cosets

Haha, what is the point of theorems and corrolarys if you can't be a little lazy?

But think about it. We know that the relation which describes the concept of a coset (specifically $a\sim b\Longleftrightarrow a^{-1}b\in H$ for left and $a\sim b \Longleftrightarrow ab^{-1}\in H$ for right) is an equivalence relation. So it partitions $G$ . So you'll know your done finding cosets when they've exhausted the elements of $G$

jmoney90 · #5 $Old$ November 4th, 2009, 06:13 PM

Quote:

Originally Posted by Drexel28 $View Post$

Haha, what is the point of theorems and corrolarys if you can't be a little lazy?

But think about it. We know that the relation which describes the concept of a coset (specifically $a\sim b\Longleftrightarrow a^{-1}b\in H$ for left and $a\sim b \Longleftrightarrow ab^{-1}\in H$ for right) is an equivalence relation. So it partitions $G$ . So you'll know your done finding cosets when they've exhausted the elements of $G$

Okay, that makes sense. But one other question I have is concerning right cosets. I remember in class the professor said lagrange only tells you the number of unique left cosets, which makes sense because that is how you do the proof: take the left cosets, then find an element in G - (the union of cosets you've taken) and use that in a coset until you've exhausted all options. But what does LaGrange tell us about right cosets, if anything?

Drexel28 · #6 $Old$ November 4th, 2009, 06:18 PM

Quote:

Originally Posted by jmoney90 $View Post$

Okay, that makes sense. But one other question I have is concerning right cosets. I remember in class the professor said lagrange only tells you the number of unique left cosets, which makes sense because that is how you do the proof: take the left cosets, then find an element in G - (the union of cosets you've taken) and use that in a coset until you've exhausted all options. But what does LaGrange tell us about right cosets, if anything?

It tells us lots of things, if you know another fact. Call $\mathcal{L}\left(H\right)$ and $\mathcal{R}\left(H\right)$ the left and right cosets of $H$ respectively.

What can you say about

$\Phi:\mathcal{L}\left(H\right)\longmapsto\mathcal{R}\left(H\right)$ given by $aH\longmapsto Ha^{-1}$ ?

jmoney90 · #7 $Old$ November 4th, 2009, 09:31 PM

Quote:

Originally Posted by Drexel28 $View Post$

It tells us lots of things, if you know another fact. Call $\mathcal{L}\left(H\right)$ and $\mathcal{R}\left(H\right)$ the left and right cosets of $H$ respectively.

What can you say about

$\Phi:\mathcal{L}\left(H\right)\longmapsto\mathcal{R}\left(H\right)$ given by $aH\longmapsto Ha^{-1}$ ?

Ah, I see what you're saying. Okay, this all makes sense now! My professor didn't explain that relation very well, or at least hasn't gotten there yet. THen again I have been sick so I haven't been attending classes lately

Drexel28 · #8 $Old$ November 4th, 2009, 09:33 PM

Quote:

Originally Posted by jmoney90 $View Post$

Ah, I see what you're saying. Okay, this all makes sense now! My professor didn't explain that relation very well, or at least hasn't gotten there yet. THen again I have been sick so I haven't been attending classes lately

So, exactly what conclusion did you draw about the mapping $\Phi$ ?

jmoney90 · #9 $Old$ November 4th, 2009, 10:28 PM

Quote:

Originally Posted by Drexel28 $View Post$

So, exactly what conclusion did you draw about the mapping $\Phi$ ?

I haven't gotten around to proving it, but to me it seems like $\Phi$ an isopmorphism, which means that the two sets are similair.

Drexel28 · #10 $Old$ November 4th, 2009, 10:40 PM

Quote:

Originally Posted by jmoney90 $View Post$

I haven't gotten around to proving it, but to me it seems like $\Phi$ an isopmorphism, which means that the two sets are similair.

An isomorphism? I think not. What binary operation were you supposing of defining on $\mathcal{L}\left(H\right),\mathcal{R}\left(H\right)$ ? But a bijection, yes.

So we can conclude that $\text{card }\mathcal{L}\left(H\right)=\text{card }\mathcal{R}\left(H\right)$

To see this another way, you can easily show that $\Psi:H\mapsto Ha$ is a bijection. Or equivalently that Lagranges theorem is equally applicable to right cosets. From this we can gather that

$|H|\left[G:H\right]_{\text{Left}}=|G|=|H|\left[G:H\right]_{\text{Right}}$

which equivalently shows that $\text{card }\mathcal{L}\left(H\right)=\text{card }\mathcal{R}\left(H\right)$ .

jmoney90 · #11 $Old$ November 4th, 2009, 10:42 PM

Quote:

Originally Posted by Drexel28 $View Post$

An isomorphism? I think not. What binary operation were you supposing of defining on $\mathcal{L}\left(H\right),\mathcal{R}\left(H\right)$ ? But a bijection, yes.

So we can conclude that $\text{card }\mathcal{L}\left(H\right)=\text{card }\mathcal{R}\left(H\right)$

To see this another way, you can easily show that $\Psi:H\mapsto Ha$ is a bijection. Or equivalently that Lagranges theorem is equally applicable to right cosets. From this we can gather that

$|H|\left[G:H\right]_{\text{Left}}=|G|=|H|\left[G:H\right]_{\text{Right}}$

which equivalently shows that $\text{card }\mathcal{L}\left(H\right)=\text{card }\mathcal{R}\left(H\right)$ .

Oh, that is clever! I even noticed it was a bijection, just didn't make that connection lol. I guess I was really focused on the idea of an isomorphism