# Thread: left and right cosets

1. ## left and right cosets

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?

2. Originally Posted by jmoney90
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?

3. Originally Posted by Drexel28
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

4. Originally Posted by jmoney90
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 $G$

5. Originally Posted by Drexel28
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?

6. Originally Posted by jmoney90
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.

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

7. Originally Posted by Drexel28
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.

$\Phi:\mathcal{L}\left(H\right)\longmapsto\mathcal{ R}\left(H\right)$ 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

8. Originally Posted by jmoney90
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 $\Phi$?

9. Originally Posted by Drexel28
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.

10. Originally Posted by jmoney90
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 $\text{card }\mathcal{L}\left(H\right)=\text{card }\mathcal{R}\left(H\right)$.

11. Originally Posted by Drexel28
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