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Math Help - left and right cosets

  1. #1
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    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?
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    MHF Contributor Drexel28's Avatar
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    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?
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    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
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    MHF Contributor Drexel28's Avatar
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    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
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    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?
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    MHF Contributor Drexel28's Avatar
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    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}?
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    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
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    MHF Contributor Drexel28's Avatar
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    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?
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    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.
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    MHF Contributor Drexel28's Avatar
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    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).
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    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
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