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Math Help - Left and Right cosets.

  1. #1
    Super Member Showcase_22's Avatar
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    Left and Right cosets.

    Let H be a subgroup of G and suppose that the number n of right cosets of H is G is finite. Prove that there are exactly n left cosets of H in G.

    (Hint: Consider the set of inverses of the elements in a right coset).
    Let |G:H|=|G|/|H|=n for n \in \mathbb{N}.

    Let H=\{h^0, \ldots , h^{n-1} \} for some h \in H.

    Then Hg=\{h^0g, \ldots , h^{n-1}g \} for g \in G.

    (Hg)^{-1}=\{(h^0g)^{-1}, \ldots , (h^{n-1}g)^{-1} \}

    This is equal to (Hg)^{-1}=\{g^{-1}(h^0)^{-1}, \ldots ,g^{-1}(h^{n-1})^{-1} \}________________(*)

    Since inverses are unique, and H is a group (closure axiom) we have thatb [MATh] (h^k)^{-1}=h^i[/tex] for 0 \leq k \leq n-1 for some 0 \leq i \leq n-1.

    Hence we have (Hg)^{-1}=\{g^{-1}h^0, \ldots , g^{-1}h^{n-1} \} =g^{-1}H

    Since |G:H|=n, we have exactly n \ g^{-1}'s to multiply H by.

    Thus, we have n left cosets of H in G.

    I think this is right (can someone check =S), but step (*) applies only to multiplicative groups. Is it a fact that addition is always commutative?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Let |G:H|=|G|/|H|=n for n \in \mathbb{N}.

    Let H=\{h^0, \ldots , h^{n-1} \} for some h \in H.


    Uh? You can't assume this unless you know H is cyclic (which isn't a given) AND h is one of its generators...


    Then Hg=\{h^0g, \ldots , h^{n-1}g \} for g \in G.

    (Hg)^{-1}=\{(h^0g)^{-1}, \ldots , (h^{n-1}g)^{-1} \}

    This is equal to (Hg)^{-1}=\{g^{-1}(h^0)^{-1}, \ldots ,g^{-1}(h^{n-1})^{-1} \}________________(*)

    Since inverses are unique, and H is a group (closure axiom) we have thatb  (h^k)^{-1}=h^i for 0 \leq k \leq n-1 for some 0 \leq i \leq n-1.

    Hence we have (Hg)^{-1}=\{g^{-1}h^0, \ldots , g^{-1}h^{n-1} \} =g^{-1}H

    Since |G:H|=n, we have exactly n \ g^{-1}'s to multiply H by.

    Thus, we have n left cosets of H in G.

    I think this is right (can someone check =S), but step (*) applies only to multiplicative groups. Is it a fact that addition is always commutative?
    .

    They gave you a big hint and that's what you should try: assume X_L\,,\,\,X_R represent the sets of left and right cosets of G\,\,\,in\,\,\,H , and define

    f: X_L\rightarrow X_R \,\,\,by\,\,\,f(gH):=Hg^{-1} . Well, now prove this function is a bijection.

    Tonio
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  3. #3
    Super Member Showcase_22's Avatar
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    I got caught up on the question before that was about cyclic groups!

    As for the function being a bijection:

    We have n lots of gH. We also have n lots of Hg^{-1} (proof as follows):

    Since we have n lots of g, we have at most n lots of Hg^{-1}. We will only have less if, for some k \in G, Hg^{-1}=Hk^{-1} \Rightarrow \ g^{-1}=k^{-1} \Rightarrow \ g=k (since inverses are unique).
    Therefore we have n lots of Hg^{-1}.

    Also, since inverses are unique, g^{-1} is unique to a g. Therefore the function is injective.

    We have a mapping from a set of n elements to a set of n elements which is injective. By the pigeonhole principle, the function is a bijection.
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    Quote Originally Posted by Showcase_22 View Post
    I got caught up on the question before that was about cyclic groups!

    As for the function being a bijection:

    We have n lots of gH. We also have n lots of Hg^{-1} (proof as follows):


    I really don't have any idea what can you possibly mean by "having lots of..."...


    Since we have n lots of g, we have at most n lots of Hg^{-1}. We will only have less if, for some k \in G, Hg^{-1}=Hk^{-1} \Rightarrow \ g^{-1}=k^{-1} \Rightarrow \ g=k (since inverses are unique).


    This is incorrect: it can perfectly well be that Ha^{-1}=Hb^{-1} without being true that a=b. You've to use the definition of the equivalence relation Ha=Hb\,\,,\iff...what?!


    Therefore we have n lots of Hg^{-1}.

    Also, since inverses are unique, g^{-1} is unique to a g. Therefore the function is injective.

    We have a mapping from a set of n elements to a set of n elements which is injective. By the pigeonhole principle, the function is a bijection.
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    Super Member Showcase_22's Avatar
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    Do you mean Ha=Hb \ \Leftrightarrow \ a=b?

    I don't think I know anything else connecting the two!
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    Quote Originally Posted by Showcase_22 View Post
    Do you mean Ha=Hb \ \Leftrightarrow \ a=b?

    No! And if you don't remember this then no wonder you're having so much problem with this question. Perhaps it is a good idea to refresh this basic stuff now.

    It is Ha=Hb\Longleftrightarrow ab^{-1}\in H ...

    Tonio

    I don't think I know anything else connecting the two!
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    Super Member Showcase_22's Avatar
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    Thanks again Tonio. I have to admit that I have never seen that before. =S

    However, doesn't that create a contradiction?

    Suppose we have g_1, \ g_2 \in G with g_1 \neq g_2.

    Since G is a group and is closed, g_1^{-1}g_2 \in G.

    This implies that Hg_1^{-1}=Hg_2^{-1}.

    This is also f(g_1H)=f(g_2H) so the function isn't injective.

    What's wrong with this argument?

    (Also, sorry i'm being the bee in your bonnet! This question is really difficult for me).
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    Quote Originally Posted by Showcase_22 View Post
    Thanks again Tonio. I have to admit that I have never seen that before. =S

    However, doesn't that create a contradiction?

    Suppose we have g_1, \ g_2 \in G with g_1 \neq g_2.

    Since G is a group and is closed, g_1^{-1}g_2 \in G.

    This implies that Hg_1^{-1}=Hg_2^{-1}.


    No! Please do pay attention: G is a group, H is a subgroup, then the equivalence relation is: for g_1,g_2\in G\,,\,\,g_1\sim g_2\Longleftrightarrow g_1g_2^{-1}\in H ,and then we get (this requires a proof, of course) that g_1g_2^{-1}\in H\Longleftrightarrow Hg_1=Hg_2
    This is VERY ,very basic stuff and you should know it and understand it before you attempt to solve your original problem.

    Tonio



    This is also f(g_1H)=f(g_2H) so the function isn't injective.

    What's wrong with this argument?

    (Also, sorry i'm being the bee in your bonnet! This question is really difficult for me).
    .
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    Super Member Showcase_22's Avatar
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    I haven't been getting anywhere with this function idea. I can't seem to prove anything with it!

    I stumbled across this:

    Suppose we have n right cosets ( |G:H|=n, G is a group and H \subset G).

    Cosets are either equal or disjoint so they partition G into a finite number of cosets (since |G| is finite).

    By Lagrange's theorem, |H| | |G| ("order of H divides order of G", it looks confusing how i've written it!) so |G|=n|H|.

    This proves the theorem (apparently).

    I understand what it's done, but I fail to see how this actually works! Is it because we have n|H| at the end and not |H|n?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Have you done Lagrange's theorem? It is easy to show that if \left[G:H\right]_L and \left[G:H\right]_R denote the number of left and right cosets respectively then |H|\left[G:H\right]_R=|G|=|H|\left[G:H\right]_L\implies \left[G:H\right]_L=\left[G:H\right]_R
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  11. #11
    Super Member Showcase_22's Avatar
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    We did it today! I decided to try using it for this question (since i've pretty much had it with that function!).

    Was what i've written correct? I'm still trying to understand it properly. =S

    Let <br /> <br />
\left[G:H\right]_R<br />
be the number of right cosets of a group G. Then, since H is a subgroup, |H| | |G| or |G|=n|H| since we know that <br /> <br />
\left[G:H\right]_R=n<br />
.

    <br /> <br />
Let \left[G:H\right]_L<br />
be the number of left cosets of the same group. Then |G|=k|H|.

    Then n|H|=k|H| so n=k so we have the same number of left and right cosets!!

    Please say that's right!!
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    We did it today! I decided to try using it for this question (since i've pretty much had it with that function!).

    Was what i've written correct? I'm still trying to understand it properly. =S

    Let <br /> <br />
\left[G:H\right]_R<br />
be the number of right cosets of a group G. Then, since H is a subgroup, |H| | |G| or |G|=n|H| since we know that <br /> <br />
\left[G:H\right]_R=n<br />
.

    <br /> <br />
Let \left[G:H\right]_L<br />
be the number of left cosets of the same group. Then |G|=k|H|.

    Then n|H|=k|H| so n=k so we have the same number of left and right cosets!!

    Please say that's right!!
    Strange way of phrasing it, but yes.
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  13. #13
    Super Member Showcase_22's Avatar
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    Thanks Drexel! (for helping me with this, and the other question I posted a while ago).
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