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Thread: Cardinality of right and left cosets.

  1. #1
    Super Member Showcase_22's Avatar
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    Cardinality of right and left cosets.

    Suppose that G is a group, H is a subgroup of G, and $\displaystyle g_1,g_2 \in G$. Show that $\displaystyle |g_1 H|=|Hg_2|$.

    (Do not assume that G is finite).
    I wanted to try and find a bijection between them but I wasn't sure quite how to do it.

    Alternatively, is there a way to prove it without using bijections?
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by Showcase_22 View Post
    I wanted to try and find a bijection between them but I wasn't sure quite how to do it.

    Alternatively, is there a way to prove it without using bijections?

    Define: $\displaystyle \phi: g_1 H \to Hg_2$ as $\displaystyle \phi(g_1 h) = hg_2$

    Show the map is a bijection.
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  3. #3
    Super Member Showcase_22's Avatar
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    Right, i'm not really sure how to show that it's a bijection but here's what i'll try:

    $\displaystyle \phi:g_1H \rightarrow Hg_2$ as $\displaystyle \phi(g_1h)=hg_2$.

    So I need to show that an inverse exists ie. $\displaystyle \phi^{-1}(hg_2)=g_1 h$

    Is this the best way of doing it or would proving that $\displaystyle g_1H\subset Hg_2$ and $\displaystyle Hg_2 \subset g_1H$ be better?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Showcase_22 View Post
    Right, i'm not really sure how to show that it's a bijection but here's what i'll try:

    $\displaystyle \phi:g_1H \rightarrow Hg_2$ as $\displaystyle \phi(g_1h)=hg_2$.

    So I need to show that an inverse exists ie. $\displaystyle \phi^{-1}(hg_2)=g_1 h$
    Not really... You have to show that the map is both one-one and onto.

    This is equivalent to doing the following steps:

    One-One: Show that $\displaystyle \phi(x) = \phi(y) \Rightarrow x = y$

    Onto: Show that $\displaystyle \forall y \in Hg_2, \exists x \in g_1H : \phi(x) = y $

    Quote Originally Posted by Showcase_22 View Post
    would proving that $\displaystyle g_1H\subset Hg_2$ and $\displaystyle Hg_2 \subset g_1H$ be better?
    This is wrong, be careful.

    The question is asking you to prove they have the same number of elements , its not asking you to prove the sets are equal.

    The argument you are using is claiming the sets $\displaystyle g_1H$ and $\displaystyle Hg_2$ are the same. But that's not necessarily true here.
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  5. #5
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    Define, $\displaystyle \phi_1 : g_1H \to H$ by $\displaystyle g_1h\mapsto h$.
    Define, $\displaystyle \phi_2 : H \to Hg_2$ by $\displaystyle h\mapsto hg_2$.
    Show that these are bijections (also that the first one is well-defined).
    Therefore, $\displaystyle |g_1H| = |H|$ and $\displaystyle |H|=|Hg_2|$, hence, $\displaystyle |g_1H| = |Hg_2|$.
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  6. #6
    Super Member Showcase_22's Avatar
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    Show $\displaystyle |g_1 H|=|Hg_2|$

    Injective: $\displaystyle \forall g_1h_1, g_1h_2 \in g_1H, \phi (g_1h_1)=\phi (g_1h_2) \Rightarrow g_1h_1=g_1h_2$

    Suppose $\displaystyle \phi(g_1h_1)=\phi(g_1h_2) \Rightarrow g_1h_1=g_1h_2$. (Need To Show: $\displaystyle h_1=h_2$).

    Since $\displaystyle g_1h_1,g_1h_2 \in g_1H$ using the property of groups that they have inverses means:

    $\displaystyle g_1h_1=g_1h_2$
    $\displaystyle g_1^{-1}g_1h_1=g_1^{-1}g_1h_2$
    $\displaystyle eh_1=eh_2$
    $\displaystyle h_1=h_2$

    Therefore the function is injective.

    Surjective: $\displaystyle \forall y \in Hg_2 \exists x \in g_1H: \phi(x)=y$

    I'm not really sure how to show it is surjective.

    I Need To Show: $\displaystyle \phi(g_1h)=hg_2$ holds for all elements in $\displaystyle Hg_2$.

    Can I define $\displaystyle \phi$ to be $\displaystyle \phi(g_1h_i)=h_i g_2$ where $\displaystyle h_i \in H$? Since all the $\displaystyle h_i$'s are different this would be a surjection (i'm wondering whether this is a reasonable assumption).

    Combining this result with the proof that it is injective would prove $\displaystyle \phi$ is bijective.
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    Lord of certain Rings
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    Thumbs up

    You are right

    Surjectivity is not an assumption.

    For the sake of clarity, let me explain the details:

    Every element of $\displaystyle Hg_2$ is of the form $\displaystyle hg_2$ for some $\displaystyle h \in H$. Now $\displaystyle \phi$ maps $\displaystyle g_1 h \in g_1H$ to this $\displaystyle hg_2$. Thus every element in the range is mapped onto by some element in the domain and hence the map is onto.
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