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Math Help - Intersection of subgroups

  1. #1
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    Intersection of subgroups

    Show that the intersection of 2 subgroups of finite index also has finite index.

    Proof.

    Let  \{S_{a} \} _{a \in I} and  \{ T_{b} \} _{b \in J}

    The intersection of them, well, since the index of both are finite, isn't the intersection  \bigcap _{c \in I \cup J } \{U _{c} \} well, then, the index has to be finite.

    But I'm not sure if I write this up correctly.

    Thanks.
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  2. #2
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    Let H,K be subgroups.
    Define (G:H) to be the cosets of H.
    Define (G:K) to be the cosets of K.
    Define [G:H] = |(G:H)| and [G:K] = |(G:K)|.

    Note that H\cap K is a subgroup.
    Define a mapping from (G:H\cap K) to (G:H)\times (G:K) by a(H\cap K)\mapsto (aH,aK).
    We need to show this mapping is well-defined.
    To show this we use a fact that if aN = bN then ab^{-1} \in N for N some subgroup.

    The next thing we notice is that this mapping is one-to-one.
    For if (aH,aK) = (bH,bK) \implies ab^{-1} \in H \text{ and }ab^{-1} \in K \implies a(H\cap K) = b(H\cap K).
    Since this mapping is one-to-one it means [G:H\cap K] \leq [G:H]\times [G:K] < \infty.
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  3. #3
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    Thanks, perfecthacker, as you can see my season is starting again, and this modern alg is hard...

    I understand the proof in general, but I do have two questions:

    How do we conclude from aN=bN implies ab^{-1} \in N? (sorry, I know should be easy, it is just that I been out of touch from Algebra for the summer... I really should know the answer for this.)

    And how does this fact show that this mapping is well-defined?

    Thank you so much!
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    How do we conclude from aN=bN implies ab^{-1} \in N?
    Since aN = bN and a\in aN it means bh=a for some h\in H. Thus, h = ab^{-1} \in H.

    And how does this fact show that this mapping is well-defined?
    Say a(H\cap K) = b(H\cap K).
    This means ab^{-1} \in H \text{ and }ab^{-1} \in K.
    Now a(H\cap K)\mapsto (aH,aK) \text{ and }b(H\cap K)\mapsto (bH,bK).
    By above aH = bH \text{ and }aK=bK.
    Thus, (aH,aK) = (bH,bK).
    The mapping is well-defined.
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