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A problem from Artin's algebra textbook

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    (a)Let H and K be subgroups of a group G. Prove that the intersection of xH and yK which are cosets of H and K is either empty or else is a coset of the subgroup H intersect K

    (b) Prove that if H and K have finite index in G then the intersection of H and K also has finite index.

    2. Relevant equations



    3. The attempt at a solution
    The intersection of xH and yK is a subgroup of both H and K, then how to continue?
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2

    jbunniii

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    This is not true in general. If xH and yK are not subgroups, then neither contains the identity, so their intersection also doesn't contain the identiy. So it can't be a subgroup.

    Moreover, in general [itex]xH \cap yK[/itex] isn't even a subSET of H or K. xH and H are disjoint unless [itex]x \in H[/itex]. Similarly for yK and K.
     
  4. Feb 28, 2012 #3
    Sorry, I made some mistakes when I wrote the post. In fact, I mean the intersection of H and K is a subgroup of both H and K...Could U give me some tips to prove it?
     
  5. Feb 28, 2012 #4

    jbunniii

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    If xH and yK have nonempty intersection, then there is an element g contained in both: [itex]g \in xH[/itex] and [itex]g \in yK[/itex].

    The cosets of [itex]H \cap K[/itex] form a partition of G, so g is contained in exactly one such coset, call it [itex]a(H \cap K)[/itex].

    If you can show that [itex]a(H \cap K)[/itex] is contained in both [itex]xH[/itex] and [itex]yK[/itex] then you're done.

    Hint: both [itex]xH[/itex] and [itex]yK[/itex] are partitioned by cosets of [itex]H \cap K[/itex].
     
  6. Feb 29, 2012 #5
    Yeah...I get it. Thanks very much. In addition, how to prove part (b), that is how can I show that both [itex]H[/itex] and [itex]K[/itex] are partitioned by finite cosets of [itex]H \cap K[/itex]... I appreciate your insightful answer!
     
  7. Feb 29, 2012 #6

    Deveno

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    the index of H in G is the number of cosets of H.

    if this number is finite, then if it just so happened that H∩K was of finite index in H, we get:

    [G:H][H:H∩K] cosets of H∩K in G in all, which would be finite.

    can you think of a way to show that [H:H∩K] ≤ [G:K]? perhaps you can think of an injection from left cosets of H∩K in H to left cosets of K in G?
     
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