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A problem from Artin's algebra textbookby AbelAkil
Tags: abstract algebra 
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#1
Feb2812, 07:22 AM

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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? 


#2
Feb2812, 01:00 PM

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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. 


#3
Feb2812, 05:26 PM

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#4
Feb2812, 07:53 PM

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A problem from Artin's algebra textbook
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]. 


#5
Feb2912, 01:56 AM

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#6
Feb2912, 03:56 AM

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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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