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Intersection of subgroups

  • Thread starter bonfire09
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  • #1
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Homework Statement



Prove that the intersection of any collection of subgroups of a group is again a subgroup

Homework Equations





The Attempt at a Solution


Fixed proof
Let [itex]H_1 and H_2 [/itex] be subgroups on G. We first see if [itex]H_1 \cap H_2[/itex] is again a subgroup. We see if [itex]a,b\in H_1 \cap H_2[/itex] then [itex] ab\in H_1 \cap H_2[/itex]. Thus [itex]H_1 \cap H_2[/itex] is closed. Automatically the identity element has to be in [itex]H_1 \cap H_2[/itex] since [itex]H_1 and H_2 [/itex] are subgroups. And if [itex]a\in H_1 \cap H_2[/itex] then it follows that [itex]a^{-1}\in H_1 \cap H_2[/itex]. Thus [itex]H_1 and H_2 [/itex] is a subgroup.

I know this argument may sound redundant and in my inductive step I noticed that I never really used my assumption but would this work as a proof?
 
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Answers and Replies

  • #2
Office_Shredder
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You can't do it by induction because they never say the intersection is a finite intersection.

However as you observe you didn't really need the inductive hypothesis, so you should be able to strip out the induction and be left with a complete proof without too much work.
 
  • #3
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Could I have two cases then. One for finite collection of subgroups and another one for a infinite number of subgroups? Or since it says any collection so I can pick an arbitrary number of subgroups on G like I did in my fixed proof where started with the simplest case and that should suffice?
 
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  • #4
pasmith
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Could I have two cases then. One for finite collection of subgroups and another one for a infinite number of subgroups? Or since it says any collection so I can pick an arbitrary number of subgroups on G like I did in my fixed proof where started with the simplest case and that should suffice?
There are two points which will solve this problem for you in a single case for both countable and uncountable collections of subgroups:

(1) Every subgroup in the collection satisfies the group axioms.
(2) An element is in the intersection if and only if it is in every subgroup in the collection.
 

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