Proving the Intersection of Subgroups is a Subgroup: A Simple Solution

In summary, to prove that the intersection of any collection of subgroups of a group is again a subgroup, one can simply show that the intersection satisfies the group axioms and that an element is in the intersection if and only if it is in every subgroup in the collection, without needing to consider separate cases for finite and infinite collections.
  • #1
bonfire09
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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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  • #2
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
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
bonfire09 said:
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.
 

Related to Proving the Intersection of Subgroups is a Subgroup: A Simple Solution

1. What is the definition of "intersection of subgroups"?

The intersection of subgroups refers to the subset of elements that are common to two or more subgroups of a larger group. In other words, it is the set of elements that belong to all of the subgroups being considered.

2. How is the intersection of subgroups calculated?

The intersection of subgroups can be calculated by finding the common elements between the subgroups. This can be done by listing out the elements of each subgroup and identifying the elements that are present in all of them.

3. What is the significance of the intersection of subgroups?

The intersection of subgroups is significant because it allows us to identify the shared properties or characteristics of different subgroups within a larger group. This can provide insight into the relationships and connections between these subgroups.

4. Can the intersection of subgroups be empty?

Yes, the intersection of subgroups can be empty if there are no common elements between the subgroups being considered. This means that the subgroups have no shared properties or characteristics.

5. How is the intersection of subgroups related to the concept of a normal subgroup?

A normal subgroup is a subgroup that is closed under conjugation, meaning that if an element of the larger group is conjugated by an element of the subgroup, the result will still be an element of the subgroup. The intersection of subgroups can help determine if a subgroup is normal by checking if the subgroup's intersection with all other subgroups is itself.

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