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

[bonfire09]

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

 

[Office_Shredder]

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.

 

[bonfire09]

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?

 

[pasmith]

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.