# Intersection of subgroups

1. Sep 24, 2013

### bonfire09

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

Last edited: Sep 24, 2013
2. Sep 24, 2013

### Office_Shredder

Staff Emeritus
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. Sep 24, 2013

### 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?

Last edited: Sep 24, 2013
4. Sep 24, 2013

### pasmith

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.