Tour:Intersection of subgroups is subgroup

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This article adapts material from the main article: intersection of subgroups is subgroup

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)
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PREREQUISITES:Definition of group and subgroup (preferably, the universal algebraic definitions)
WHAT YOU NEED TO DO:
  • Try proving that in any group, an intersection of subgroups is again a subgroup. Here, by intersection we mean the intersection of the underlying subsets.
  • Check out the proof below after you've tried.


Statement

Verbal statement

The intersection of any arbitrary collection of subgroups of a group is again a subgroup.

Symbolic statement

Let H_i\! be an arbitrary collection of subgroups of a group G\! indexed by i \in I. Then, \textstyle\bigcap_{i \in I} H_i is again a subgroup of G\!.

Note that if the collection I\! is empty, the intersection is defined to be the whole group. In this case, the intersection is clearly a subgroup. It should be noted that the case of an empty intersection is covered in the language of the general proof.


Proof

Given: Let H_i\! be an arbitrary collection of subgroups of a group G\! indexed by i \in I. Let us denote H = \textstyle\bigcap_{i \in I} H_i. Here, e\! denotes the identity element of G.\!

To prove: We need to show that H is a subgroup. In other words, we need to show the following:

  1. e \in H
  2. If g \in H then g^{-1} \in H
  3. If g, h \in H then gh \in H

Proof: Let's prove these one by one:

  1. Since e \in H_i for every i,\! e \in H.
  2. Take g \in H. Then g \in H_i for every i \in I. Since each H_i\! is a subgroup, g^{-1} \in H_i for each i \in I. Thus, g^{-1} \in H.
  3. Take g, h \in H. Then g, h \in H_i for every i,\! so gh \in H_i for every i \in I. Thus gh \in H.
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