# Tour:Intersection of subgroups is subgroup

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

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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 be an arbitrary collection of subgroups of a group indexed by . Then, is again a subgroup of .

Note that if the collection 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 be an arbitrary collection of subgroups of a group indexed by Let us denote Here, denotes the identity element of

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

- If then
- If then

*Proof*: Let's prove these one by one:

- Since for every
- Take . Then for every Since each is a subgroup, for each Thus,
- Take Then for every so for every Thus

This page is part of the Groupprops guided tour for beginners. Make notes of any doubts, confusions or comments you have about this page before proceeding.PREVIOUS: Introduction three (beginners)|UP: Introduction three (beginners)|NEXT: Union of two subgroups is not a subgroup