# Intersection of subgroups is subgroup

This article gives the statement, and possibly proof, of a basic fact in group theory.
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## 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.

## Related facts

For examples, see the article Intersection of subgroups

### The related notion of join of subgroups

Given a collection of subgroups, their join is defined as the smallest subgroup containing all of them; equivalently, it is the intersection of all subgroups containing them.

This is closely related to the notion of the subgroup generated by a subset. The subgroup generated by a subset is the intersection of all subgroups containing that subset.

Notice that although the union of subgroups is not a subgroup, the fact that an intersection of subgroups is a subgroup tells us that there is a smallest subgroup containing any given collection of subgroups. This is analogous to the fact that the greatest lower bound property on a totally ordered set yields the least upper bound property.

### Other facts about intersections of subgroups

A subgroup property is termed:

• intersection-closed if the intersection of an arbitrary nonempty collection of subgroups with the property also has the property.
• finite-intersection-closed if the intersection of a finite nonempty collection of subgroups with the property also has the property.
• strongly intersection-closed if it is intersection-closed and also true for the whole group as a subgroup of itself. Thus, it is preserved on taking intersections of possibly empty collections.
• strongly finite-intersection-closed if it is finite-intersection-closed and also true for the whole group as a subgroup of itself.

There are some basic results of importance about intersection-closedness:

### Intersections of subsets other than subgroups

Subset property Proof that it is closed under arbitrary intersections
twisted subgroup intersection of twisted subgroups is twisted subgroup
1-closed subset intersection of 1-closed subsets is 1-closed subset
symmetric subset intersection of symmetric subsets is symmetric subset

### Analogues in other algebraic structures

For any variety of algebras, an intersection of subalgebras is a subalgebra. The proof is exactly the same as that for groups. In fact, the result holds in a slightly greater generality than varieties of algebras. For instance, an intersection of subfields is a subfield, although fields do not form a variety of algebras.

## 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 $gh \in H.$