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 $gh \in H.$