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

Other facts about combining subgroups in different ways

• Union of two subgroups is not a subgroup unless they are comparable: If we have two subgroups of a group, neither of which is contained in the other ,their union is not a subgroup.
• Directed union of subgroups is subgroup: In particular, the union of an ascending chain of subgroups of a group is again a subgroup.

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:

• Normality is strongly intersection-closed
• Characteristicity is strongly intersection-closed
• Invariance implies strongly intersection-closed

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 $gh \in H.$

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 62, Section 2.4 (Subgroups generated by subsets of a group), Proposition 8, (considers the case of a nonempty collection)^{More info}. Also, Page 48, Exercise 10(a) and 10(b) (10(a) asks for the special case where there are only two subgroups being intersected)
• Groups and representations by Jonathan Lazare Alperin and Rowen B. Bell, ISBN 0387945261, Page 3, Proposition 1, ^{More info}
• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, Page 8, Section 1.3 (Intersections and joins of subgroups), Proposition 1.3.2, ^{More info}
• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, Page 48, 3.3.4, ^{More info}
• Algebra by Serge Lang, ISBN 038795385X, Page 9, ^{More info}
• A First Course in Abstract Algebra (6th Edition) by John B. Fraleigh, ISBN 0201763907, Page 75, Exercise 54, (only the finite case)^{More info}
• Topics in Algebra by I. N. Herstein, Page 46, Problem 1, (only the finite case, hinting at the infinite case)^{More info}