Intersection of subgroups is subgroup

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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 Hi be an arbitrary collection of subgroups of a group G indexed by i∈I. Then, ⋂i∈IHi is again a subgroup of G.

Proof

Given: Let Hi be an arbitrary collection of subgroups of a group G indexed by i∈I. Let us denote H=⋂i∈IHi.

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

  1. e∈H
  2. If g∈H then g−1∈H
  3. If g,h∈H then gh∈H

Proof: Let's prove these one by one:

  1. Since e∈Hi for every i, e∈H
  2. Take g∈H. Then g∈Hi for every i∈I. Since each Hi is a subgroup, g−1∈Hi for each i∈I. Thus, g−1∈H.
  3. Take g,h∈H. Then g,h∈Hi for every i, so gh∈Hi for every i∈I. Thus gh∈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