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 be an arbitrary collection of subgroups of a group indexed by . Then, is again a subgroup of .
Proof
Given: Let be an arbitrary collection of subgroups of a group indexed by . Let us denote .
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 .
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