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Contents 1 Statement 1.1 Verbal statement 1.2 Statement with symbols 2 Related facts 2.1 Conditions for the existence of two incomparable subgroups 2.2 Tightness 2.3 Related facts in group theory 2.4 Analogues in other algebraic structures 3 Proof 3.1 Proof idea 3.2 Proof details 4 References 4.1 Textbook references

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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Statement

Verbal statement

The union of two subgroups, neither of which is contained in the other, is not a subgroup.

A more general statement is true: any subgroup contained in a union of two subgroups, must be contained in one of them.

Statement with symbols

Suppose G is a group and H and K are subgroups of G. Then, the following are true:

• Any subgroup contained in is contained either in H or in K
• If is a subgroup, then either H is contained in K or K is contained in H. Note that if H is contained in K, , and if K is contained in H, .

Related facts

Conditions for the existence of two incomparable subgroups

This result is of interest only when we have examples of a group G and subgroups H and K of G such that neither is contained in the other. Here are some examples:

• In the group of integers under addition, the subgroups generated by any two integers, neither of which is a divisor of the other, are not comparable. This is because subgroup containment relation in the group of integers equals divisibility relation on generators. In particular, the subgroups of multiples of 2 and multiples of 3 have the property that neither is contained in the other.
• In the direct product of two nontrivial groups, neither direct factor is contained in the other.

On the other hand, in a cyclic group of prime power order, any two subgroups are comparable. In fact, the only finite groups where any two subgroups are comparable are the trivial group and the cyclic groups of prime power order; the only infinite groups with the property are the quasicyclic groups.

A group where any two normal subgroups are comparable in termed a normal-comparable group: there are more examples of normal-comparable groups.

Tightness

It is possible for a union of three subgroups to be a subgroup; for instance, the Klein four-group (which is a direct product of the cyclic group of order two with itself) is a union of three of its subgroups. This is more generally related to the notion of partition of a group: a way of expressing a group as a union of proper subgroups any two of which intersect in the trivial subgroup.

In general, the elementary Abelian group of order p² (defined as the direct product of two cyclic groups of order p) is a union of (p + 1) subgroups of order p, any two of which intersect trivially.

We do have a fairly tight restriction on how a group may be a union of three subgroups and somewhat weaker restrictions on other unions of finitely many subgroups:

• Union of three subgroups is the whole group implies they have index two and form a flower arrangement
• Union of n subgroups is the whole group iff the group admits one of finitely many groups as quotient
• There is no group that is a union of seven proper subgroups but not a union of fewer proper subgroups
• B. H. Neumann's lemma

Related facts in group theory

• Directed union of subgroups is subgroup
• Union of all conjugates is proper
• Every group is a union of cyclic subgroups
• Every group is a union of maximal among Abelian subgroups
• Cyclic iff not a union of proper subgroups

Analogues in other algebraic structures

• A closely related result in commutative algebra is the prime avoidance lemma. The prime avoidance lemma states that a prime ideal is contained in a union of finitely many ideals, at most two of which are not prime, if and only if it is contained in one of them. In fact, the prime avoidance lemma for two ideals is precisely this statement, and the case of two ideals is the starting point for the argument.
• Union of two subquasigroups is not a subquasigroup unless they are comparable: The analogous result holds for subquasigroups of a quasigroup. The key reason why the result holds is that it continues to be true that if two of the elements h,k,hk are in the subquasigroup, so is the third.
• Union of two incomparable submonoids may be a submonoid: For monoids, we can havetwo submonoids, neither of which is contained in the other, but their union is a submonoid. Note that it is still true that in most cases, the union is not a submonoid; however, the possibility cannot be ruled out. For instance, the union of the non-negative integers and the non-positive integers is the group of integers. Both the non-negative integers and the non-positive integers form submonoids, and neither is contained in the other.

Proof

Proof idea

The proof is based on the idea that if two of the elements h,k,hk belong to a subgroup, so does the third.

Proof details

Given: A group G, subgroups H,K,L such that

To prove: L is contained in H or L is contained in K. In particular, if H is not contained in K and K is not contained in H, then is not a subgroup.

Proof: Suppose L is contained neither in H nor in K. Then, we can find and (clearly and ).

Then consider the elements h,k,hk. We know that so or :

• If , then clearly since , we get , a contradiction.
• If , then clearly since , we get , a contradiction.

So we've got a contradiction, indicating that our original assumption was false. So or .

For the second part of the statement, observe that if is a subgroup, we can take , to get that or , yielding or .

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 48, Exercise 8, (states only the particular case of proving that the union is a subgroup iff one is contained in the other)^{More info}