Normal implies permutable

From Groupprops
Jump to: navigation, search
DIRECT: The fact or result stated in this article has a trivial/direct/straightforward proof provided we use the correct definitions of the terms involved
View other results with direct proofs
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) must also satisfy the second subgroup property (i.e., permutable subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about normal subgroup|Get more facts about permutable subgroup

Statement

Verbal statement

Any Normal subgroup (?) of a group is a Permutable subgroup (?).

Symbolic statement

Let G be a group and H a normal subgroup of G. Then H is a permutable, or quasinormal, subgroup of G. In other words, HK = KH for any subgroup K of G (or equivalently, the Product of subgroups (?) HK is a subgroup for any subgroup K \le G).

Property-theoretic statement

The subgroup property of being normal is stronger than the subgroup property of being permutable.

Definitions used

Normal subgroup

Further information: Normal subgroup A subgroup H of a group G is a normal subgroup if for any g \in G, gH = Hg (viz, the left cosets are the same as the right cosets).

Permutable subgroup

Further information: Permutable subgroup, permuting subgroups A subgroup H of a group G is a permutable subgroup if for any subgroup K \le G, HK = KH (or equivalently, HK is a group). In other words, H and K are permuting subgroups for every K. Here HK denotes the product of subgroups:

HK = \{ hk : h \in H, k \in K \}, \qquad KH = \{ kh: k \in K, h \in H \}

Proof

Hands-on proof

Given: Let H be a normal subgroup of G.

To prove: H is permutable in G.

Proof: Let K be any subgroup of G. For every g \in K, Hg = gH. Now we have:

HK = \bigcup_{g \in K} Hg

and

KH = \bigcup_{g \in K} gH

Since Hg = gH \forall g \in K, we conclude that HK = KH.

Notice that the above proof does not anywhere use the fact that K is a subgroup.

Converse

Permutable subgroups need not be normal

Further information: Permutable not implies normal

A permutable subgroup need not be normal. There are many counterexamples for finite groups.

References

Textbook references

  • Topics in Algebra by I. N. Herstein, More info, Page 53, Problem 3 (the notion of permutable subgroup is not explicitly introduced, but what we're asked to show is implicitly the same)
  • Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Page 75, Exercise 5 of Section 5 (Restriction of a homomorphism to a subgroup) (the notion of permutable subgroup is not explicitly introduced, but what we're asked to show is implicitly the same)