Normal not implies characteristic: Difference between revisions

From Groupprops
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===Example of a direct product===
===Example of a direct product===


Let <math>G</math> be any nontrivial [[group]]. Then consider <math>K = G \times G</math>, viz., the [[external direct product]] of <math>G</math> with itself. The subgroups <math>G_1 := G \times \{ e \}</math> and <math>G_2 := \{ e \} \times G</math> are [[direct factor]]s of <math>K</math>, and are hence both normal in <math>K</math>. Note also that they are distinct, since <math>G</math> is nontrivial.
Let <math>H</math> be any nontrivial [[group]]. Then consider <math>G = H \times H</math>, viz., the [[external direct product]] of <math>H</math> with itself. The subgroups <math>H_1 := H \times \{ e \}</math> and <math>H_2 := \{ e \} \times H</math> are [[direct factor]]s of <math>G</math>, and are hence both normal in <math>G</math>. Note also that they are distinct, since <math>H</math> is nontrivial.


However, the exchange automorphism:
However, the exchange automorphism:
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<math>(x,y) \mapsto (y,x)</math>
<math>(x,y) \mapsto (y,x)</math>


exchanges the subgroups <math>G_1</math> and <math>G_2</math>. Thus, neither <math>G_1</math> nor <math>G_2</math> is invariant under all the automorphisms, so neither is characteristic. Thus, <math>G_1</math> and <math>G_2</math> are both normal subgroups of <math>K</math> that are not characteristic.
exchanges the subgroups <math>H_1</math> and <math>H_2</math>. Thus, neither <math>H_1</math> nor <math>H_2</math> is invariant under all the automorphisms, so neither is characteristic. Thus, <math>H_1</math> and <math>H_2</math> are both normal subgroups of <math>K</math> that are not characteristic.


Note that this example also shows that [[direct factor]] does not imply [[characteristic subgroup]].
Note that this example also shows that [[direct factor]] does not imply [[characteristic subgroup]].
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===Particular cases of this example===
===Particular cases of this example===


* When <math>G</math> is [[cyclic group:Z2|the cyclic group of order two]], <math>K</math> is the [[particular example::Klein four-group]]. In particular, this gives a counterexample where the ambient group is an [[abelian group]]. More generally, we can start with any nontrivial abelian group <math>G</math>.
* When <math>H</math> is [[cyclic group:Z2|the cyclic group of order two]], <math>G</math> is the [[particular example::Klein four-group]]. In particular, this gives a counterexample where the ambient group is an [[abelian group]]. More generally, we can start with any nontrivial abelian group <math>H</math>.


==GAP implementation==
==GAP implementation==

Revision as of 02:38, 10 January 2010

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about characteristic subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not characteristic subgroup|View examples of subgroups satisfying property normal subgroup and characteristic subgroup

Statement

Verbal statement

A Normal subgroup (?) of a group need not be a Characteristic subgroup (?).

Statement with symbols

It is possible to have a group G and a normal subgroup H of G that is a normal subgroup of G but not a characteristic subgroup of G.

Related facts

Converse

Further information: Characteristic implies normal

The converse statement is indeed true. That is, every characteristic subgroup is normal.

Stronger formulations using the same examples

Other related facts

Related group properties

There are some groups in which every normal subgroup is characteristic. Further information: Group in which every normal subgroup is characteristic

Proof

Example of a direct product

Let H be any nontrivial group. Then consider G=H×H, viz., the external direct product of H with itself. The subgroups H1:=H×{e} and H2:={e}×H are direct factors of G, and are hence both normal in G. Note also that they are distinct, since H is nontrivial.

However, the exchange automorphism:

(x,y)(y,x)

exchanges the subgroups H1 and H2. Thus, neither H1 nor H2 is invariant under all the automorphisms, so neither is characteristic. Thus, H1 and H2 are both normal subgroups of K that are not characteristic.

Note that this example also shows that direct factor does not imply characteristic subgroup.

Particular cases of this example

GAP implementation

Implementation of the generic example

Before using this generic example, you need to define G for GAP, choosing any nontrivial group (double semicolons have been used here to suppress GAP's output for the first three commands, which depends on the specific choice of G -- you can use single semicolons instead).

gap> K := DirectProduct(G,G);;
gap> G1 := Image(Embedding(K,1));;
gap> G2 := Image(Embedding(K,2));;
gap> IsSubgroup(K,G1);
true
gap> IsSubgroup(K,G2);
true
gap> IsNormal(K,G1);
true
gap> IsNormal(K,G2);
true
gap> IsCharacteristicSubgroup(K,G1);
false
gap> IsCharacteristicSubgroup(K,G2);
false

References

Textbook references

  • Topics in Algebra by I. N. HersteinMore info, Page 70 (Problem 7(a))
  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347More info, Page 137 (Problem 6)