Characteristic of normal implies normal: Difference between revisions

Revision as of 14:32, 4 April 2008

This article describes a computation relating the result of the Composition operator (?) on two known subgroup properties , to another known subgroup property
View a complete list of composition computations

For applications of this term/fact/idea, refer: Category:Applications of characteristic of normal implies normal

Statement

Property-theoretic statement

Characteristic * Normal $\leq$ Normal

Here, $*$ denotes the composition operator.

Verbal statement

Every characteristic subgroup of a normal subgroup is normal.

Symbolic statement

Let $H \leq K \leq G$ such that $H$ is characteristic in $K$ and $K$ is normal in $G$ , then $H$ is normal in $G$ .

Related facts

Characteristicity is transitive: A characteristic subgroup of a characteristic subgroup is characteristic.
Left transiter of normal is characteristic: Characteristicity is the weakest, or most general property, for which the above statement is true. This is made precise in the statement that characteristicity is the left transiter for normality.
Automorph-permutable of normal implies conjugate-permutable: This statement has many corollaries; for instance, 2-subnormal implies conjugate-permutable

Applications

For a complete list of applications, refer:

Category:Applications of characteristic of normal implies normal

Proof

Hands-on proof

Given: groups $H \leq K \leq G$ such that $H$ is characteristic in $K$ and $K$ is normal in $G$ .

To Prove: For any $g \in G$ , the map $c_{g} : x \mapsto g x g^{- 1}$ takes $H$ to within itself.

Proof: First, notice that since $K ◃ G$ , $c_{g} (x) \in K$ for every $x \in K$ . Thus, $c_{g}$ restricts to a function from $K$ to $K$ . Since this function arises by restricting an automorphism of $G$ , it is an endomorphism of $K$ .

Further, since $c_{g^{- 1}} \circ c_{g}$ is the identity map, and $K$ is invariant under both, the restriction of $c_{g}$ to $H$ is actually an invertible endomorphism, viz an automorphism. Call this automorphism $σ$ .

Since $H$ is characteristic in $K$ , $σ$ takes $H$ to within itself. But since $σ$ is the restriction of $c_{g}$ to $K$ in the first place, we conclude that $c_{g}$ in fact takes $H$ to itself.

Using the function restriction formalism

In terms of the function restriction formalism:

The following is a restriction formal expression for the subgroup property of normality:

Inner automorphism $\to$ Automorphism

The following is a restriction formal expression for the subgroup property of characteristicity:

Automorphism $\to$ Automorphism

We now use the composition rule for function restriction to observe that the composition of characteristic and normal implies the property:

Inner automorphism $\to$ Automorphism

Which is again the subgroup property of normality.

References

Textbook references

Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347^{More info}, Page 135, Page 137 (Problem 8(a))
A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613^{More info}, Page 28 (Characteristic and Fully invariant subgroups, 1.5.6(iii))
Topics in Algebra by I. N. Herstein^{More info}, Page 70 (Problem 9)

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@@ Line 35: / Line 35: @@
 ===Hands-on proof===
-Given groups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>K</math> and <math>K</math> is normal in <math>G</math>. We need to show that for any <math>g \in G</math>, the map <math>c_g : x \mapsto gxg^{-1}</math> takes <math>H</math> to within itself.
+''Given:'' groups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>K</math> and <math>K</math> is normal in <math>G</math>.
-First, notice that since <math>K \triangleleft G</math>, <math>c_g(x) \in K</math> for every <math>x \in K</math>. Thus, <math>c_g</math> restricts to a function from <math>K</math> to <math>K</math>. Since this function arises by restricting an automorphism of <math>G</math>, it is an endomorphism of <math>K</math>.
+''To Prove'': For any <math>g \in G</math>, the map <math>c_g : x \mapsto gxg^{-1}</math> takes <math>H</math> to within itself.
+''Proof:'' First, notice that since <math>K \triangleleft G</math>, <math>c_g(x) \in K</math> for every <math>x \in K</math>. Thus, <math>c_g</math> restricts to a function from <math>K</math> to <math>K</math>. Since this function arises by restricting an automorphism of <math>G</math>, it is an endomorphism of <math>K</math>.
 Further, since <math>c_{g^{-1}} \circ c_g</math> is the identity map, and <math>K</math> is invariant under both, the restriction of <math>c_g</math> to <math>H</math> is actually an invertible endomorphism, viz an automorphism. Call this automorphism <math>\sigma</math>.