# Difference between revisions of "Characteristicity is transitive"

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|
This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about characteristic subgroup |Get facts that use property satisfaction of characteristic subgroup | Get facts that use property satisfaction of characteristic subgroup|Get more facts about transitive subgroup property

## Statement

Let $H$ be a characteristic subgroup of $K$, and $K$ a characteristic subgroup of $G$. Then, $H$ is a characteristic subgroup of $G$.

## Related facts

### Close relation with normality

A normal subgroup is a subgroup that is invariant under all inner automorphisms.

Below, we take $H \le K \le G$, with $H$ the bottom group, $K$ the middle group, and $G$ the top group.

Statement Change in assumption Change in conclusion
Normality is not transitive $H$ normal in $K$, $K$ normal in $G$ $H$ not normal in $G$
Characteristic of normal implies normal $K$ normal in $G$ $H$ normal in $G$
Left transiter of normal is characteristic $H$ in $K$ such that if $K$ is normal in $G$, $H$ is normal in $G$ $H$ is characteristic in $K$

### Generalizations

Balanced implies transitive: Any subgroup property that can be expressed as a balanced subgroup property is transitive. Characteristicity is a special case. Other special cases include:

Property Balanced with respect to ... Proof
Fully invariant subgroup endomorphisms Full invariance is transitive
Central factor inner automorphisms Central factor is transitive
Transitively normal subgroup normal automorphisms Transitive normality is transitive
Injective endomorphism-invariant subgroup injective endomorphisms Injective endomorphism-invariance is transitive

### Analogues in other algebraic structures

Characteristicity is transitive in Lie rings Lie ring $\leftrightarrow$ group, characteristic subring of a Lie ring $\leftrightarrow$ characteristic subgroup
Derivation-invariance is transitive Lie ring $\leftrightarrow$ group, derivation of a Lie ring $\leftrightarrow$ automorphism of a group, derivation-invariant Lie subring $\leftrightarrow$ characteristic subgroup
Characteristicity is transitive for any variety of algebras

### Generalizations in the one-of-its-kind sense of the statement

Property Meaning Proof
Second-order characteristic subgroup no subgroup equivalent in the second-order theory of groups Second-order characteristicity is transitive
Monadic second-order characteristic subgroup no subgroup equivalent is the monadic second-order theory of groups Monadic second-order characteristicity is transitive
Purely definable subgroup definable in the pure theory of groups Pure definability is transitive

## Definitions used

### Characteristic subgroup

Further information: Characteristic subgroup

A subgroup $H$ of a group $G$ is termed a characteristic subgroup if whenever $\sigma$ is an automorphism of $G$, $\sigma$ restricts to an automorphism of $H$.

This is written using the function restriction expression:

Automorphism $\to$ Automorphism

In other words, every automorphism of the whole group restricts to an automorphism of the subgroup.

### Transitive subgroup property

Further information: Transitive subgroup property

A subgroup property $p$ is termed transitive if whenever $H \le K \le G$ are groups such that $H$ satisfies property $p$ in $K$ and $K$ satisfies property $p$ in $G$, $H$ also satisfies property $p$ in $G$.

## Proof

### Hands-on proof

Given: A group $G$ with a characteristic subgroup $K$. $H$ is a characteristic subgroup of $G$. $\sigma$ is an automorphism of $G$.

To prove: $\sigma(H) = H$ and $\sigma$ restricts to an automorphism of $H$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $\sigma(K) = K$, and $\sigma$ restricts to an automorphism of $K$, that we call <math\sigma'[/itex]. definition of characteristic subgroup $K$ is characteristic in $G$, $\sigma$ is an automorphism of $G$. direct
2 $\sigma'(H) = H$, and $\sigma'$ restricts to an automorphism of $H$ definition of characteristic subgroup $H$ is characteristic in $K$ Step (1) direct
3 $\sigma(H) = H$ and $\sigma$ restricts to an automorphism of $H$. Steps (1), (2) [SHOW MORE]

### Function restriction expression metaproperty satisfaction

This proof of a subgroup property satisfying a subgroup metaproperty relies on the nature of a function restriction expression for the subgroup property.

This proof method generalizes to the following results: balanced implies transitive

The idea behind this proof is to observe that characteristicity can be written as the balanced subgroup property:

Automorphism $\to$ Automorphism

In other words, every automorphism of the big group restricts to an automorphism of the subgroup. Any balanced subgroup property is transitive, and this gives the proof.

## References

### Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 137, Problem 8(b), More info
• Groups and representations by Jonathan Lazare Alperin and Rowen B. Bell, ISBN 0387945261, Page 17, Lemma 4, More info
• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, Page 28, Section 1.5 (Characteristic and Fully invariant subgroups), 1.5.6(ii), More info
• Nilpotent groups and their automorphisms by Evgenii I. Khukhro, ISBN 3110136724, Page 4, Section 1.1, (passing mention)More info