Proving transitivity

This is a survey article related to:subgroup metaproperty satisfaction
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A subgroup property $p$ is termed a transitive subgroup property if whenever $H\leq K\leq G$ are groups such that $H$ satisfies property $p$ in $K$ and $K$ satisfies property $p$ in $G$, then $H$ satisfies property $p$ in $G$.

A subgroup property $p$ is termed a t.i. subgroup property if it is transitive as well as identity-true: every group satisfies the property as a subgroup of itself.

Also refer:

• Disproving transitivity: A survey article on methods to prove that a given subgroup property is not transitive.
• Using transitivity to prove subgroup property satisfaction: A survey article discussing how to use the fact that a subgroup property is transitive to establish that certain subgroups have the property.
• All transitive subgroup properties
• All subgroup properties that are not transitive

Quick discussion on transitivity and subordination

Given a subgroup property $p$, we define the subordination of $p$ as the following property: $H$ has the property in $G$ if there exists an ascending chain of subgroups:

$H=H_{0}\leq H_{1}\leq H_{2}\leq \dots \leq H_{n}=G$,

such that each $H_{i-1}$ satisfies property $p$ in $H_{i}$. By definition, any group satisfies the subordination of $p$ in itself, since we can take $n=0$ and take a chain of length $0$.

The subordination of any property is a t.i. subgroup property (it is both transitive and identity-true) and a t.i. subgroup property equals its own subordination.

Here are some quick points on the subordination operator:

1. The subordination operator is an ascendant operator: If $p$ is a subgroup property, $p$ is stronger than its subordination.
2. The subordination operator is a monotone operator: If $p,q$ are subgroup properties such that $p$ is stronger than $q$, then the subordination of $p$ is stronger than the subordination of $q$.
3. If $p$ and $q$ are subgroup properties such that $q$ is t.i., then the subordination of $p$ is stronger than $q$. In fact, the subordination of $p$ is the strongest t.i. subgroup property among those weaker than $q$.

The basic proof idea: express the subgroup property in a form that makes it obvious

The idea is to express the subgroup property using a formalism that makes it obvious that it is transitive.

The most typical idea is that of a balanced subgroup property. We discuss the idea for function restriction expressions first, and then discuss some other, more general, variants.

Balanced subgroup properties in the function restriction formalism

Further information: Balanced subgroup property (function restriction formalism), balanced implies transitive

Suppose $a$ and $b$ are properties of functions from a group to itself. The property $p$ with function restriction expression $a\to b$ is defined as follows: $H$ satisfies $p$ in $G$ if every function from $G$ to itself satisfying $a$ restricts to a function from $H$ to itself satisfying $b$. (A bunch of examples is available at the function restriction formalism chart).

A balanced subgroup property with respect to the function restriction formalism is a subgroup property having an expression where both the left and right sides are equal. Such an expression is termed a balanced expression. For instance:

• The property of being a characteristic subgroup is a balanced subgroup property, because it can be expressed as:

Automorphism $\to$ Automorphism

• The property of being a central factor is a balanced subgroup property, because it can be expressed as:

Inner automorphism $\to$ Inner automorphism

• The property of being a fully characteristic subgroup is a balanced subgroup property, because it can be expressed as:

Endomorphism $\to$ Endomorphism

The easy but important fact is that any balanced subgroup property is transitive. In fact, for properties that have function restriction expressions, being t.i. (transitive and identity-true) is equivalent to being balanced, something that follows from either the left tightness theorem or the right tightness theorem. Note that a transitive subgroup property may have another function restriction expression that is not balanced; however, either left tightening or right tightening yields a balanced expression.

The idea of balance in other formalisms

For any formalism that involves restricting/extending functions, relations, or other constructs, the properties having balanced expressions are transitive. Here are some examples:

1. Function extension expressions: A function extension expression $a\leftarrow b$ defines a property as follows: $H$ has the property in $G$ if every function on $H$ satisfying property $b$ on $H$ can be extended to a function on $G$ satisfying property $a$ on $G$. A balanced expression is where $a=b$, and such properties are t.i.. Examples include:
   • AEP-subgroup, where both sides are automorphism.
   • EEP-subgroup, where both sides are endomorphism.
2. Subgroup intersection restriction expressions: Suppose $a$ and $b$ are two subgroup properties. The subgroup property with subgroup intersection restriction expression $a\to b$ is defined as follows: a subgroup $H$ of a group $G$ has this property if whenever $K$ satisfies property $a$ in $G$, $H\cap K$ satisfies property $b$ in $H$. Again, if $a=b$, the subgroup property we get is transitive.
   • Large subgroup is a subgroup whose intersection with every nontrivial subgroup is nontrivial. Further information: largeness is transitive
   • Normality-large subgroup is a subgroup whose intersection with every nontrivial normal subgroup is nontrivial. Note that this has a balanced expression, because since normality satisfies transfer condition, the intersection is nontrivial and normal in the subgroup. Further information: Normality-largeness is transitive
3. Subgroup intersection extension expressions: Suppose $a$ and $b$ are two subgroup property. The subgroup property with subgroup intersection extension expression $a\leftarrow b$ is defined as follows: a subgroup $H$ of a group $G$ has this property if whenever $K\leq H$ satisfies property $b$ in $H$, there exists a subgroup $L$ of $G$ such that $H\cap L=K$, with $L$ satisfying $a$ in $G$. Again, if $a=b$, the property is t.i.. Examples include:
   • CEP-subgroup, where both $a$ and $b$ are the property of being a normal subgroup.
4. Equivalence relation expressions: Suppose $a,b$ are rules that specify, for every group, an equivalence relation on it. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
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Properties that are functions of the index, order, or related things

Index and its multiplicativity

If $S$ is a multiplicatively closed collection of cardinals, and we define a subgroup property $p$ as saying that a subgroup has property $p$ if and only if its index is in $S$, then $p$ is transitive. This follows from the fact that index is multiplicative. Some examples:

• Subgroup of finite index: A subgroup of finite index in a subgroup of finite index again has finite index.
• Subgroup of odd index
• Hall subgroup: A Hall subgroup of a Hall subgroup is again a Hall subgroup. A Hall subgroup is a subgroup whose order and index are relatively prime. The reasoning behind this is a little more complicated, but it essentially follows from multiplicativity of the index. Further information: Hall satisfies transitivity

Complements and piecing them together

Factors in a particular kind of product

If a certain internal product notion is associative, the property of being one of the subgroups featuring in that product is transitive. Here are some examples:

• Direct factor: This follows from the fact that direct product is associative. Further information: Direct factor is transitive
• Retract: This follows from the fact that semidirect products are associative in a weak sense. Further information: Retract is transitive
• Base of a wreath product: This follows from the fact that wreath product is associative, again in a weak sense. Further information: Base of a wreath product is transitive
• Free factor: Further information: Free factor is transitive
• Regular retract: Further information: Regular retract is transitive
• Central factor: Further information: Central factor is transitive

Complements to quotient-transitive properties are transitive

Further information: Quotient-transitive and stronger than normality implies complementary property is transitive

Suppose $p$ is a subgroup property stronger than normality, that is also a quotient-transitive subgroup property. In other words, if $H\leq K\leq G$ are such that $H$ satisfies $p$ in $G$ and $K/H$ satisfies $p$ in $G/H$, then $K$ satisfies $p$ in $G$.

Consider the following property $q$: $H\leq G$ has property $q$ if and only if $H$ has a normal complement in $G$ satisfying property $p$. Then, $q$ is a transitive subgroup property.

Closure under certain equations/expressions/operations

This is a generalization of the notion of balance with respect to restriction/extension formalisms. Here, we require the subgroup to be closed under some operations, or to admit solutions to certain equations. The crucial point is that we can iterate on the closure condition to prove transitivity.

Verbal subgroup

Further information: Verbal subgroup, verbality is transitive

A verbal subgroup is defined by a collection of words, and is defined as the subgroup generated by all elements of the group that equal that word when evaluated at some elements of the group. For instance, the commutator subgroup of a group is a verbal subgroup with the word being $xyx^{-1}y^{-1}$. All members of the derived series as well as of the lower central series are verbal subgroup.

First, we note that we can expand the collection of words to all products involving the words and their inverses, so there exists an enlarged collection of words so that every subgroup element equals the value of one or more of the words at some suitable element of the group. For instance, in the commutator subgroup example, we expand to include all products of finite length of commutators. Now, if we have a verbal subgroup $H$ (with word collection ${\mathcal {V}}$) of a verbal subgroup $K$ (with word collection ${\mathcal {W}}$) of a group $G$, we can take as the new collection of words the following: we substitute for every possible letter in the collection of words ${\mathcal {V}}$for $H$ in $K$, the words in ${\mathcal {W}}$. This ability to substitute words into letters establishes transitivity.

Other examples

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Effect of logical operators

Conjunction

Further information: Transitivity is conjunction-closed

Another interesting case where we may easily be able to prove transitivity is where the given property is the conjunction (AND) of two existing properties. If both properties are transitive, then their conjunction is also transitive. Note that in some cases, one of the properties may actually be a group property evaluated on the subgroup as an abstract group. In this case, the other property being transitive is sufficient.

In some cases, one of the properties may be transitive and the other one may fail to be transitive; however, the transitive property may provide the right additional conditions that force transitivity of the conjunction. This may happen because the conjunction is stronger than the left transiter or the right transiter of the second property. Here are some examples:

• Conjugacy-closed normal subgroup: This is the conjunction of conjugacy-closed subgroup (transitive) and normal subgroup (not transitive). However, if $H\leq K\leq G$ and $K$ is conjugacy-closed normal in $G$, then it is clear that $H$ being normal in $K$ implies that $H$ is normal in $K$.
• Simple normal subgroup: Here, the condition of being simple, which is a group property, forces the conjunction to be transitive, even though normality is not transitive.
• Direct factor: This is the conjunction of retract (transitive) and normal subgroup (not transitive).
• Normal Hall subgroup: This is the conjunction of normal subgroup (not transitive) and Hall subgroup (transitive).

Disjunction

The disjunction (OR) of transitive subgroup properties is not necessarily transitive. To see this, suppose $p$ and $q$ are both transitive subgroup properties. Suppose $H\leq K\leq G$ are groups such that $H$ satisfies $p$ or $q$ in $K$ and $K$ satisfies $p$ or $q$ in $G$. It is not necessary that $H$ satisfies $p$ or $q$ in $G$. This is because it may be the case that $H$ satisfies $p$ but not $q$ in $K$ and $K$ satisfies $q$ but not $p$ in $G$, and we can then conclude nothing about $H$ in $G$.

Effect of subgroup property modifiers

Modifiers that create transitive properties by definition

In some cases, the way a subgroup property is defined already makes it clearly a t.i. subgroup property, and hence, transitive. There are three idempotent operators that take any subgroup property and output a t.i. subgroup property:

• Subordination as well as its transfinite versions.
• Left transiter
• Right transiter

Transfer condition operator

Further information: Transfer condition operator preserves transitivity, Subhomomorphism relation between transfer condition operator and composition operator

The transfer condition operator takes as input a subgroup property $p$ and outputs the property $T(p)$ defined as follows: a subgroup $H$ of a group $G$ satisfies $T(p)$ in $G$ if, for any subgroup $K$ of $G$, $H\cap K$ satisfies property $p$ in $K$.

If $p$ is a transitive subgroup property, so is $T(p)$. Here are some examples:

• Transfer-closed characteristic subgroup: Further information: Characteristicity is transitive, Transfer-closed characteristicity is transitive
• Transfer-closed fully invariant subgroup: Further information: Full invariance is transitive, Transfer-closed full invariance is transitive