Disproving transitivity

A subgroup property $$p$$ is termed a survey article about::transitive subgroup property if whenever $$H \le K \le G$$ 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 survey article about::t.i. subgroup property if it is transitive as well as identity-true: every group satisfies the property as a subgroup of itself.

This article discusses techniques used to prove that a given subgroup property is not transitive.

Also refer:


 * Proving transitivity: A survey article on methods to prove that a given subgroup property is 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.

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:

$$H = H_0 \le H_1 \le H_2 \le \dots \le H_n = G$$,

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
Suppose we want to prove that a given subgroup property $$q$$ is not transitive. One approach is as follows: find an identity-true subgroup property $$p$$such that $$p$$ is stronger than $$q$$ but the subordination of $$p$$ is not stronger than $$q$$. Let's now see how such $$p$$ can be chosen.

Using normality
The subordination of the property of being a normal subgroup is commonly called the property of being a subnormal subgroup. Thus, one way of showing that a given property $$q$$ is not transitive is to show that every normal subgroup of a group satisfies $$q$$, but there are subnormal subgroups that do not satisfy $$q$$.

Here are some examples of subgroup properties that are satisfied by normal subgroups, but not satisfied by all subnormal subgroups, and hence are not transitive:


 * All properties of being a $$k$$-subnormal subgroup for finite $$k$$ (in other words, all the subgroups of a fixed subnormal depth). These properties are not transitive because there exist subgroups of arbitrarily large subnormal depth. For instance, 2-subnormality is not transitive, 3-subnormality is not transitive.
 * The property of being a hypernormalized subgroup.
 * Any property that is between normality and the property of being a subnormal-to-normal subgroup is not transitive. In other words, a subgroup property that is true for normal subgroups but is not true for any non-normal subnormal subgroup cannot be transitive.

Using maximality
In a finite group, every subgroup is a submaximal subgroup -- there exists an ascending chain of subgroups starting from the subgroup and ending at the whole group such that each is a maximal subgroup of its successor. Thus, we have the following:

If an identity-true subgroup property $$q$$ is satisfied by every maximal subgroup, but there is a subgroup of a finite group not satisfying it, then the subgroup property is not transitive. Examples include, for instance:


 * Pronormality is not transitive: This follows from the fact that maximal implies pronormal. Similarly, we have weak pronormality is not transitive, paranormality is not transitive, polynormality is not transitive, weak normality is not transitive.

Looking at specific examples
This section discusses some general ideas for how to come up with specific examples to prove that a particular subgroup property is not transitive.

First, eliminate the kinds of groups where the property is transitive
To make the search for a counterexample efficient, it is helpful to first remove from consideration all groups where the property is transitive for obvious reasons.

General strategies:


 * 1) Suppose $$\alpha$$ is a trim subgroup property: it is always satisfied by the whole group as a subgroup of itself, and by the trivial subgroup. Then, the trivial group cannot provide a counterexample. A group of prime order cannot provide a counterexample, and a group whose order is of the form $$p^2$$ or $$pq$$ (where $$p,q$$ are primes) cannot provide a counterexample. Thus, to start looking for groups where $$\alpha$$ is not transitive, the place to start looking is groups of order $$p^3$$, $$p^2q$$, or $$pqr$$, where $$p,q,r$$ are primes.
 * 2) Find properties $$\beta$$ stronger than $$\alpha$$ that are transitive, and eliminate all groups where the subgroups with property $$\alpha$$ also have property $$\beta$$: For instance, the property of being a central subgroup (or more generally, a central factor) is transitive, and it is stronger than the property of being a normal subgroup. Thus, to look for examples of groups where normality is not transitive, we should look at groups where it is not true that all normal subgroups are central (or, where all normal subgroups are central factors). In particular, we should not look at abelian groups. We should also not look at simple groups, quasisimple groups, and characteristically simple groups.
 * 3) Try to find the precise left transiter for the property, and then show that the property does not imply its left transiter: The left transiter of a property $$\alpha$$ is the unique property $$\beta$$ such that $$H$$ satisfies $$\beta$$ in $$K$$ if and only if, for every $$G$$ containing $$K$$, whenever $$K$$ satisfies $$\alpha$$ in $$G$$, $$H$$ satisfies $$\alpha$$ in $$G$$. Showing that $$\alpha$$ is not transitive is then equivalent to finding a subgroup $$H$$ of a group $$K$$ that satisfies $$\alpha$$ but not $$\beta$$. Once we have such a pair, we find a $$G$$ containing $$K$$ such that $$K$$ satisfies $$\alpha$$ in $$G$$ but $$H$$ does not satisfy $$\alpha$$ in $$G$$. An example of this is showing that normality is not transitive. One way of doing this is to show that left transiter of normal is characteristic, and combine this with the fact that normal not implies characteristic.

Selecting the better examples to begin with
As remarked above, we should start with groups of order $$p^3$$ when trying to prove that a trim subgroup property $$\alpha$$ is not transitive. There are some general rules for what groups to prefer:


 * The cyclic group of order $$p^3$$ (and more generally, any cyclic group) is a very bad place to start looking. This is because subgroups of a cyclic group satisfy some very strong subgroup properties, and countering transitivity is in general extremely hard.
 * The other abelian groups of order $$p^3$$ (and more generally, of order $$p^n$$) are good places to start looking if there is no reason to believe that the property is transitive for all abelian groups. This includes the elementary abelian group and the product of the cyclic group of order $$p^2$$ and the cyclic group of order $$p$$.
 * Among the non-abelian $$p$$-groups, the quaternion group (exists only for $$p = 2$$) is a very bad place to start looking. This is because it has only one subgroup of order $$p$$, and this subgroup satisfies some very strong subgroup properties.
 * The dihedral group of order eight, and the two non-abelian groups for odd primes (prime-cube order group:U3p and prime-cube order group:p2byp) are good places to start looking.

Among groups of order $$p^2q$$ and higher orders, here are some general ideas:


 * Looking at the direct product of a group of order $$p^2$$ and a group of order $$q$$ is in general a bad idea. Rather, it is better to look at a group where at least one of the Sylow subgroups is not normal.
 * A good starting point is the alternating group of degree four. Also useful is the symmetric group of degree four.

Some illustrations

 * Suppose we are looking at the property of being a normal subgroup. Since normality is a trim subgroup property, it is useless to look at groups whose order is a product of zero, one or two primes. Thus, we can start by looking at groups of order $$p^3$$. Among these, it is useless to look at abelian groups. Thus, we need to look at groups of order $$p^3$$ that are non-abelian. For $$p=2$$, there are two such groups up to isomorphism. One of them, the quaternion group, fails to provide a counterexample. The other, the dihedral group, does provide a counterexample.
 * Another approach to proving that normality is not transitive begins by proving that the left transiter of normal is characteristic. Then, a simple construction shows normal not implies characteristic. The smallest example of the general construction is where the big group $$K$$ is a Klein four-group and the subgroup $$H$$ is a subgroup of order two. We now need to find a bigger group containing $$K$$ in which $$K$$ is normal but $$H$$ is not. This can be done explicitly by taking the semidirect product of $$K$$ with an automorphism that does not preserve $$H$$. There are many ways of doing this, yielding different examples of the fact that normality is not transitive. These include the dihedral group of size eight, alternating group of degree four, and symmetric group of degree four.
 * Consider the property of being an intermediately characteristic subgroup: a subgroup that is characteristic in every intermediate subgroup. Since this is a trim subgroup property, it suffices to start by looking at groups of order $$p^3$$. Setting $$p=2$$, we try the quaternion group and the dihedral group of order eight. As usual, the quaternion group fails to provide a counterexample, whereas the dihedral group does.
 * Consider the property of being an intermediately fully characteristic subgroup: a subgroup that is fully characteristic in every intermediate subgroup. Since this is a trim subgroup property, it suffices to start by looking at groups of order $$p^3$$. $$p=2$$ fails to yield a counterexample, but for odd $$p$$, the prime-cube order group:p2byp provides a counterexample.
 * Consider the property of being a permutably complemented subgroup: a subgroup that has a permutable complement in the whole group. Since this is a trim subgroup property, it suffices to start by looking at groups of order $$p^3$$. Once again, the smallest counterexample is furnished by the dihedral group with eight elements.