Normality is not transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
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Statement

There can be a situation where $H$ is a normal subgroup of $K$ and $K$ is a normal subgroup of $G$ but $H$ is not a normal subgroup of $G$.

Partial truth

Transitivity-forcing operator

• A group in which every normal subgroup of a normal subgroup is normal is termed a T-group. Note that abelian groups and Dedekind groups are T-groups, and any nilpotent group that is a T-group is also a Dedekind group.
• A group $K$ has the property that whenever $K$ is normal in $G$, every normal subgroup of $K$ is normal in $G$ (in other words, transitivity holds with $K$ as the middle group) if and only if $K$ is a group in which every normal subgroup is characteristic.
• There is no nontrivial group $G$ such that whenever $G$ is a normal subgroup of a normal subgroup of some group, $G$ is normal in that group. In fact, the general example we construct here shows precisely that. Further information: every nontrivial normal subgroup is potentially 2-subnormal-and-not-normal

Left transiter

While normality is not transitive, it is still true that every characteristic subgroup of a normal subgroup is normal. Characteristicity is the left transiter of normality -- it is the weakest property $p$ such that every subgroup with property $p$ in a normal subgroup is normal. For full proof, refer: Characteristic of normal implies normal, Left transiter of normal is characteristic

Right transiter

While normality is not transitive, every normal subgroup of a transitively normal subgroup is normal. Being transitively normal is the right transiter of being normal. Properties like being a direct factor, being a central subgroup, and being a central factor imply being transitively normal.

Subnormality

The lack of transitivity of normality can also be remedied by defining the notion of subnormal subgroup. Subnormality is the weakest transitive subgroup property implied by normality. More explicitly, a subgroup $H$ is subnormal in a group $G$, if we can find a chain of subgroups going up from $H$ to $G$, with each subgroup normal in its successor.

A special case of this is the notion of 2-subnormal subgroup, which is a normal subgroup of a normal subgroup. Another special case is the notion of a 3-subnormal subgroup, which is a normal subgroup of a normal subgroup of a normal subgroup.

There are also related notions of hypernormalized subgroup, 2-hypernormalized subgroup, ascendant subgroup, descendant subgroup, and serial subgroup.

Corollaries

• Normality is not a finite-relative-intersection-closed subgroup property, because finite-relative-intersection-closed implies transitive.

Related facts

Making normality transitive

For simplicity, we assume $H \le K \le G$, with $H$ the bottom group, $K$ the middle group, and $G$ the top group.

Statement	Change in assumption	Change in conclusion
Characteristic of normal implies normal	$H$ a characteristic subgroup of $K$	$H$ is normal in $G$
Left transiter of normal is characteristic	$H$ in $K$ is such that for any possible $G$ with $K$ normal in $G$, $H$ is normal in $G$	$H$ is characteristic in $K$
Equivalence of definitions of transitively normal subgroup	$H$ is normal in $K$; spelling out conditions for $K$ in $G$ such that ...	$H$ is normal in $G$
Central factor implies transitively normal	$K$ a central factor of $G$	$H$ is normal in $G$
Direct factor implies transitively normal	$K$ a direct factor of $G$	$H$ is normal in $G$

For particular kinds of groups

For simplicity, we refer below to the three groups as $H \le K \le G$, with $H$ the bottom group, $K$ the middle group, and $G$ the top group, such that $H$ is normal in $K$ and $K$ is normal in $G$, but $H$ is not normal in $G$.

Stronger formulation	Additional restrictions introduced	Additional comments/examples
Normality is not transitive for any nontrivially satisfied extension-closed group property	$H,K,G$ all satisfy a group property $\alpha$ closed under taking extensions	$\alpha$ could be the property of being a finite $p$-group, any fixed prime $p$; or solvability, or finiteness
Conjunction of normality with any nontrivial finite-direct product-closed property of groups is not transitive	$H$ and $K$ both satisfy a group property $\alpha$ that is closed under finite direct products	Abelian normal subgroup of abelian normal subgroup need not be normal
Every nontrivial normal subgroup is potentially 2-subnormal-and-not-normal	We are given $H \le L$ and must find $K,G$ with $G$ containing $L$
Normality is not transitive for any pair of nontrivial quotient groups	We are given nontrivial groups $A,B$ and must ensure $K/H \cong A, G/K \cong B$

The extent of lack of transitivity

Stronger formulation	Meaning of formulation	How "normality is not transitive" is a special case
There exist subgroups of arbitrarily large subnormal depth	For any natural number $n$, there exists a group $G$, subgroup $H$, such that the shortest subnormal series for $H$ in $G$ has length $n$. In other words, the minimum length of a chain from $H$ up to $G$, with each subgroup normal in the next, is $n$	Case $n = 2$
Descendant not implies subnormal	There exist subgroups for which there is a descending chain from whole group to subgroup, each normal in predecessor, of countable length (so intersection of all members is subgroup) but no finite chain
there exist subgroups of arbitrarily large descendant depth
Ascendant not implies subnormal	There exist subgroups for which there is an ascending chain from subgroup to whole group, each normal in successor, of countable length (so union of all members is whole group) but no finite chain
there exist subgroups of arbitrarily large ascendant depth
Normal not implies left-transitively fixed-depth subnormal	We can have a normal subgroup $H$ of $K$ such that for every $k$, there exists a group $G$ containing $K$ as $k$-subnormal but $H$ is not $k$-subnormal	Case $k = 1$
Normal not implies right-transitively fixed-depth subnormal	We can have a normal subgroup $K$ of $G$ such that for every $k$, there exists a group $H$ that is $k$-subnormal in $K$, not in $G$	Case $k = 1$

Analogues in other algebraic structures

Statement	Analogy correspondences	Additional comments
Ideal property is not transitive for Lie rings	Lie ring $\leftrightarrow$group, ideal of a Lie ring $\leftrightarrow$ normal subgroup
Normality is not transitive for field extensions	field extension $\leftrightarrow$ group (namely, its Galois group), normal field extension $\leftrightarrow$ normal subgroup in Galois correspondence	A normal field extension of a normal field extension need not be normal. In fact, by the fundamental theorem of Galois theory, this corresponds directly to the fact that a normal subgroup of a normal subgroup need not be normal.
Normality is not composition-closed	normal monomorphism $\leftrightarrow$ normal subgroup	A composite of normal monomorphisms need not be normal.

General tricks

• Disproving transitivity
• Using dihedral groups as counterexamples

Proof

(Also see #List of counterexamples of small order).

Generic example

A natural example is as follows. Take any nontrivial group $H$, and consider the square, $K = H \times H$ (the external direct product of $H$ with itself). Now, consider the external semidirect product of this group with the group $\mathbb{Z}/2\mathbb{Z}$ (the cyclic group of two elements) acting via the exchange automorphism (the automorphism that exchanges the coordinates). Call the big group $G$.

(Note that $G$ can be described more compactly as the external wreath product of $H$ with the group of order two acting regularly.)

Let $H_1, H_2$ be the copies of $H$ embedded in $K$ as $H \times \{ e \}$ and $\{ e \} \times H$. We then have:

• $H_1$ is normal in $K$: In fact, $H_1$ is a direct factor of $K$, and is thus normal.
• $K$ is normal in $G$: $K$ is the base of a semidirect product, and is thus normal. Equivalently, any inner automorphism of $G$ is the composite of an inner automorphism in $K$ and the exchange automorphism, both of which preserve $K$.
• $H_1$ is not normal in $G$: The exchange automorphism is an inner automorphism of $G$, and it exchanges $H_1$ and $H_2$ -- in particular, it does not preserve $H_1$. Thus, $H_1$ is not normal in $G$.

Note that both $H_1$ and $H_2$ are copies of $H$, and hence either can be viewed as the Base of a wreath product (?) in $G$.

SIDENOTE: This example is not, in some sense, an extreme example of normality not being transitive. In fact, the property of being the base of a wreath product is transitive, and any base of a wreath product is a 2-subnormal subgroup, which implies that applying this construction iteratively does not yield subgroups of subnormal depth more than two. Even further, base of a wreath product implies right-transitively 2-subnormal, or equivalently, any 2-subnormal subgroup of the base of a wreath product is 2-subnormal in the whole group.

Specific realizations of this generic example

The smallest case of this yields $H_1$ a group of order two, and $G$ a group of order eight. In fact, here $G$ is the dihedral group of order eight and $H$ is a cyclic group of order two, with $H_1$ and $H_2$ being subgroups of order two generated by reflections. Here's how this relates to the usual definition of the dihedral group:

$G = \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$

$H_1 = \langle x \rangle, \qquad H_2 = \langle a^2 x \rangle, \qquad K = \langle x, a^2 \rangle$.

Here, $H_1$ and $H_2$ are both normal in $K$, which is normal in $G$, but neither $H_1$ nor $H_2$ is normal in $G$.

For more on the subgroup structure of the dihedral group, refer subgroup structure of dihedral group:D8, Klein four-subgroups of dihedral group:D8, and non-normal subgroups of dihedral group:D8.

GAP implementation

Implementation of the generic example

Here is an implementation of the generic example, with any nontrivial group $H$. Note that you need to define $H$ for GAP before executing the commands in this example! Double semicolons have been used to suppress the output here, since the output depends on the choice of $H$ (you can use single semicolons instead to display all the outputs).

We first construct the groups $H_1, H_2, K, G$ using the wreath product command:

gap> G := WreathProduct(H,SymmetricGroup(2));;
gap> H1 := Image(Embedding(G,1));;
gap> H2 := Image(Embedding(G,2));;
gap> K := Group(Union(H1,H2));;

Next, we check that $H_1$ and $H_2$ are subgroups of $K$ and $K$ is a subgroup of $G$:

gap> IsSubgroup(G,K);
true
gap> IsSubgroup(K,H1);
true
gap> IsSubgroup(K,H2);
true

Finally, we check that $H_1, H_2$ are both normal in $K$ and $K$ is normal in $G$, but $H_1$ and $H_2$ are not normal in $G$.

gap> IsNormal(G,K);
true
gap> IsNormal(K,H1);
true
gap> IsNormal(K,H2);
true
gap> IsNormal(G,H1);
false
gap> IsNormal(G,H2);
false

The implementation in some special cases

Here is the implementation when $H$ is cyclic of order two:

gap> G := WreathProduct(H,SymmetricGroup(2));
<group of size 8 with 2 generators>
gap> H1 := Image(Embedding(G,1));
<group with 1 generators>
gap> H2 := Image(Embedding(G,2));
<group with 1 generators>
gap> K := Group(Union(H1,H2));
<group with 3 generators>
gap> IsSubgroup(G,K);
true
gap> IsSubgroup(K,H1);
true
gap> IsSubgroup(K,H2);
true
gap> IsNormal(G,K);
true
gap> IsNormal(K,H1);
true
gap> IsNormal(K,H2);
true
gap> IsNormal(G,H1);
false
gap> IsNormal(G,H2);
false

List of counterexamples of small order

Big group	Order of big group	Violation of normality being transitive
dihedral group:D8	$8$	Klein four-subgroup is normal, has normal subgroup of order two that is not normal in the whole group.
alternating group:A4	$12$	The normal Klein four-group comprising the identity and three double transpositions has a normal subgroup of order two that is not normal in the whole group.
SmallGroup(16,3)	$16$
SmallGroup(16,4)	$16$
M16	$16$
dihedral group:D16	$16$
semidihedral group:SD16	$16$
quaternion group:Q16	$16$

References

Textbook references

• Groups and representations by Jonathan Lazare Alperin and Rowen B. Bell, ISBN 0387945261, Page 8, ^{More info} Also, Page 6 (first mention), and Page 17 (further explanation)
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 91, Section 3.2 (More on cosets and Lagrange's theorem), Example (3), (example of the dihedral group)^{More info} Also, Page 135, with justification of the related fact that characteristic of normal implies normal
• An Introduction to Abstract Algebra by Derek J. S. Robinson, ISBN 3110175444^{More info}, Page 66
• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, Page 17, Exercise 1.3.15, ^{More info} Also: Page 28, Page 63
• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, Page 236, Miscellaneous Problems (Chapter 6), Exercise 4, (starred problem)^{More info}

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