Commutator of a normal subgroup and a subgroup not implies normal

Statement
It is possible to have a group $$L$$, a fact about::normal subgroup $$K$$ of $$G$$, and a subgroup $$G$$ of $$L$$, such that the fact about::commutator of two subgroups $$[K,G]$$ is not a normal subgroup of $$L$$.

Opposite facts

 * Normality is commutator-closed: If both subgroups are normal, the commutator is also normal.
 * Subgroup normalizes its commutator with any subset: In particular, both the subgroups normalize their commutator.
 * Commutator of two subgroups is normal in join: In particular, if the subgroup generated by the two subgroups is the whole group, the commutator is normal.
 * Commutator of a normal subgroup and a subset implies 2-subnormal: The commutator, even though not necessarily normal, must be a 2-subnormal subgroup.

Other related facts

 * Commutator of a 2-subnormal subgroup and a subset implies 3-subnormal
 * Commutator of a 3-subnormal subgroup and a finite subset implies subnormal
 * Commutator of a 3-subnormal subgroup and a subset not implies subnormal
 * Commutator of a 3-subnormal subgroup and a finite subset not implies 4-subnormal

A generic example
Let $$G$$ be a nontrivial perfect group. Consider the square $$K = G \times G$$ (the external direct product of $$G$$ with itself). Let $$L$$ be the external semidirect product of $$K$$ with the cyclic group of order two acting by the coordinate exchange automorphism $$(a,b) \mapsto (b,a)$$.

In short $$L$$ can also be defined as the external wreath product of $$G$$ and a cyclic group of order two acting regularly.

Then, define $$G_1 = G \times \{ e \}$$, i.e., the first direct factor of $$K$$, and define $$G_2 = \{ e \} \times G$$, i.e., the second direct factor. Observe that:


 * 1) $$K$$ is normal in $$L$$ by construction.
 * 2) $$G_1$$ is a subgroup of $$L$$, but is not normal in $$L$$, because the coordinate exchange automorphism sends $$G_1$$ to $$G_2$$.
 * 3) $$G_1$$ is a direct factor of $$K$$, hence normal in $$K$$. Thus, $$[G_1,K] \le G_1$$.
 * 4) On the other hand, since $$G_1 \le K$$, $$[G_1,G_1] \le [G_1,K]$$. But since $$G_1 \cong G$$ and $$G$$ is perfect, we have $$G_1 = [G_1,G_1]$$, so $$G_1 \le [G_1,K]$$.
 * 5) Thus, by steps (3) and (4), we get $$G_1 = [G_1,K]$$. Step (2) tells us that $$G_1$$ is not normal in $$L$$. Thus, we have found a normal subgroup $$K$$ and a subgroup $$G_1$$ such that the commutator $$[G_1,K]$$ is not normal.

Particular cases of this example
Any simple non-Abelian group is an example of a nontrivial perfect group, so the smallest particular case of this generic example is to set $$G$$ as the alternating group of degree five.

A more generic example
In the above example, we assumed that $$G$$ was perfect. We can relax this slightly to assuming only that $$G$$ is non-Abelian. In this more general case, we obtain that $$[G_1,K] = [G_1,G_1]$$, which is a nontrivial subgroup of $$L$$ contained inside $$G_1$$. This subgroup is not normal because the coordinate exchange automorphism sends it to a corresponding subgroup of $$G_2$$.

This more generic example allows us some nilpotent and solvable groups:


 * Setting $$G$$ as the symmetric group on three letters yields a group $$L$$ of order $$6^2 \cdot 2 = 72$$. This group is a solvable group, since $$G$$ itself is solvable.
 * Setting $$G$$ as the dihedral group of order eight yields a group $$L$$ of order $$8^2 \cdot 2 = 128 = 2^7$$. This is a group of prime power order, and in particular, is a nilpotent group.