# Commutator of a normal subgroup and a subgroup not implies normal

## Statement

It is possible to have a group $L$, a Normal subgroup (?) $K$ of $G$, and a subgroup $G$ of $L$, such that the Commutator of two subgroups (?) $[K,G]$ is not a normal subgroup of $L$.

## Proof

### 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

Further information: Alternating group:A5, A5 is simple, A5 is the simple non-Abelian group of smallest order

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.