# Retract not implies normal complements are isomorphic

## Statement

Suppose $G$ is a group and $H$ is a Retract (?) of $G$. In other words, $H$ has a Normal complement (?) in $G$: there exists a normal subgroup $N$ of $G$ such that $NH = G$ and $N \cap H$ is trivial.

Then, there may exist another normal complement $M$ to $H$ in $G$ such that $M$ is not isomorphic to $N$.

## Proof

### Example of the dihedral group

Further information: dihedral group:D8

Consider the following example:

• $G$ is the dihedral group of order eight:

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

• $H$ is the subgroup $\{ x, e \}$, i.e., a two-element cyclic subgroup.
• There are two normal complements to $H$ in $G$. The subgroup $N = \{ x, x^2, x^3, e \}$ is cyclic of order four and the subgroup $M = \{ e, x^2, ax, a^3x \}$ is a Klein-four group.