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 $M=\{e,x^{2},ax,a^{3}x\}$ is a Klein-four group.