Complement to normal subgroup is isomorphic to quotient group

Statement

Suppose $G$ is a group, $N$ is a normal subgroup and $H$ is a subgroup such that $N$ and $H$ are permutable complements: $NH = G$ and $N \cap H$ is trivial. Then, $G/N \cong H$ .

In particular, any two permutable complements to $N$ are isomorphic to each other.

Related facts

Caveats

Note the following:

If $N$ is a normal subgroup of $G$ , it is not necessary that $N$ has a permutable complement in $G$ . For instance, if $G$ is the cyclic group of order four and $N$ is a subgroup of order two, $N$ does not have a permutable complement. A normal subgroup that does have a permutable complement is termed a complemented normal subgroup. A subgroup that occurs as the permutable complement to a normal subgroup is termed a retract.
Any lattice complement to $N$ in $G$ is also a permutable complement to $N$ in $G$ . This is because any normal subgroup is permutable: its product with any subgroup is a subgroup. Further information: Equivalence of definitions of complemented normal subgroup

Other related facts

Complements to abelian normal subgroup are automorphic: If $N$ is an abelian normal subgroup of $G$ and $H, K$ are two permutable complements to $N$ in $G$ , then there is an automorphism of $G$ that is the identity map on $N$ and sends $H$ to $K$ . In particular, $H$ and $K$ are automorphic subgroups.
Complements to normal subgroup need not be automorphic: If $N$ is a normal subgroup of and $H,K$ are two permutable complements to $N$ in $G$ , there need not be an automorphism of $G$ sending $H$ to $K$ .
Retract not implies normal complements are isomorphic: This states that if $N$ and $M$ are two permutable complements to a subgroup $H$ , and both $N$ and $M$ are normal, this does not imply that $N$ is isomorphic to $M$ . In other words, interchanging the role of which subgroup is the normal one renders the result false.
Schur-Zassenhaus theorem (normal Hall implies permutably complemented and Hall retract implies order-conjugate): This states that every normal Hall subgroup, and hence every normal Sylow subgroup, has a complement, and that any two such complements are conjugate subgroups in the whole group.
Every group of given order is a permutable complement for symmetric groups: Any group of order $n$ occurs as a permutable complement to $\operatorname{Sym}(n-1)$ in $\operatorname{Sym}(n)$ , via the embedding obtained by Cayley's theorem. This is a far cry from the fact that any two permutable complements to a normal subgroup are isomorphic.
Semidirect product is not left-cancellative for finite groups: We can have finite groups $G,H,K$ such that $G \rtimes H \cong G \rtimes K$ , but $H$ is not isomorphic to $K$ . Note that the isomorphism between $G \rtimes H$ and $G \rtimes K$ must not map the normal subgroup $G$ to the normal subgroup $G$ , otherwise we would contradict the result of this page.