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.

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.

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.