Schur–Zassenhaus theorem

Examples

If we do not impose the coprime condition, the theorem is not true: consider for example the cyclic group ${\displaystyle C_4}$ and its normal subgroup ${\displaystyle C_2}$. Then if ${\displaystyle C_4}$ were a semidirect product of ${\displaystyle C_2}$ and ${\displaystyle C_4/C_2\cong C_2}$ then ${\displaystyle C_4}$ would have to contain two elements of order 2, but it only contains one. Another way to explain this impossibility of splitting ${\displaystyle C_4}$ (i.e. expressing it as a semidirect product) is to observe that the automorphisms of ${\displaystyle C_2}$ are the trivial group, so the only possible [semi]direct product of ${\displaystyle C_2}$ with itself is a direct product (which gives rise to the Klein four-group, a group that is non-isomorphic with ${\displaystyle C_4}$).

An example where the Schur–Zassenhaus theorem does apply is the symmetric group on 3 symbols, ${\displaystyle S_3}$, which has a normal subgroup of order 3 (isomorphic with ${\displaystyle C_3}$) which in turn has index 2 in ${\displaystyle S_3}$ (in agreement with the theorem of Lagrange), so ${\displaystyle S_3/C_3\cong C_2}$. Since 2 and 3 are relatively prime, the Schur–Zassenhaus theorem applies and ${\displaystyle S_3\cong C_3⋊C_2}$. Note that the automorphism group of ${\displaystyle C_3}$ is ${\displaystyle C_2}$ and the automorphism of ${\displaystyle C_3}$ used in the semidirect product that gives rise to ${\displaystyle S_3}$ is the non-trivial automorphism that permutes the two non-identity elements of ${\displaystyle C_3}$. Furthermore, the three subgroups of order 2 in ${\displaystyle S_3}$ (any of which can serve as a complement to ${\displaystyle C_3}$ in ${\displaystyle S_3}$) are conjugate to each other.

The non-triviality of the (additional) conjugacy conclusion can be illustrated with the Klein four-group ${\displaystyle V}$ as the non-example. Any of the three proper subgroups of ${\displaystyle V}$ (all of which have order 2) is normal in ${\displaystyle V}$; fixing one of these subgroups, any of the other two remaining (proper) subgroups complements it in ${\displaystyle V}$, but none of these three subgroups of ${\displaystyle V}$ is a conjugate of any other one, because ${\displaystyle V}$ is abelian.

The quaternion group has normal subgroups of order 4 and 2 but is not a [semi]direct product. Schur's papers at the beginning of the 20th century introduced the notion of central extension to address examples such as ${\displaystyle C_4}$ and the quaternions.

Proof

The existence of a complement to a normal Hall subgroup H of a finite group G can be proved in the following steps:

1. By induction on the order of G, we can assume that it is true for any smaller group.
2. If H is abelian, then the existence of a complement follows from the fact that the cohomology group H²(G/H,H) vanishes (as H and G/H have coprime orders) and the fact that all complements are conjugate follows from the vanishing of H¹(G/H,H).
3. If H is solvable, it has a nontrivial abelian subgroup A that is characteristic in H and therefore normal in G. Applying the Schur–Zassenhaus theorem to G/A reduces the proof to the case when H=A is abelian which has been done in the previous step.
4. If the normalizer N=N_G(P) of every p-Sylow subgroup P of H is equal to G, then H is nilpotent, and in particular solvable, so the theorem follows by the previous step.
5. If the normalizer N=N_G(P) of some p-Sylow subgroup P of H is smaller than G, then by induction the Schur–Zassenhaus theorem holds for N, and a complement of N∩H in N is a complement for H in G because G=NH.