Schur-Zassenhaus's theorem

The Schur-Zassenhaus theorem is a mathematical theorem in group theory . The sentence named after Issai Schur and Hans Julius Zassenhaus reads:

• For a finite group and a normal subgroup with there is a subgroup with and . The group is therefore the semi-direct product of and . ${\displaystyle G}$ ${\displaystyle N\triangleleft G}$ ${\displaystyle \operatorname {ggT} (|N|,[G:N])=1}$ ${\displaystyle U\leq G}$ ${\displaystyle G\,=\,NU}$ ${\displaystyle N\cap U\,=\,1}$ ${\displaystyle G}$ ${\displaystyle G=N\times _{\theta }U}$ ${\displaystyle N}$ ${\displaystyle U}$

The subgroup in the above sentence is generally not clearly defined, but it can be shown that every two such subgroups are conjugated . ${\displaystyle U}$

Examples

• The cyclic group has the normal divisor . Since the numbers and are relatively prime, the Schur-Zassenhaus theorem can be applied. is apparently the only subgroup that fulfills the statement of the sentence. Since the group is Abelian, the semi-direct product in this case is even direct. ${\displaystyle G=\mathbb {Z} /6\mathbb {Z} =\{{\overline {0}},{\overline {1}},\ldots ,{\overline {5}}\}}$ ${\displaystyle N=\{{\overline {0}},{\overline {2}},{\overline {4}}\}}$ ${\displaystyle |N|=3}$ ${\displaystyle [G:N]=2}$ ${\displaystyle U=\{{\overline {0}},{\overline {3}}\}}$ ${\displaystyle G}$

• The symmetric group has the normal divisor . Because of and can the Schur-Zassenhaus theorem be applied, apparently the three subgroups fulfill the proposition of the theorem. ${\displaystyle G=S_{3}=\{e,d,d^{2},s_{1},s_{2},s_{3}\}}$ ${\displaystyle N=\{e,d,d^{2}\}}$ ${\displaystyle |N|=3}$ ${\displaystyle [G:N]=2}$ ${\displaystyle U_{i}=\{e,s_{i}\},\,i=1,2,3}$

• The cyclic group has the normal divisor . Here and are not coprime, which is why the theorem is not applicable. In fact, there is no subgroup that satisfies the proposition, because one would have to have an element of order 2, but the only element of order 2 is and that is already in . This example shows that the coprime numbers of and in the above sentence cannot be dispensed with. ${\displaystyle G=\mathbb {Z} /4\mathbb {Z} =\{{\overline {0}},{\overline {1}},{\overline {2}},{\overline {3}}\}}$ ${\displaystyle N=\{{\overline {0}},{\overline {2}}\}}$ ${\displaystyle |N|=2}$ ${\displaystyle [G:N]=2}$ ${\displaystyle U\subset G}$ ${\displaystyle {\overline {2}}}$ ${\displaystyle N}$ ${\displaystyle |N|}$ ${\displaystyle [G:N]}$

• If there is any group, the example shows that the coprime condition is not necessary for the existence of a representation as a semidirect, even direct, product. ${\displaystyle N}$ ${\displaystyle G=N\times N}$

Individual evidence

1. ↑ Rowen B. Bell, JL Alperin: Groups and Representations , Springer-Verlag, Graduate Texts in Mathematics, Volume 162, ISBN 0-387-94526-1 (Chapter 9: The Schur-Zassenhaus Theorem)

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• Andreas Nickel: Basic knowledge of algebra (PDF; 544 kB), University of Regensburg, p. 6