The Schur–Zassenhaus theorem is a theorem in group theory which states that if is a finite group, and is a normal subgroup whose order is coprime to the order of the quotient group /, then is a semidirect product (or split extension) of and /.

Zassenhaus theorem implies Gis a semidirect product of Nand H: N\His trivial since (a;b) = 1, so G= NH˘=NoHwhere Hacts on Nby conjugation. Here are two cases where the Schur-Zassenhaus theorem has proofs using no hard work.

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Understanding the proof and meaning of the Butterfly lemma (Zassenhaus) (Lang's Algebra, pp. 20--21)

Zassenhaus's Lemma is a result in group theory. Hans Zassenhaus published his proof of the lemma in 1934 to provide a more elegant proof of Schreier's Theorem. He was a doctorate student under Emil Artin at the time. In this article, group operation is …

Zassenhaus 定理是一个群论中的定理，它是一个有关正规群列的同构关系的命题，在群列的很多证明中发挥着重要作用。 假设 A , A ∗ , B , B ∗ {\displaystyle A, A^*, B, B^*} 是群 G {\displaystyle G} 的子群，且 A ∗ ⊴ A , B ∗ ⊴ B {\displaystyle A^* \trianglelefteq A…

2024年9月16日 · The conjugacy part of the Schur-Zassenhaus theorem states that, if $G = HN = KN$, with $H \cap N = K \cap N = 1$ (and $(|H|, |N|) = (|K|, |N|) = 1$), then $H$ and $K$ are $G$-conjugate. My question is whether we can drop the assumption that …

COROLLARY (Zassenhaus). Let (R , +, •) be a finite near-field with identity 1 . Then (R , +) is abelian . Proof. Suppose (-l)x = x for some x in R . If x i- 0 , then there exists a y in R such that xy = 1 . Thus (-l)xy = xy = 1 and hence 1+1 = 0. It follows that w + w = w(l + 1) = 0 for each w in R. Consequently (R , + ) is abelian.

Zassenhaus引理 (蝴蝶引理)。 给定 G 的四个子群 A\triangleleft A^* 和 B\triangleleft B^* ，则 A(A^*\cap B)\triangleleft A(A^*\cap B^*) ， (A\cap B^*)B\triangleleft(A^*\cap B^*)B ，且存在同构 \frac{A(A^*\cap B^*)}{A(A^*\cap B)}\simeq\frac{(A^*\cap B^*)}{(A^*\cap B)(A\cap B^*)}\simeq\frac{(A^*\cap B^*)B}{(A\cap B^*)B} 。

Let’s put the Schur{Zassenhaus theorem to work. We ask, out of idle curiosity, whether pj#Gimplies pj#Aut(G). The answer, of course, is no: try G= Z=(p). As we now show, this counterexample essentially explains all the others. Corollary 4. Fix a prime p. For a nite group Gwith order divisible by p, the following n