Question

State & prove " Zassenhaus Lemma "

Subject :- Algebra

Standard :- Msc 3rd Sem ​

Answer 1

Step-by-step explanation:

I will later give the proof of the theorm actually I forgot the proof.

Answer 2

Answer:

We claim that ∗(A∩B∗)◃(A∗∩B∗): that is, if c∈A∩B∗ and x∈A∗∩B∗, then xcx−1∈A∩B∗. Now xcx−1∈A because c∈A, x∈A∗, and A◃A∗; but also xcx−1∈B∗, because c,x∈B∗. Hence, (A∩B∗)◃(A∗∩B∗); similarly, (A∗∩B)◃(A∗∩B∗). Therefore, the subset D, defined by D=(A∩B∗)(A∗∩B), is a normal subgroup of A∗∩B∗, because it is generated by two normal subgroups.

Using the symmetry in the remark, it suffices to show that there is an isomorphism

$\frac{A(A∗∩B∗)}{A(A∗∩B)} \rightarrow \frac{A∗∩B∗}{D}$

Define ϕ:A(A∗∩B∗)→(A∗∩B∗)/D by ϕ:ax→xD, where a∈A and x∈A∗∩B∗. Now ϕ is well-defined: if ax=a'x', where ax=a′x′, where a′∈A and x′∈A∗∩B∗, then (a′)−1a=x′x−1∈A∩(A∗∩B∗)=A∩B∗⊆D. Also, ϕ is a homomorphism: axa′x′=a′′xx′, where a′′=a(xa′x−1)∈A (because A◃A∗), and so ϕ(axa′x′)=ϕ(a′′xx′)=xx′D=ϕ(ax)ϕ(a′x′). It is routine to check that ϕ is surjective and that kerϕ=A(A∗∩B). The First Isomorphism Thoerem completes the proof.

Hope it help you.

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