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From group theory we know that a homomorphism $\phi: G \to \operatorname{Sym}(S)$, where S is a set, then $\operatorname{Sym}(S) \cong \Sigma_n $. Its kernel is given as $\bigcap_{s \in S}G_s$, which is the intersection of all the stabilizer subgroups of $G$.

Now, the pure braid group $P_n$ is defined as the kernel of the map $\phi: B_n \to \Sigma_n$. Is it fair to say, that the pure braid group is such an intersection?

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    Yes. The first paragraph is correctly rephrased like this. Sorry, if I didnt make it clear enough.2012-06-09

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Assuming my comment is correct, there is no restriction in the hypothesis on the group $G$. The statement is true for any $G$ at all. In particular it is true for $G=B_n$.

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    Right. So it was obvious. I didn't read it anywhere in this manner and I guess that was due to your point. Thanks.2012-06-09