Pure braid group, stabilizer

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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Can you define $B_n$? –  math-visitor Jun 8 '12 at 11:21
$B_n$ is the braid group. If $C_n$ is the space of unordered n -tuples of distinct points in the complex plane, then the braid group $B_n$ is the fundamental group of $C_n$. Details e.g. here: planetmath.org/BraidGroup.html –  Hamurabi Jun 8 '12 at 13:02
Your first paragraph is hard to read, I cannot tell what is the hypothesis and what is the conclusion. Is it supposed to say "if $\phi : G \to Sym(S)$ is a homomorphism, where $S$ is a set of cardinality $n$ and $Sym(S) \approx \sigma_n$, then the kernel of $\phi$ is given as $\bigcap_{s \in S} G_s$''? –  Lee Mosher Jun 9 '12 at 13:26
Yes. The first paragraph is correctly rephrased like this. Sorry, if I didnt make it clear enough. –  Hamurabi Jun 9 '12 at 13:49

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 $G=B_n$.