Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
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
add comment

1 Answer

up vote 1 down vote accepted

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 $G=B_n$.

share|improve this answer
    
Right. So it was obvious. I didn't read it anywhere in this manner and I guess that was due to your point. Thanks. –  Hamurabi Jun 9 '12 at 13:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.