Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Let me just check I've got this straight. "Left and right cancellable" should mean something like "is injective and surjective" and "invertible" should mean "possesses an inverse". Right?
I don't understand this question. Are you asking whether there's a non-orientable handlebody bounded by N_3? If by N_3 you mean the non-orientable surface of Euler characteristic -2, the answer is yes. (Just take a solid Klein bottle and attach a 1-handle.) Please clarify.
Right. I assumed from the "simple" flavour of your question that you weren't interested in answering your question by looking for infinite quotient/finite overgroups (which is the usual way of approaching these things). Another possibility is that if your kernel satisfies some sort of "small cancellation" condition then you can sometimes prove that the quotient is infinite. But I've no idea how you make that work on a Coxeter group.
What do you mean by "for many covers of figure-eight"? It seems clear that it's true for all finite covers of the wedge of two circles. (The number of edges in the maximal tree is m-1.)
Could you make it clear which part of this is actually a question, please? More question marks might help! For instance, "Or the traditional way of browsing periodical section of your library is still a better way to get a glimpse of current development in math." reads like a statement, although I think it's intended as a question.
Your "equiconjugate" condition is sometimes called a Frattini embedding: H is said to be Frattini embedded in G. (Though the Frattini subgroup is something completely different, confusingly.)
I don't understand what you're getting at with 1. I don't see any reason to think that all characteristic subgroups can be described using definable subsets; and characteristic subgroups don't behave very well with respect to direct products, free products, quotients etc.
SixWingedSeraph, I admit that I find that solution unproblematic. I shouldn't have been so hasty to rule diagrams out of sentences altogether! "The commutative diagram <diagram> has no punctuation" would have been a nice way of dealing with Simon's last example. BTW: I accidentally attributed Simon's example to Sam above. Sorry!
