matt grime

S_n is the permutation group on n objects. It has n! elements. A_n is the subgroup of even elements. If you don't know what that is then you need to learn it, from your notes and understand why it has n!/2 elements. You also should immediately see that there are two cosets of A_n in S_n and thus that A_n is normal so the quotient group exists and has order.....?

They arent' differentials either. Differentials are something else beyond the topic of this question.

a+ar^2+ar^3+ar^4+ar^5 = 9(a+ar+ar^2) divide by a use the sum formla (r^6-1)/(r-1)=3(r^3-1)/(r-1) cancel the r-1 let y=r^3 y^2-1=3y-3 solve

Singularity is very important, and shouldn't be gotten rid of - you know its geometric meaning. Then there are also differential considerations (top volume forms).

Here are some other important facts that show why S_3 amd Q and A5 are important. p, q, r will be distinct primes. Every group of order p is cyclic - hence C2xC2 is the first possible non-cyclic group. Every group of order p^2 is abelian, hence S_3 is the first possible non-abelian group. Combining these two, Q is the first possible non-abelian p-group. Every group of order p^a.q^b is solvable. (Burnside) Every group of order pqr is solvable. Hence 60 is the smallest possible order for a simple group. A5 is simple and of order 60. What is amazing is that the smallest order for a counter example actually has a counter example. Well, I think that's quite astounding. Q gives us the quaternionic algebra. This is the first division algebra that isn't a field you will meet, too.

What are the orders of An and Sn? Hence what is the order of Sn/An? How many groups are there of that order?

Yep, that's what F_p is. Another way of creating an infinte field of characteristic p is to take (p), 'the' maximal ideal in the ring of algebraic integers, A, containing p, and let F= A/(p).

0*0 = 0, as can be deduced from the basic axioms of a ring. The only problems you will come up with are consistency ones, and the philosophical ones, and the sheer "why on earth would you even want to do that" ones....

S_3 is the smallest non-abelian group, that is why it is used - it will often be the simplest counter example. And Q is the smallest non-abelian p-group. Hence it's use. Others are: A_5 smallest simple group. S_6 only permutation group to possess an outer automorphism. Yes, there are infinite fields of non-zero characteristic - the algebraic closure of F_p

No. Even in the extended complex plane you may not multiply infinity by 0. Define what you mean by inverse before you carry on.

Shed load is just a bog standard english phrase. I hope what i got across was that there is nothing special about Z_2, or the fact that these are matrices, really. It's just the standard - defining 1/0 takes a whole lot of effort and is of dubious worth in this algebraic setting (as opposed to the analytic setting, where 1/0 = infinity is a useful thing for the one point compactification of the complex plane).

Eh? No. It's a ring with zero divisors. You cannot divide by zero without introducing a whole shed load of inconsistencies and special cases. Something that posseses multiplicative inverses and still has any hope of being a ring is either a field or a division ring (algebra). In no case may you have zero divisors, and you can never invert zero. You never invert 0 inside the field of reals or complexes (and still havae it remain a field). Your question is exactly the same as asking to do so with R or C or Q instead of Z_2.

In terms of "really, wow, who'd've thought that": All automorphisms of S_n are inner unless n=6. (Why 6, is what should spring to mind). That Dynkin diagrams not only parametrize (simple) Lie algebras, but things about representation types of quivers too. If anyone wants the full explanation let me know. That the outer automorphism just a few lines back is how (some?) of the exceptional simple groups arise. Important ones: Did someone mention Bolzano Weierstrass? That every sequence lying in a closed bounded subset of R^n has a convergent subsequence. Fundamental theorem of Calculus. F(x)= int_0^x f(t)dt => F' = f (assuming f continuous) and its variations.

You might change your mind if you saw the full banach space version of that.

So let's add some more detail: S4 to Z_2 (written additively to match your convention). there's the trivial homomorphisms everything gets sent to 0 suppose now that (12) gets sent to 1. Since (123) must get sent to 0 ((123) has order 3, so its image has order divisible by 3) but (123) = (12)(23) so it must be that (23) gets sent to 1 as well. By similar logic (14) must be sent to 1, and similalry we see all 2-cycles must be sent to 1. This describes the homomorphism: s in S_4 is sent to 1 if it is an odd permutation 0 otherwise.

Matrix inversion injective? Z_2 is a field, look at the usual results about matrices over s field and you'll see that they are independent of the field. The matrices, over any field, with non zero determinant form a group under multiplication. Sorry, I've no references for sliding, but it is a simple observation that since there are only 2^n possible vectors of length n, then once you've worked out how to multiply a given part of the matix by some of the columns you can do the rest by inspection.

1) There is no guarantee that (over the reals, say) a matrix with entries 0 or 1 has an inverse where the entries are even integers, never mind binary digits. Unless you are doing operations mod 2, which you ought to be if you were properly going to call them boolean. If we are operating of Z_2, then there are several speed up techniques for operations, such as "sliding". This is useful for dealing with matrix representations of very large groups such as the monster where a single matrix fills up a single floppy disk, and there are as many matrices as there are bytes on that disk. 2) Cannot possibly hold by considering the linear spans of a columns/rows. In particular given an nxm, A, matrix with n>m, there will never be more than m linearly indendent vectors in the image, and the best you can do is find an mxn, B, matrix such that AB has a block diagonal decomposition as an mxm identitiy matrix and an (n-m)x(n-m) null matrix.

Actually I was wondering if you were about to use p-adics.... don't forget that sqrt(2) could be the limit of: [math] \frac{\text{floor}(10^n \sqrt{2})}{10^n}[/math]

Shame there's no such thing as an infinite integer then isn't it?

You do know that transpositions generate Sn? They must get sent to elements of order 1 or 2. so....

No. Let F(x) and F(y) be any two points in F©. each point in the segment l(t)=x+t(y-x) is in C, so l'(t)=F(l(t)) is in F©, and is a straight line, by the assumption F is affine, and l'(0)=F(x), l'(1) = F(y) so it is a straight line segment lying wholly in F© with end points as required.

yes, that's it - an affine map sends straight lines to straight lines - the end points are correct and it is a line. I can't see anything else you need to show, though like I say, this (version of affine) isn't my stuff (hence the stupid reference to something else completely useless).

we used convexity of C because we required the line segment from x to y to be in C, and hence its image to be in F© Forget spec, that was me being stupid.

Ok, just subsitute the word point every where where you see the word vector then - was taking as an example R^n which is affine and thinking in those terms. However since an affine map is simply one that preserves lines in some affine space then you see how the proof was completed above? As I said I'd like to clarify what you mean by affine here. Spec(k[x,y,...,z])? Actually forget that, that was silly. Been a while since I thought about these things. Anyway, essentially an affine space is just a vector subspace of a vector space translated along. In essence we can pick an origin and a basis, however for any choice of point as origin in the space we must always get the same results, is that about right?

Erm, I'm not sure what you're getting at - this is linear algebra: a vector isn't "something with magnitude and direction", it's an element of a vector space. v is an element of the vector space, it is the image of 0 under the affine transformation. Point and vector are intrchangable, it just means an element of the vector space.