matt grime

c0= c(0+0) as usual

Given that G has two distinct normal subgroups, there are two decomp series 1<N<G 1<M<G by jordan holder it follows that {N,G/N} = {M,G/M} implying that G/M=N, and G/N=M that is G= M(semi-direct product)N and G =N(semidirect product)M i've not thought if the conclusion thus semidirect product must be direct, follows easily or not (i'm not really that much into my groups more their representations) actually it does follow easily, doesn't it.

And r5 etc are? I'll guess the number of Sylow5 supgroups and so on. Usually one looks at the action of G by conjugation on the sylow subgroups has no fixed points to show there is no normal Sylow subgroup (does that imply there are no normal subgroups at all?) Perhaps you mean show it is not simple? Or that the non-commutative groups of order 540 are simple.

Have you looked up the probability stuff? Seriously, this is a large area of research over the last 30 years.

Then saying "why they are" was a bad choice of words.

I don't have time to check that link, but, ed84c, like I said mathematics has been used to model these things. Check out Markov chains and queue processes. And chaos still doesn't explain why weather patterns are the way they are. Chaotic systems display topological trnasitivity and sensitive dependence on initial conditions and are a purely mathematical definition. It doesn't explain why the patterns come as they do it explains why we can't predict them accurately, it explains properties about the model of the weather. Coffee table mathematics books have a lot to answe for.

suppose that T were a submodule of M. pi(T) is a submodule of S, hence it is 0 or S, which implies T is either alpha(S) or M.

Chaos theory doesn't explain weather patterns. Models of weather patterns are chaotic systems. Transpiration, convection, and so on explain weather patterns. Just as the fact that too much traffic in too small a road system explains congestion. Mathematics has been used to model congestion, and has been used to create ways to control flows and so on, but it isn't causal. You may wish to learn about queue theory, markov chains and so on to understand the techniques involved.

If f and g are two differentiable functions then d/dx{f(x)g(x)} = f'(x)g(x)+f(x)g'(x) we also know that integrating "undoes" differentiating, so we get f(x)g(x) = int ( f'g+fg')dx or, rearranging fg - int f'gdx = int fg'dx so if we spot an integral of the form int uvdx, and let u=f, g'=v then we can use this identity ot replace it with anothe, hopefully simpler integral and an evaluation of fg. here log(4x) = log4 +logx, so we just want to know how to integrate logx let f=logx, g'=1, then int logx dx = xlogx - int x*(1/x)dx = xlogx -x

The characteristic is always a prime (if it's non-zero), you don't need to state that, so they don't. You need odd characteristic because squaring in charactersitc two does odd things.: (X+r)^"=X^2+r^2

At least that's one thing cleared up then. (I certainly have neither time, experience, nor inclination to give you a full assesment of it. Try asking in some place where the readership may contain a combinatorialist who cares about the 4-colour thorem. sci.math.research for instance. You almost certainly won't get any proper evaluation from this site.)

your idea appears to be look at some node, recolour, proceed. However, you do not appear to show that this can be done in a consistent way that terminates with 4 colours. it is up to you to explain where I am wrong in this assesment, which i may be.

Did you rearrange the equation and think about how many prime factors are on each side if N is not a perfect square? On one side it's an even number, and on the other it's odd.

The answers all follow with a littel thught and care about the definitions,. Try and think about them a little harder. Though 4 makes no sense. 2 is very easy: if a is of order pq then a^q is of order..... 5 is standard and can be found in any galois theory text., for 6 one needs only to mention the word euler and possibly totient...

The flaw is simple. the idea is that given some node where there are intetsections, that you can recolour the segments in a constistent fashionl the constistency you claim isn't in anyway proven to be consistent. it is an obvious observation, You might wish to prove tha your method shows that a some concrete diagram can be recoloured in 4 colours, HINT: you proof isn't valid in showing that nay arbitrary diagram can be recloured.

the reasons there are no solutions are quite trivial, and i suggest you try and figure them out yourself: rearrajngee and think about prime decomposition.

no, there are no solutions, in fact N can be any non-perfect square and there are no solutions.

The obvious flaw is that you do not show that your recolouring can be done in a consistent way without colouring two touching segments with the colour.

please forgive me if i suggest that you go away and check that. since a quick inspection tells me that the points of intersection you claim to have foudn for the two curves aren't correct either. 6/x = 2x-x^2 and you're claiming (-3, -2) is one of them.... let's check lhs is even, rhs is odd.... nope, that working. and (2,3), well, lhs is 3 at x=2. granted, but the rhs is zero when x=2 you need to solve x^3-2x+6=0 which has 3 roots (possibly with repeats, but I doubt it considering its derivative)

you need to find the three points where those two curves meet. then you know three points that satisfy y=ax^2+bx+c, and thus you can find a,b,c

sort of, every metric gives a topology on the space. a topological space is a generalization of the ideas of continuity to spaces where distance might not make sense.

the discrete topology is one in which all sets are declared to be "open", though if you don't know what a topology on a space is this is meaningless. i don't propose explaining it here since there are lots of references out there that can explain it better than an off the cuff one here and now. Anyway, here is a continuous bijection that is not invertible: let S be the subset of R given by [-1,0]u(1,2] let f(x)=x if x is in [-1,0] and x-1 if x is in (1,2] then this map is continuous but not with continuous inverse onto the interval [-1,1]

How abstract do you wish to be? It may help if you recall that a function is continuous iff the inverse image of an open set is open. Thus all you need to do is find an invertible continuous function that sends an open set to a closed set. Of course if you're allowed to play around with the topology then it's very easy. Let S be any set with the discrete topology, and let S' be a copy of S with any other topology, then the identity map from S to S' is continuous (the inverse image of an open set is open) but its inverse is not continous, otherwise S' would have the discrete topology.

In all honesty it is far better to do it with integer factorizations: n(3n-4) = 3*5 though in this case you need to consider -3*-5 as an option too. I say this because this techique allows you to solve equations like: find all primes, p such that p+9=n^2 where n is an integer, and we do not even need to consider anything as high faluting as a formula.

Let n be an integer such that 3n^2 - 4n=15, find n. Can be done in two ways. one you can use the quadratic formula, but better, factor the left and right and conclude something using uniqueness of prime factorization