Math Is Fun Forum

Then I don't understand what you mean by practical applications. In my mind, a unifying theory that brings all sorts of groups together, in some sense, is of immense value. Most of the really important stuff in math that I can bring to mind falls into that category.

Maybe you mean, Cayley has no applications in physics?

In the hope that Ricky's introduction to groups on the main page may have stimulated some interest in this subject, I'll plod on (not hopefully, though).

I ended my last with the remark that groups can talk to each other.
That was, of course, a bit of hyperbole.. So what did I mean? Here....
To each group G we can associate its underlying set, which I earlier
denoted |G|. Now we know that functions on sets take elements of
one set to elements of another thusly: f: X → Y. Let's quickly remind ourselves about such functions. (By the way, I use the terms function and map interchangably, I hope that's OK with you)

If a function f: X → Y is such that for every y in Y there is at
least one x in X such that f(x) = y, we say that f is surjective
(or onto). And if for each y in Y there is at most one x in X
for which f(x) = y we say f is injective (or one-to-one). Just note that for a surjection there are no elements in the codomain which are not mapped onto by element(s) of the domain, while for an injection there are no elements in the domain which share an image in the codomain

Now, given a function on |G|, we may have that h: |G| → |H|, where |H| is the underlying set of some group H. But groups carry a notion of structure, built on the allowable group operation and the axioms which provide the "rules of manipulation" of group objects. So we require our function h to be rather more that a function on sets.

Let's have now f: G → H. If f preserves the operational relationship
between elements of G and H and is surjective, it is called a
homomorphism. Specifically, let the operation on G be × and that on H be +, we will have f(g × g) = f(g) + f(g) for all g in G and all f(g)
in H. Here's a famliar example. log(xy) = log(x) + log(y). This is in fact an example of an injective function, that is if f: G → H is injective f is said to be an isomorphism: a one-to-one, operation-preserving map.

OK. (You at the back, are you paying attention?) One of the most important theorems in group theory follows. Remember that both Ricky and I talked about the permutation (a.k.a the symmetric) groups? Some cool fella called Cayley came up with this:

Every group of order n is isomorphic to a subgroup of the symmetric group S_n (order n!)

The proof is straightforward, but rather messy. Anyway, think about it in the conext of the theorem of Lagrange that I gave earlier. I'll show it if you want.

Umm. What's going on here? I thought this section was supposed to be about mathematically cool facts we have discovered?

By all means chat about drunks vs moby's, but surely this is the wrong place to do it.

Hey! I am also a very keen amateur musician. Is there a way of recording directly onto my hard drive? My computor has a "mic" socket. Do I need special software?

Oh dear

Ah, but there's a neat trick here.

Let the group 2 = {0, 1}, and let P: X → 2 be the parity map on X.  Define P: x |→ 0 if x is odd, P: x |→ 1 if x is even. Now {±1} is isomorphic to the group 2 = {0, 1} so we may replace, say 0 with -1 and 1 with 1. Thus odd/even and -/+ are the same thing, up to isomorphism, as the usual parity rules of arithmetic easily show

Yes, but Mikau's original point was about imaginary numbers, not complex numbers

I don't think I quite agree, but you may be right. But note the two following bits of trivia (just for fun):

.........

see the pattern?

Note also that the 4th power of all of these guys is +1!!. Is this true of all integer powers of i? Of course!

That's where I left off last time. Let me continue (even though there
doesn't seem a lot of interest. Do you all know this stuff already? Why no challenges? Too easy? Huh?)

Using the notation in my self-quote, if g ~ h ~ k, we may form an
equivalence class {g, h, k}, for which we can elect a class
representative, say g. Then we write that class as [g], it being
understood that g, h and k are in [g].

Note that, by the property of transitivity above, any element g of G
can only be in one class: if g is in [g] and g is in [j], then [g] =
[j]. This is thus a partition of G (strictly of the underlying set |G|) into
disjoint subsets. Here's the nifty bit. Ah no, wait, let's get concrete
(but remember the caveat I gave when introducing the permutation
groups).

I talked about the normal subgroup 3Z of Z, that being the
elements of Z exactly divisible by 3. We saw (at least I did!) that 1 ~
4 ~ 7 (for 1 + 6 = 4 + 3......) and {2 ~ 5 ~ 8, and we (!) noted that
not only did elements of 3Z differ by a factor of 3, so did the
elements of the sets {1, 4, 7...} and 2, 5, 8....}. In fact I think I
presented that the wrong way round, but you get the picture.

So we have 3 equivalence classes: [1], [2] and [3]. But wait! The
subgroup 3Z has as elements {...-6, -3, 0, 3, 6,...} so let's run the
election for class representative again: [0], [1] and [2], let's say.
So if we gather these guys together we have an identity in [0]. We also know that any member of [1] plus any member of [2] is a member of [3] etc. so we have a closed operation too. Inverses are similarly easily checked.

In other words, the collection of equivalence classes is itself a group
which satisfies all the usual group axioms. Its called the quotient
group. It is written G/~. That's the nifty bit I referred to earlier.

To sum up. For any group G, the set comprising a normal subgroup H of G together with its cosets forms a group, the quotient group G/~. This is an incredibly important construction, and not only in group theory.

Got to run now, but next up (if you want it): groups talk to each other! Wow!

Sorry, I don't mean to be rude, but you do need to be expert on the complexes to even get a hint at what the Riemann hypothesis is all about.

I'm not going to pursue this, because I detest number theory, but maybe this will help you.

Years and years ago, mathematicians stumbled upon a neat trick. It was well known in the 17th centuary that you cannot define the square root of a negative number. But these guys found that if you "pretended" such a thing exists, held you breath and plodded on with your calculation regardless, everything came out as expected in the end.

This led to the introduction of the imaginary unit i, that
being defined as follows: it is always the case that i² = -1.
(Therefore, equivalently i can be defined as √-1, but it's best not to start there)

Now a complex number is simply a number which has two parts, real and imaginary (I assume we all know what a real number is). It is written thus: z = x + iy, where x and y are real numbers.

Perhaps confusingly, x is called the real part Re(z) and y the imaginary part Im(z). For the most part, compex numbers can be manipulated just like real numbers, as long as we remember to replace i² with -1 wherever it appears. However, division requires some work, but it is well worth it, as it throws us a very useful gadget. Let me explain.

First, take it from me that multiplication works as follows:

as is easily checked.
It is less easy to derive the following:

.
This suggests that

.
Now c² + d² = (c + id)(c - id), so that, for
any z = x + iy, one says there is a complex conjugate z* = x
- iy such that zz* = x² + y²
So, now we have the complex conjugate in our possession, we can do
this. Note that z = x + iy can be written z = Re(z) + Im(z),
which suggests we can think of z as a point on a plane with coordinates
Re(z) and Im(z). Moreover, the existence of the complex
conjugate allows us to think of this plane as a vector space with a
notion of distance between points - a so-called metric space. Why is
this?

First, note that in general, for two complex numbers z ≠ w we
will have zw ≠ wz, as is easily shown. But it is the case that
(zw) = (wz)*. Translated into the language of vector spaces we
will say that the inner product (w·z) = (z·w)*
Now recall that in a real metric space, Pythagoras says that the
distance between two points a and b is given by √(a² +
b²). This implies that the "length" of a one-dimensional vector x is found by
√|x²|. But suppose we had no complex conjugate, then on the
complex plane we would have √(ix²) = √-x², an imaginary
length! But if we may have a x* for each x, we may have
√(x·x*) as the "length" of this vector in the complex plane.

I don't know why I made that detour, just general education, I guess. But as has been pointed out to you, unless z is an even,  negative real, the zeros of Riemann's zeta function occur (by hypothesis) when Re(z) = ½. That's all I know, or care about the subject

I've run out of puff for now, let me know if you need more

My poor young friend, you have a very warped view of the way the academic world works. For your information, it is not a market place. If you really have something (my doubts are even stronger that Ricky's) the custom among academics is to share and be generously congratulated by those with the ability to recognize the worth of your work.

This may be an acceptable motivation in your street, but I can absolutely assure you that you would have to travel a long way to find an academic who shared it.

This is just silly. The 3 problems you claim to be solving (Fermat, Golbach and Riemann) are all problems in number theory. Do you really think that proofs in number theory carry over to, say topology, manifold theory, Lie algebras, differential forms, exterior algebra, Clifford algebra...? Because they don't.

Hmm. To do it justice would require a detour into the theory of abstract linear vector spaces. Let me just say this.

In a nice, homely 3-dimensional vector space we will say that the inner product x·y can be defined as |x| |y| cos θ. We say that x and y are orthogonal when x·y = 0 which implies that θ = 90°. And as John E. correctly said, this simply means that x and y have zero projection along each other.

But our most familiar vector spaces are infinite-dimensional, which makes the issue of mutual perpendicularity problematical. So we turn the deinition of orthogonal round, and say two vectors are orthogonal if they have zero mutual projection, and x·y = 0 defines this condition.

Note that this need not imply they are mutually perpendicular. To see this consider two vectors which are not perpendicular. Suppose x has some projection along y. Then x·y ≠ 0. But if we subtract from y the projection of x, we find that x·y = 0, but we haven't rotated any vector, so they are orthogonal but not perpendicular!

I hope that is clear.

I just checked on this

for real x, cosh x = cos ix, and not as you wrote (how do you get thetas and stuff in here, without using LaTex?)

Euler's Attic! First it is alliterative (as long as you pronounce "Euler" correctly).  Second it conveys a sense of something higher. Lastly, it suggests it contains stuff that Euler used, or may have used, at some time.