Math Is Fun Forum

Well you guys are no fun! Here.

Let

be a right inverse. We know that ae = a. Note first that

.

by associativity

This suggests that last term is a right inverse

But

by assumption, so I have

TraLa!

No problem, math is fun, right?

But why should I? Where did it come from, other than the taylor series I showed?

Why? It could be, by your reasoning, that

Anyway, let's not fall out in public. Others seem to have lost interest anyway.

Yes you are right. Why didn't I spot that? However, we may not assume that, if e is a right (resp. left) identity then it is also a left (resp. right) identity.

However, the proof is trivial. Let ab = c and ae = a. Then, because we insist on associativity of the group operation, we have that  (ae)b = a(eb) = c, which implies that eb = b, so e is also a left identity.

The proof that inverses are identically left and right is a bit more tedious, but equally trivial. If anyone wants a crack, remember that (a^-1)^-1 = a.

Whence the even famouser

.

The 5 fundamental transcendentals, all in bed together (but no action!!)

Umm. Maybe we should agree what we mean by an ideal in general terms? Like in any ring?

But you're being really naughty here. What for the love of all that's holy is F*? Where do a and b come from? What do you by a unit? You really must define your terms. I'm not just being nit-picky here, I truly have no idea what you're talking about

Yes,
OK, I see I need to build from the bottom. I am happy to try that.
Ricky, George - waddya think? Where do I start? Sets? Define abstract
algebra? Any way this to Ricky

I hope you have solved this because I'm really rusty on this, so I'm
sort of feeling my way. But I suspect
you need to take advantage of the fact that every PID is also a UFD.
But I'm a bit confused by your notation. The way I learned it, F[x,y]
means a polynomial ring in 2 variables, x and y, with coefficients
taken from the ring F. But you have <x,y> generated by {xg + yh},
g and h in F[x, y]. I don't quite see what's going on here. Please tell.

Anyway, this is as close as I can get. Let F be a field and F[x, y] be the ring of polynomials over. Let <x,y> be generated by (xg + yh).
The ideal I is generated by the polynomials f(xh + yg) = x and
k(xh + yg) = y. Since no element of F[x, y] divides both f and g except for elements of the field F, and elements of the field are not contained in the ideal, the ideal cannot be principal.

And I tell you now, I am not in the slightest convinced by that. What did you you finally get?

Certainly not! Here's your homework.

Is 1/3 rational or not?

Is 1/3 + 1/3 +1/3 equal to 1?

Is 1 rational or not?

What is the decimal represenation of 1/3?

Of 1/3 + 1/3 + 1/3?

See where it goes?

Well,
first I hadn't intended by question as a challange, I wanted to know
for myself. I've got it figured now.
Second, Ricky, I can't see what you did there. You assumed the result I
wanted to see shown in order to find what I wasn't looking for! Or have
I missed something?
Anyway, with the help of some textbook, I see it.
Note first that z = x + iy here. Note also that e^z is analytic
throughout the complex plane: it is an entire function. We expect there
to be a taylor series therefore.

Recall that the taylors for the circular trigs are

and with

applied to the above gives

which reduces to

We now want to let x = z = x + iy. Notice that even powers of i are all negative 1 or i, odd powers are positive 1 or i, alternating.
Remembering that

we will easily find that

gathering terms

which is merely to say that

It then follows that

All I need do now is remind you the expansion for e^-x is

and you should easily see (by comparing the appropriate taylors) that

And as we know e^z is an entire function, we can assume the same is true here.

Now that's something that really does deserve to be called cool!

Ricky, as you know I'm new here, so I may have mistaken the level of my pitch. Overpitched, you think? Your intro to symmetric groups was a lot more elementary than mine (nonetheless quite correct). I'm sure other members are grateful for that.

Nevertheless, I shall continue on my course. Remember, I am quite happy to expand, explain, illustrate and generally dance the fan-dango.

After Ricky's nice concrete introduction to groups, let me burden you all with a degree of abstraction. First let's have a definition:

A group G is a set which has

a well-defined, associative and closed operation
an identity e
an inverse.

Ricky covered all this, so let's unpack it a little. In group theory, the group operation is always referred to as "group multiplication". But don't be fooled: group theorist don't always mean multplication in the arithmetic sense; it might mean addition (they mean something like "a multiplicity of group elements") There is a convention which I'll expailn later.

The operation, group multiplication, is said to be "closed" if the result of the operation is still an element of the group (are the primes then a group?).

The identity e is as Ricky described: 0 when the operation is addition, 1 when it is multiplication.

The inverse is again as Ricky said, respectively -x and 1/x.

Now some notational conventions. Some authors use the centre dot to denote the group operation thus a·b, but most use juxtaposition thus ab. Don't assume this means arithmetic product, though, we might be in an additive group.

Similarly most authors use x^-1 for the inverse, to be interpreted according to the operation in question

With this and the previous posts in mind, we see that if a, b are in G, ab = c implies c is in G

We also see that ae = a , and aa^-1 = e. Note that these last equalities refer to what are called right identity and inverse, respectively. The left identity and inverse follow follow from the group axioms. (Any body want to try the simple proof?)

Now, perhaps the most important distinction we can make between different sorts of groups is this: if the operation on the elements h, g of a group G is commutative (i.e. gh = hg) the group is said to be abelian, otherwise non-abelian. I referred earlier to a notational convention: for abelian groups the operation is said to be addtion.

Now non-commutivity need not throw us into a panic - think matrix multiplication for example, but the notation can be misleading at first. ab = c need not imply that ba = c, but it is always the case that ab = c means a = cb^-1 = b^-1c, by my axioms above.

One more word on commutivity; it, or the lack of it, is a property of the group elements, not of the operation (associativity would not follow otherwise).

Now there is another, even more abstract, way to describe a group, which I think is super cool: a group G is a set of sets, these being the underlyng set written |G|, the Cartesian set |G|×|G|×|G|....(this is where our operation lives) and an axiom set (the set of rules that tell us about identity, inverse etc.)

Roughly speaking, the order of a group G is the number of elements in the underlying set |G|. There's an interesting theorem of my buddy Lagrange, but first I need to tell you this.

Consider the groups G and H. If every element of H is also an element of G clearly G = H iff every element of G is also an element of H. But if there are are elements in G not in H we say H is a subgroup of G. As H is a group in its own right it must share the identity e with G, must have an inverse and must be closed under the group operation. Evidently the operation on G and H must be the same.

OK. Lagrange says that the order of any subgroup H of G divides the order of G. This is less easy to prove than it looks, but it's a really cool result.

I'm going to close (for now) with some examples of groups and things which aren't groups:

Z, the integers, is an additive abelian  group

The even integers are an additive abelian group; the odds are not a group, neither are the primes

S_n, the set of permutations on n objects is a group, non-abelian for n > 2

The set of rotations in 3-space are a non-abelian group.

Tired of typing, more another time if anyone wants (we've scarecly started, it just keeps getting better!) But I am happy to answer questions, even happier to be corrected!