Pure Math General

[Anonymous No.4680279]

Pure math general. Ask stupid questions! Win girls' hearts with your prowess in analysis! Convince each other that we will be not be unemployed after we finish school! Brag about your superior intellect!

To start off with a problem for people to attempt. Feel free to ignore it:
Show that the group <span class="math">\langle x^{-1}yxy^{-1}xy = yx^{-1}yx[/spoiler] is nonabelian.
I'll give a hint of necessary.

[Anonymous No.4680293]

I'm going to start University in the autum (fall), I'm really intrested in groups, especially galois and the idea of 5< degree polynomials not always having expressible roots. To this end, ive been trying to teach myself group theory, i already have a good grasp of sets and formal logic, but im finding it hard to visualise/grasp exactly what a group is and how it is used/ manipulated, care to explain the basics of group theory?
also, when would i expect to study group theory, post grad id assume
cheers

[Anonymous No.4680300]

Can't you simplify that to xxy=yy since the positive and negative powers cancel out?

[Anonymous No.4680308]

How do I get more interested in Math? I already study it.

I don't hate it but I don't love it or anything. I like it when I solve questions easily and do great on tests but I hate it when I am frustrated because I can't solve the fucking problem.

I want to have good math skills because it's so important for a lot of university majors that are important.

[Anonymous No.4680309]

>>4680293
Group theory is a part of a standard abstract algebra course. Galois theory is slightly more advanced and may be missing from your undergrad algebra series, but it is basic enough that most algebra classes will go over it by the end of the course.

A group itself is just an abstract object defined by a few simple axioms. They can be visualized, however, by picturing what are called "group actions". A Rubik's cube is one of the more common examples. You can do nothing to it (the identity action), and any move you can do, you can do the opposite to move it back (so every "element" has an inverse). Thus, rotating slices of a rubik's cube can be imagined as applying elements of a group to it.

[Anonymous No.4680319]

>>4680300
xy=y^2 can be shown if the group is abelian. Which it is not.

>>4680308
Read about things you are interested in? There are a lot of ideas in math that are interesting, but not taught in "general" math courses. Godel's theorems, cantors ordinals/cardinals, halting problem, etc.

[Anonymous No.4680329]

>>4680293
>post grad id assume
try your second year if you don't go to a shit tier uni.

A group is a set of things, anything really, that is:

Closed under an operation, such as multiplication. Meaning, if a and b are in a group, a*b is in a group.

Associative.
a(bc) = (ab)c

Has an identity (that is in the group)
a*I=a

and an inverse (that is in the group):
a*a'=I

That's it.

[Anonymous No.4680333]

>>4680309
I understand that there are certain operations, I mapped out the caley table for D3 (the rotational symmetries for a square i think? tho it was a while ago) and i understand Z groups as analogous to clock faces ie. mod(n). But how exactly are they useful? are they like sets but for groups of specific operations instead of specific (or uncountable w/e) number like sets. such that you can say that for example, a certain group of operations over a certain interval would produce an output that can be usefully described, like set builder notation? I'm sure im missing bits as always happens with autodidactism, but i really want to grasp this, What do you reccomend i learn in order to grasp this? Im getting into real analysis atm, right direction? :(
also any free sources for learning about abstract algebra is anyone has any would be great

[Anonymous No.4680345]

>>4680333
They are plenty useful, but as with a lot of abstract algebra, most of the uses require a decent amount of sophistication.

One natural place where they pop up is when you consider automorphisms. For example, Given any set X, the bijections <span class="math">X \to X[/spoiler] form a group. This is called the symmetric group, and is one of the most important classes of groups.

Another example is Fermats little theorem. If you consider the integers mod p, for a prime p, then the nonzero elements give a multplicative group of size p-1. This immediately gives the result that <span class="math">x^p = x[/spoiler] mod p.

Anyway, real analysis is pretty much the exact wrong direction (although it is important as well). I don't know free sources, but I like Dummit&Foote's algebra text.

[Anonymous No.4680388]

bumpin

[Anonymous No.4680850]

bump

[Anonymous No.4680864]

>>4680279
Can't you start it off by supposing that it is abelian and find a contradiction?

It's been a while since I've taken Group Theory, so yeah. The obvious route(for me, anyway) would be to suppose that it is and find a contradiction somewhere.

[Anonymous No.4680870]

>>4680864
Although that is correct, it doesn't really suggest an approach to the problem.

For a hint, try showing that there is a homomorphism onto a nonabelian group (the hard part, though, is finding that group and homomorphism).
For example, lets say we want to show that the trefoil group <span class="math">G = \langle x,y: x^2 = y^3[/spoiler] is nonabelian. We do this by defining the homomorphism <span class="math">\varphi:G \to S_3[/spoiler] given by <span class="math">x \mapsto (1,2),\quad y \mapsto (1,2,3)[/spoiler]. Then it is easily checked that this is a well defined surjective homomorphism onto the nonabelian group <span class="math">S_3[/spoiler], so the group <span class="math">G[/spoiler] is also nonabelian.

[Anonymous No.4680913]

>>4680870
Using the approach you suggested.
Let us denote <span class="math">G=\langle x,y: x^{-1}yxy^{-1}xy=yx^{-1}yx \rangle [/spoiler] and <span class="math">D_5[/spoiler] is the 10-element dihedral group, which is represented by the symmetries of a regular pentagon <span class="math">ABCDE[/spoiler]. Further <span class="math">r[/spoiler] and <span class="math">s[/spoiler] are the reflections through <span class="math">A[/spoiler] and <span class="math">B[/spoiler], respectively.
Then the homomorphism <span class="math">\varphi: G\to D_5[/spoiler] is defined by <span class="math">x\to r[/spoiler] and <span class="math">y\to s[/spoiler], which is a well-defined surjective homomorphism from <span class="math">G[/spoiler] onto the non-abelian <span class="math">D_5[/spoiler], so G is non-abelian.

[Anonymous No.4680935]

>>4680913
Almost, but I think you made a small mistake somewhere. The relationship given gives something like <span class="math">sr^4 = r^2[/spoiler] if you just plug r and s in, which clearly isn't right.

[Anonymous No.4680940]

OP, did you say you were a girl?
I'm into pure math. I can provide a proof if you want.

[Anonymous No.4680951]

>>4680935
I think it is correct, <span class="math">r^{-1}=r[/spoiler], <span class="math">s^{-1}=s[/spoiler]. Thus we need is <span class="math">rsrsrs=srsr[/spoiler], that is <span class="math">(rs)^5=1[/spoiler], which is true.

[Anonymous No.4680963]

>>4680300
He specifically stated that the group is non-abelian

[Anonymous No.4680998]

>>4680951
I see. Usually <span class="math">r[/spoiler] is taken to mean the rotation, so <span class="math">r^5 = 1[/spoiler]. However, the correct answer is taking <span class="math">x \mapsto rs[/spoiler] and <span class="math">y \mapsto s[/spoiler], or something similar with that notation. If r is interpreted as a reflection then it that answer is indeed correct.

[Anonymous No.4681000]

>>4680998
Yup, looking back I missed that he said they were both reflections.

Good job then. Did you do what I did and add the condition <span class="math">x^2 = y^2 = 1[/spoiler] to find that it mapped to <span class="math">D_5[/spoiler]?

[Anonymous No.4681006]

>>4680998
http://www.youtube.com/watch?v=4rnsmkLUOsw

[Anonymous No.4681008]

>>4681000
Yes, that's exactly what I did. I concluded that <span class="math">(rs)^5=1[/spoiler] is needed in that case, so the simplest choice was <span class="math">D_5[/spoiler].

[Anonymous No.4681009]

<div class="math">\frac{\mathrm{d} \phi}{\mathrm{d} s}=\kappa =\sqrt{x(s)^2+y(s)^2}</div>Prove that, given <span class="math">x(0)=y(0)=y'(0)=0[/spoiler], there exists a positive real number <span class="math">B[/spoiler] such that <span class="math">x(B)=y(B)=y'(B)=0[/spoiler]. (ie, that the curve (whose curvature at a point is equal to the point's distance from the origin and which crosses the origin) is periodic for some period B)

[Anonymous No.4681012]

I'm new to some of the notation in abstract algebra.

So OP, in what sense has a group actually been defined here? Are we to assume the existence of a group generated by x and y? That is, do we assume all product combinations of x and y are well-defined, that all such combinations have inverses, etc?

It seems like the equation you provided just gives us an extra assertion about the way that group (the one generated by x and y) would act in this problem, if we were allowed to assume its existence in the first place. And then, starting from that equation, by some kind of manipulation and observation, we would generate an example where A x B does not equal B x A...

Sorry if this whole post sounds retarded.

[Anonymous No.4681020]

>>4681012
It is a group presentation:
http://en.wikipedia.org/wiki/Presentation_of_a_group

Basically, it is the quotient of the free group on the given generators (left side) quotiented by the smallest normal subgroup containing the relations (left side). Alternatively, it is the "most general" group with 2 generators satisfying the given relations.

For any presentation such a group always exists. However, the group may have to be the trivial group. For example, the presentation <span class="math">\langle x,y,z : x = y^2 = y^3 = z = xz \rangle[/spoiler] is legal, but the only group satisfying those conditions is the group with one element.

[Anonymous No.4681062]

>>4681020
Interesting, thanks.

>>4681009
>the curve (whose curvature at a point is equal to the point's distance from the origin and which crosses the origin
Now, why are we allowed to assume such a curve or family of curves exists? Is that just given in the problem? Are we supposed to assume anything about the curve being "nice" ? Ever since Weierstrass' stuff, I don't know what to believe anymore about badly-behaved functions.

[Anonymous No.4681083]

The best I can figure so far on this curve problem is that, if the curve isn't some exotic fractal craziness, then the curving has to continue in the same direction once it has gotten started (to prevent a non-origin point of zero curvature), and the farther away it gets from the origin, the more "turned back around" it has to get and face the origin again and come back toward it.

[Anonymous No.4681089]

Does anyone know any algebraic geometry? Do you ever get any intuition about this shit? I'm trying to learn the basics and its really freaking hard mostly because I can't seem to get my head around the basic concepts. Schemes are really confusing. Maybe its my lack of algebra background (I really study physics, not math).

[Anonymous No.4681092]

>>4681020

Then affairs seem rather easy. Assume the group were Abelian. Then all x and y and their inverses would commute, leaving xy = yy or x = y, implying the group were trivial, in contradiction to x and y being two generators.

[Anonymous No.4681094]

>>4681083
Yes, I forgot to mention that the curve is given to be smooth. Symmetry provides that the curve returns to the origin and the curvature of the curve provides that it has zero curvature at the origin. The real question is how the curve approaches the origin. If it does so by an irrational rotation, then it cannot be periodic. Otherwise it must be. There is also a more general case of the problem but I think it is involved in the proof of this one...

[Anonymous No.4681107]

>>4681094
>The real question is how the curve approaches the origin. If it does so by an irrational rotation, then it cannot be periodic
Excellent precise point, I hadn't even thought of that. If the re-approach is rotated by an irrational fraction of 2pi, then the curve can't be periodic and will probably end up getting arbitrarily close to any point you want within some certain distance from the origin... How fun.

[Anonymous No.4682482]

a + (b + c) = (a + b) + c = (a + c) + b