will be starting my first lesson on group theory next week what

easy as shit. group theory is one of the first things you learn, when you start studying math

>>8676913

No. Groups are an extremely simple idea. Let's get a head start on your lesson.

A group is two things, put together, which satisfy a few basic properties. A group is a set (call it G), together with ONE binary operation (call it #) which operates on the elements of G. So this group, for example, can be written as (G,#), which is simply a notation which specifies the set and the one binary operation involved. Say you have any three elements a,b,c of G (any or all of which might be equal, or not). Then if the following four properties are satisfied, you have a group:

For all a, b, c in G,

a # b is also in G. Closure. The set is closed under the operation #, you'll never go outside of G by applying # to elements of G.

a # ( b # c ) = ( a # b ) # c . Associativity. When the operation is applied more than once, in some ambiguous order, then it does not matter in what order the instances of the operation are applied (it may easily happen, however, /that the order of the operands is still important, and must not be disturbed/).

There exists some e in G such that a # e = e # a = a. Identity. An identity element relative to the operation #, which leaves all elements unchanged when operated upon in connection with another element.

There exists some a-inverse, or a^-1, such that a # a^-1 = a^-1 # a = e. Inverses.

These four basic properties describe groups (as you could have easily gleaned by instead reading wiki), but depending on the text you read, one or two other properties concerning binary operations may also be explicitly listed. To understand algebraic structures, it is assumed that you have at least a naive understanding of "set" and "operation", and mashing the two together. Still, you are obliged to have a precise mathematical definition for what a binary operation /is/, and this is readily accomplished by mapping the cartesian product of a set with itself, onto that set.

>>8676949
thanks for taking the time to write this!!

I think I understand everything there apart from the associativity bit

If it does not matter what order the instances of the operation are applied, why is it that the operands order are still important and sometimes shouldn't be disturbed?

thanks again buddy

>>8676949

For example, consider "two times negative three equals negative six", or

2 x -3 = -6.

What we have here are three integers. A binary operation is performed on the first two integers, and they give back... a third integer. Put another way, capturing all the same information, we've just sent an ordered pair of integers, to another integer, or we've mapped the former to the latter, which we might write like this:

(2,-3) --> -6.

Notice how this treatment does away with explicitly listing the /arithmetic operation itself/. Still, the mapping is accomplished by multiplying the two components of the pair together, to give the product on the RHS. But the point is that this captures precisely, for our purposes of closed operations and such, what a binary operation /is/: a map from (in this case) Z x Z --> Z, where the "x" here is not to be confused with ordinary multiplication, and instead simply denotes Cartesian product, ordered pairs.

>>8676913
Don't be fooled, OP. Your prof will likely present the material to you in such a way that it really *seems* complicated, but the mechanics of Group theory are awfully simple, at least at the undergraduate level. Do yourself a favor and download a few pdf's on beginner's abstract algebra and read the chapters corresponding to lecture diligently -- textbooks will usually give you a few extra theorems in your toolbox, but be sure to cite the author if you use them out of no where.

Group theory was the first class that made me realize I ought to be studying analysis instead. Lol. GL OP

>>8676960

Because some types of algebraic structures are /commutative/, and some are not! :^)

Commutativity is the property when the /order of the operands does not matter/. This is quite different from saying that the /order of applying the operation does not matter/, which is what associativity says (and which incidentally required three dummy variables a,b,c in order to express the idea-since we had to be able to write the operation down twice, in order to illustrate that the order of operation does not matter).

Suppose that a group is /commutative/. In such a group, then yes, the order of the operands does not matter. This can be expressed thus:

For all a, b, in G,

a # b = b # a

A group with the property that it is /also/ commutative, is referred to as an /Abelian Group/. Notice that this is a "special" type of group, and not part of the definition for groups in general that I gave above: hint: this scratches the surface of the nomenclature-game of college algebra. Several different types of algebraic structures are given special jargon-names/terms of art, and then you are asked to prove/describe what properties such-and-such an algebraic structure does or does not have. So if a teacher were to try to trip you up with a simple question like "are all groups commutative?", you would correctly reply "No."

Now, back to commutativity itself, just to make the point that it isn't always a feature, two exercises. Consider ordinary division, and look up matrix multiplication. Ask yourself: are these operations commutative?

>>8676985
division surely isn't commutative because 6/5 =/= 5/6 right?

and neither is matrix multiplication i'm sure of that

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just go through the first couple sections of this and you'll be fine

>>8677052
Matrix multiplication is not commutative, yes.

Division shouldn't really be considered since division is just a special case of multiplication where you are multiplying by inverses. So what you actually wrote:

6/5 = 6*5^-1 = 5^-1*6 = 6/5

So actually division is commutative since it is just multiplication (which is commutative in our usual sense).

>>8676913
>are the basics a confusing concept to your head around?

You haven't researched the subject? Not at all?

>>8676913
I was good until cosets. Still don't have the intuition for what they really are, but I don't think undergrad texts give the full picture (things like orbits, conjugacy classes, and outer automorphisms).

It was one of my favorite math courses though, and every time I revisit it I come away with new intuition I didn't have before.

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was group theory really that easy for everyone else

i studied for almost a year before actually taking the class and i still struggled with most of the proofs

>>8676949
I've asked this before on /sci/, so apologies if I just didn't get it last time.

Say you have something like the Rubik's cube in OPs pic. What would be the operation #?

From what I understand, a group can also be defined as a set (call it [math] G [/math] again) along with a family of function ([math] \mathfrak{F}[/math]), so that the group can be written as [math](G, \mathfrak{F})[/math].

Then the axioms of the group are closure
[eqn] \forall x\in G, f\in \mathfrak{F}: f(x) \in G [/eqn]
identity
[eqn] \exists I\in\mathfrak{F}:\forall x\in G: x = I(x) [/eqn]
inverse
[eqn] \forall f\in\mathfrak{F}:\exists g\in\mathfrak{F}: f\circ g = I[/eqn]
(where [math] \circ [/math] is the composition operation)
and associativity:
???

So what's this final axiom look like in groups like the Rubik's one, or the dihedral group

>>8678797

No problem. But I'm going to ignore your decent notation suggestion for the moment, to suggest something else.

Basically, the case of the Rubik's cube presumably (I have to check this, I haven't checked this!) works out just like a group, in the sense that the possible, legal positions of the cube are the set, while the various possible turns (?) are the operations on the cube. I notice that my sketch-phrase is quite different from what Wiki presents,and so therefore take my suggestion with a grain of salt-I haven't thought hard about it!

I am dubious about your "family of function" description for a group's operation, and I invite you to describe where you got that idea from. The idea should be related in some sense to my insistence on binary operations, which is standard about this point from an undergrad point of view.

I suspect that the operation for the cube would be the only thing that comes to mind: the /turn/. Turning, permuting the cube any which-way, as is legal, according to the operation. The trick is to describe this, and as the wiki makes clear, there are many cases.

https://en.wikipedia.org/wiki/Rubik%27s_Cube_group

>>8678877
How is a turn an operation?
It doesn't take two members of the set (the Rubik's cube state) and combine them. A turn takes one Rubik's cube state and transforms it into another using one of the allowed moves.
This transition can be put together with the other transitions into a family of transitions.
Groups like the Rubik's cube group or symmetry groups led me to believe using these transformations is an equivallent way of describing groups

Also that wiki does help. So more rigorously it looks like the family of functions is the group G, with the composition of two turns being the operation...however
that operation doesn't seem closed..
ie [math] F \circ F [/math] is not in the set of moves.
My confusion continues.

With permutations the members of the underlying set are essentially functions acting on some other set, and multiplication is the composition of functions.

So say you have a permutation group of 5 elements, a member might look like (2 1 3 5 4), which says "swap the elements in positions 1 and 2, and swap the elements in 4 and 5".

If you apply this to a set of 5 elements you get :
(2 1 3 5 4){abcde}={baced}

Multiplication is then:
(3 5 1 2 4)(2 1 3 5 4){abcde}
=(3 5 1 2 4){baced}
={cdbae}
=(3 4 2 1 5){abcde}

So,
(3 5 1 2 4)(2 1 3 5 4)=(3 4 2 1 5)

In terms of a Rubik's Cube, the members of the group are the states of the cube, but you interpret them as permutations acting on the original solved state.

This group is generated by members representing a half turn of each side, in the sense that every element can be written (non-uniquely) as a product of just those members.

>>8678930
>In terms of a Rubik's Cube, the members of the group are the states of the cube, but you interpret them as permutations acting on the original solved state.
hokay. that's cool as fuck.