Group Associativity

IE for a*(b*c) = (a*b)*c to be true is it required there be 3 different members in a group?

>>8515380
No. ({0}, +) is a group, although it's not particularly interesting to say that 0+(0+0) = (0+0)+0

>>8515394
How am I supposed to infer associativity from it if the definition uses 3 differently marked members (a,b,c)? Why can I just put the same number as all of them

Also, take [math] \mathbb{Z}_{2} [/math]. The three elements are general variables, so they don't need to be distinct.

>>8515380
i was wondering just this the other day,
turns out its just 3 arbitrary elements.
they dont have to be unique

>>8515398
You can just run the calculation.
LHS is 0+(0+0) = 0 + 0 = (0+0) + 0 = RHS
so done

>>8515398
it does not exclude a=b=c

>>8515398
This is just basic usage of variables, man.

Having a different variable does not mean the numbers can't be the same. You've solved linear equations before, right? When your teacher gave you
x+y = 2
x - y = 0
you didn't assume that x and y had to be different numbers because the names were.

the set {a,b} with binary operation *
such that
a*a=a
b*a=b
a*b=b
b*b=a
is associative because
(a*a)*a=a*(a*a)=a
(a*a)*b =a*(a*b) =b
aba=b
abb=a
baa=b
bab=a
bba=a
bbb=b
it is also the smallest non trivial(ish) group

>>8515408
Yeah, I didn't realize it applies here considering various rules about groups of numbers where each variable has to mean a different member.

>>8515398
>CS majors

>>8515430
ayy i was spotted

>>8515394
>>8515399

This (these) poster(s) are correct, OP. A group need not have at least three elements in order to be judged associative, or shown to be associative.

Language which provides a (the) definition of a group, particularly the bit about associativity, pick whatever textbook or wiki article you like, always runs something like this:.

"A group (G,#) satisfies the following properties. FOR ALL (read also: any whatsoever) a,b and c in G...
.
.
.
3) ASSOCIATIVITY: a # ( b # c ) = ( a # b ) # c
..."

Notice that in such a treatment, the only stipulation is that a,b and c are things in G. Nowhere in a proper definition is it stipulated that /no two of a,b, and c are equal to each other/. In particular, it might happen that a = b = c, a = b =/= c, etc. As used here, they're just dummy variables for elements of G which are used to illustrate the required behavior of G under its operation ( # in this case).

So why the three variables, with the three distinct glyphs? Because this is the minimal number required to illustrate the meaning of associativity itself /in its most general form/. From here, we can easily work backwards and consider the situations where b = c, and so on. But if we had only written something like

a # ( a # a ) = ( a # a ) # a

in our general definition, then this would have been /more/ confusing, and indeed not the definition of a group itself. Then we would only start with the weaker requirement that taking some a over-and-over, always the same thing in all the same spots, the thing associates. The above version is /stronger/ because it entails both itself (when the things are all unequal, which is common but not every case) /AND/ the situations where two arguments are equal, which can be easily dealt with as trivialities, exercises, caveats, etc.

cont.

hoe do you check the associativity of a binary operation on a finite set if you already have the table?
i mean, the axioms let's you "play sudoku" on the table to get what a given group should look liek if it existed, but pricing associativity seems like a computational nightmare

Related question to OPs topic:

If I have the group A in relation to *, and I only know that a,b,c are parts of it, as well as neutral e

Can I know for sure that there's x that is part of A so that a*x = b? or a y so that y*a=b?

>>8515441
Isn't sudoku only relevant if there's reduction (not sure if that's the right translation from my language) qualities? Not just associativity

>>8515442
a*(a^-1*b) = b

>>8515436
Was answered already as you said but this really explains it well, thanks.

>>8515442
to solve the equation
a+x=b where b is unknown
act with -a on both sides. (-a exists if A is a group)
-a+a+x =-a+b
e+x=-a+b
x=-a+b

>>8515448
shiet that's clever senpai

>>8515436

cont.

What is essential about the property of associativity is /not/ that it requires three hypothetical operands. What is essential, the /point/ of speaking of associativity at all, is that when you have this property of associativity about some algebraic structure, for example groups which have it by definition, then if you're just doing the same operation over-and-over again in some train of computations, then /it does not matter in what order you choose to do the operations (which all consist of the same one type of operation, whatever type that might be). And the simplest, most compact and most general way of describing or illustrating this property is, again, to write something like

a # ( b # c ) = ( a # b ) # c

All that is really going on here is to illustrate that when a choice of order of performing the operations exists, then that choice does not matter. It follows immediately from the irrelevance of choice when a choice (read: entailing at least two operations) exists, that if you have some finite, yet arbitrarily large chain of operations to be performed (and thus an arbitrarily large yet finite number of possible choices-sequences in which to perform those operations), /it still doesn't matter what order the operations are performed in/. So you can just specify some standard if you like in such a situation.

The above is a very simple example, but being able to make general statements and observations of these kinds is why we speak of the properties in general, in the first place.

Associativity says, "The order in which this one operation is performed (when it is performed many times over some one computation) /does not matter/.

Commutativity says, "The order in which the operands are placed with respect to the (binary) operation /does not matter/.

That these properties variously hold and don't hold with respect to the most elementary, fundamental arithmetic binary operators, is of great interest for doing algebra.

>>8515455
does x = -a*b = x*a = b apply to every group?

>>8515465
But associativity doesn't prove you can place the numbers themselves in any location and still get the same result, right? Just the operation

>>8515471
>>8515465
Yeah so that's commutativity, got it

>>8515473

Yes. It just happens that illustrating the essential feature of commutativity,

a @ b = b @ a,

requires only two operands and thus only one (binary) operator (on one side of the equation). Since associativity is to do with choice among the operations themselves, then we need a bare minimum of two operators (to choose from) on one side of our equation-illustration, which in turn entails a bare minimum of three hypothetical operands in the string. But that has already been discussed at length by others in the thread, and myself.

Notice how every operator that has been under discussion is a /binary/ operator, requiring two operands. This is common in algebra, exactly because the most elementary arithmetic operations are themselves all binary operations. But one can also think of unary operators, n-ary operators, etc.

>>8515471

Ah, I think you're getting onto something interesting! /The two properties are different/! Here are some exercises for you to do:

1) Look up the definition of an Abelian group. Are all groups Abelian groups? Why or why not?
2) give an example of an elementary arithmetic binary operation which is not associative with respect to the integers. Show that it is not associative.

>>8515501
1) all natural numbers in relation to +
2) (5-3) - 4 = -2
5-(3-4) = 6

did i do this right senpai

>>8515522

Hm, needs work.

1) What do you mean by your answer? What about the other questions? If that thing you mentioned is an abelian group, is it the only one? Are there others?

2) okay, fine, two equations. what do the two equations tell you about the operation as it relates to the integers? Do you need to check that associativity also fails in other cases? Why or why not?

>>8515574
Oh oops, I answered as if you asked for an albelian group

Not all groups are abelian because being commutative isn't a requirement for groups
like if the binary action is - for every a*b the result is a

2) you don't because associativity is supposed to apply to every a b c
and I'm not sure how to answer question 1