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I am reading Alan Beardon's "Algebra and Geometry" and it says:

(2) for all f, g and h in G, f * (g * h) = (f * g) * h.

(2) is called the associative law, and it says that f * g * h is uniquely defined regardless of which two operations * we choose to do first. The point here is that * only combines two objects at a time, and we have to apply it twice (in some order) to obtain f * g * h. There are exactly two ways to do so and the two must always yield the same results.

The issue I have is that I do not see how is it that there are only two ways to combine two elements out of three. For as I understand I could make six different combinations (if I do not assume commutative property):

(1) ( f * g ) * h

(2) ( f * h ) * g

(3) ( g * f ) * h

(4) ( g * h ) * f

(5) ( h * g ) * f

(6) ( h * f ) * g

And even if I assume that there is a commutative property, that I have not seen stablished in the definition so far, there would still be three different combinations:

(1) ( f * g ) * h

(2) ( f * h ) * g

(3) ( g * h ) * f

Can you please explain to me how is it that there are only two ways to apply a binary operation over three elements and thus defining the associative axiom shall be expressed as f * (g * h) = (f * g) * h and not as ( f * g ) * h = ( f * h ) * g = ( g * h ) * f ?

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First, I am a little confused regarding your question: you first point out $6$ different ways to combine $f$, $g$, and $h$. You then say that even if you assume the commutative property there would be three left ... but then when you propose to define the associative property differently, why do you pick only two of those three, and add a different one? Indeed, do you see how weird your proposed definition is, with one of the terms having the parentheses in a different place than the other two?

Anyway, just focusing on two of those: no, we do not have $(f*g)*h=(f*h)*g$ as part of the associative property, because that would require a commutative property as well, and that is something different from the associative property.

I think you are confusing the "ways in which we can perform the order of the operations" and the "ways in which we can order the operands".

Associativity is about the order of the operations. That is, if you have $f * g * h$, then we have two choices for the order of the operations: do we first apply the left *, and then the right - this would be $(f * g) * h$ - or do we first apply the right *, and then the left - which would be $f * (g * h)$? Well, if the * is associative, then it turns out that the order of the operations doesn't matter, i.e. that $f *(g*h) = (f *g)*h$.

You, however, seem to be worrying about the number of ways in which you can order the operands ... and yes, given three operands $f$, $g$, and $h$, you can have many ways to order them, as you point out. However, whether that always gives the same result is not what the associative property is about ... the order of the operands is what the commutative property is about.

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    $\begingroup$ Downvoters: Please explain your downvotes to me .... $\endgroup$ – Bram28 Nov 9 at 2:53
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    $\begingroup$ I'm confused why this was downvoted. I think it's the best answer as it addresses OP's confusion $\endgroup$ – infinitezero Nov 9 at 12:09
  • $\begingroup$ @infinitezero Thanks! Yes, that's exactly what I was trying to do. $\endgroup$ – Bram28 Nov 9 at 12:15
  • $\begingroup$ Aaah! ok! I thought that the associativity property was something like this "if you have three elements, and need to make a binary operations on them, it does not matter which two you make first, they will always yield the same result" but from your explanation that is more the definition of a group that it is both associative and commutative. Because if they are associative it does not mean that (f * g) * h = (f * h) * g, it just means that (fg)*h=(gh)*f and (fh)*g=(hg)*f and (gf)*h=(fh)*g. What do you think? $\endgroup$ – César D. Vázquez Nov 9 at 23:10
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The thing is, when we write $a\cdot b\cdot c$, you can't first evaluate $a\cdot c$ and then compute $ac \cdot b$, because we don't know if the binary operation is commutative or not. So you don't have six possible way to compute $abc$, you only have two, which are $(ab)c$ and $a(bc)$.

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  • $\begingroup$ How is it that I can't evaluate a * c and then (a * c) * b ? I thought that by definition you can only apply a binary operation to two elements and that if you have three elements you necessarily need to apply a binary operation to one pair and then to another pair. I'm sorry but I do not understand I think you are contradicting the author when it says "The point here is that * only combines two objects at a time, and we have to apply it twice (in some order) to obtain f * g * h." $\endgroup$ – César D. Vázquez Nov 9 at 21:59
  • $\begingroup$ @CésarD.Vázquez You can only apply binary operation to neighboring elements. $\endgroup$ – 魏小淇 Nov 9 at 22:15
  • $\begingroup$ I do not understand. If you have three numbers (a, b, c) and you need to perform a binary operation on them three, wouldn't I necessary make a ab * c or ac * b or any of the other possible combinations. I have the intuition that your answer is the kind of answer that I need but I do not really think I understand. Can you develop further your reasons or recommend something to read? $\endgroup$ – César D. Vázquez Nov 9 at 22:21
  • $\begingroup$ @CésarD.Vázquez Let's consider an example. Suppose you have three square matrices $A, B, C$, and you want to compute $ABC$, would you first compute $AC$ and then compute $(AC)B$? $\endgroup$ – 魏小淇 Nov 9 at 22:38
  • $\begingroup$ The issue I think is that I thought that the associativity property was about "If you have to make a binary operations over any three elements, it does not matter which two you choose first, any way you choose first will always yield the same result." I did not think of the associativity property that a * b * c meant that I could only make either a * b or b * c. I thought that for me to say that the set (1, 2, 3) is associative under (+) operation, it meant that it does not matter if I do ( 1 + 2 ) + 3 or (1 + 3) + 2, or (2 + 1) + 3, or (...) it will always yield the sam result. $\endgroup$ – César D. Vázquez Nov 9 at 22:47
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The definition of associative requires the order to stay the same, without commutative law. In your first set $(f*g)*h\ne (g*f)*h$, unless $f$ and $g$ are commutative. Also $(g*h)*f\ne f*(g*h)$ without commutative. For your second question, those combination are derivable from using both associative and commutative together, so they are not needed for the definition of associative.

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  • $\begingroup$ I understand that I must assume the commutative property for (𝑓∗𝑔)∗ℎ = (𝑔∗𝑓)∗ℎ and that if I assume it the possible "different" combinations that I must do would be reduced by half (thus 3 different combinations instead of 6). What I do not understand is why ( f * h ) * g is not included in the definition of associativity. Why is it granted that if f * (g * h) = (f * g) * h holds ( f * h ) * g = f * (g * h) = (f * g) * h will also hold $\endgroup$ – César D. Vázquez Nov 9 at 22:04
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    $\begingroup$ @CésarD.Vázquez $(f*h)*g=f*(h*g)$ by associative law, However without commutative $(f*h)*g\ne f*(g*h)$. $\endgroup$ – herb steinberg Nov 9 at 23:56
  • $\begingroup$ Great! Thanks. I think my real confusion was that I thought that for a group to be associative under a certain operation meant that no matter how you combined three elements into a two step binary operation they will always be equivalent (Which would be complying with the six cases I list up there). But from your answer I see that that idea I had about what the associative property is, is more of the associative and the commutative property together. $\endgroup$ – César D. Vázquez Nov 10 at 17:39
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Groups abstract the idea of transformations that can be applied to some object. Group elements are individual transformations, and the group operation represents performing the transformations in sequence.

Importantly, group operations are typically not commutative. If $L$ means "turn 90 degrees left" and $F$ means "walk forward one foot," then $(LF)$ and $(FL)$ are different operations.

To take another example, we have e.g.

$\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix}2 & 3 \\ 0 & 3 \end{pmatrix},$ but $\begin{pmatrix}2 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix}2 & 2 \\ 0 & 3 \end{pmatrix}.$

On the other hand, obviously $(fg)h = f(gh)$ if $f$, $g$, and $h$ are transformations and the operation is applying them in sequence, because $f(gh)$ means "first do $f$, then (do $g$, then do $h$)," whereas $(fg)h$ means "(first do $f$, then do $g$), then do $h$," and these are of course the same thing. In other words, a nonassociative operation cannot correspond to combining transformations by applying them in sequence.

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The author says that there are two ways to understand $f * g * h$ specifically. He never states that there are "two ways to combine two elements out of three" in general.

((By the way, there are in fact 12 possibilities for 3 elements and 2 identical operators.))

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  • $\begingroup$ Are you sure? So the associativity property is not about "You can perform a binary operation over three elements in any way you like and you will get the same results" it would only say that the combinations fg * h = f * gh and fh * g = f * hg and gf * h = fh * g but not that fg * h = gf * h $\endgroup$ – César D. Vázquez Nov 9 at 22:59
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    $\begingroup$ @CésarD.Vázquez: Exactly. Equation (2) means what you see, no more, no less. This is mathematics :) $\endgroup$ – g.kertesz Nov 11 at 13:05

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