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Prove associativity in the group $(A,\cdot)$ where $$\begin{matrix}A&=&\left\lbrace1,-1,x,-x,x^2,-x^2\right\rbrace&&\text{with}&x^3=1\end{matrix}$$ and $\cdot$ product.

The statement is as it appears here.

Instead of making the table of elements of $(A, \cdot)$, can I say that associativity is fulfilled by inheritance of the product of functions?

We have done exercises where we had directly demonstrated through the inheritance, but here I do not know because I do not know what the $x$ is (I am sure that it is a function). Or should I prove the associativity for all the elements?

Thank you!

manooooh
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    Inheritance of what "product of functions"? What do you mean when you say you are "sure" that $x$ is a function? Certainly nothing in the problem statement seems to indicate that $x$ is a function... – Eric Wofsey Jul 23 '18 at 00:44
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    Right. You can make this work out but you need to specify the domain and codomain of the functions, taking into account the need to satisfy the relation $x^3 = 1$. – Qiaochu Yuan Jul 23 '18 at 00:45
  • Ok. The statement says nothing about the domain and codomain, you are right. Thanks! – manooooh Jul 23 '18 at 00:47
  • What do you mean by "inheritance"? I haven't heard the word before in mathematical English; I wonder if it's just my ignorance, or perhaps a mistranslation from another language? – bof Jul 23 '18 at 07:45
  • @bof it's probably my lack of learning English (along with the help of Google Translator haha). For example take the following case: demonstrate the associativity of the group $({-1,0,1},\cdot)$ with the usual product. Without any fear we can say that it fulfills the associativity because the product is an heritage/inheritance of the product in $\mathbb Z$, since ${-1,0,1}\subseteq\mathbb Z$. If you still do not understand me, please let me know. – manooooh Jul 23 '18 at 08:04
  • Note that in the original statement we don't know in which set $x$ is. – manooooh Jul 23 '18 at 08:18
  • @manooooh No, formally this is wrong. What is "$-$" in your context? It is just a symbol. Lets use another one, say $&$, and rewrite your set: $A={1, &1, x, &x, x^2, &x^2}$. Do you see how broken your definition now is? You need more rules to understand this structure. Is $&x$ actually $(&1)\cdot x$? Formally your $\cdot$ is simply not defined on all possible elements. So how can you conclude anything? You can't say that $-1$ is "inherited" from $\mathbb{Z}$, because is $x$ an element of $\mathbb{Z}$? It is not, so what does $-x$ even mean?And even if it was do we know what $\cdot$ is? – freakish Jul 23 '18 at 10:03
  • @freakish thank you for your answer. So are you saying the statement we have is incomplete (we need more info about the group $(A,\cdot)$)? This statement was given in an exam... Btw I think the meaning of "product" in this context is clearly algebraic. With all your comments I know now in this particular excersise we can not use heritage of any set because... we don't know it! Are you agree with the use of heritance in the case $({-1,0,1},\cdot)$? I wait your comments. – manooooh Jul 23 '18 at 14:31
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    @manooooh If ${-1,0,1}$ is understood as a subset of $\mathbb{Z}$ and $\cdot$ as the usual integer multiplication then yes, the "inheritance" works here perfectly fine. As for your $A$ problem: I'm very surprised that someone gave something like this at an exam without further clarification. This sounds really incompetent. But if you have to work with this kind of things then I guess you have to make some implicit assumptions, like $-x=(-1)\cdot x$. You can work with that. I'll gather everything as a full answer soon. – freakish Jul 23 '18 at 14:44
  • @freakish thank you again. "If $({-1,0,1},\cdot)$ is understood as a substet of $\mathbb Z$ (...)" and what another set could be if it has only integer numbers?! Even if we work with reals I think the statement holds. "And $\cdot$ as the usual integer multiplication (...)" yes, I mentioned that (btw is quite ambiguous say "with the usual product" but... I am not a student of Logic, so in my course we don't prove in most of times formally, unfortunately :(). So yes, I am studying with "implicit assumptions". I will wait your full answer :). – manooooh Jul 23 '18 at 14:56
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    @manooooh are you familiar with rings? $0, 1, -1$ mean something similar but different there. No longer numbers anyway. Give me some time pls to gather thoughts. :) – freakish Jul 23 '18 at 14:59
  • @freakish no, we don't study rings. I hear something about that but I never have done excersises using this structure. Take your time!! – manooooh Jul 23 '18 at 15:01

1 Answers1

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Since you are starting your adventure with maths this may come to you as confusing. I will try to explain everything as simply as I can.

First of all neither $-1$ nor $1$ is a number. These are just symbols representing numbers. So for example if I do $x:=1$ then now $x$ and $1$ is the same number even though represented by different symbols. This is important. In higher maths you have to distinguish between symbols and the meaning behind them.

So first thing you need to do in the definition

$$A=\{1,-1,x,x^2, -x,-x^2\}$$

is to give a meaning to every symbol there. We are not going to talk about ={}, symbols (yes, everything in maths is actually well defined), I will assume the meaning is clear.

The important symbols are: $1$, $x$, $-$ and $()^2$. There's a hidden meaning behind them. For example typically $x^2$ is a shortcut for $x\cdot x$. But what does $-x$ mean? We don't know that, you didn't tell us. Or what does $-1$ mean?

You said in comments that $1,-1$ are numbers and $\cdot$ is the usual number multiplication. Fair enough, then the "inheritance" works, meaning you don't have to check associativity between pairs from that subset. But do not be fooled: $1,-1$ does not always mean the usual two integers. They can mean something very different (see: rings). Actually often $1$ is used as a symbol for the neutral element in any group.

But what is $x$? Is that a number as well? But which one, you said $x^3=1$. Is that a complex root of $1$? And what does $-x$ mean? Is that simply $(-1)\cdot x$?

You see, maths is a science of formalism. While intuition plays a crutial role you cannot do anything outside of formalism. And formally your question as it stands is not well defined. Let me give you an example: assume that I define

$$(-x)\cdot(-x)=x$$ $$(-x)\cdot x = 1$$ $$x\cdot(-x)=x^2$$

You didn't specify how $\cdot$ works on that pair so I did it somewhat randomly. And you can check that associativity doesn't hold with that definition.

Now let's do some additional assumptions in order to actually solve that. What is safe to assume is the following:

$$(-1)^2=1$$ $$(-x)=(-1)\cdot x=x\cdot(-1)$$ $$(-x^2)=(-1)\cdot x=x\cdot (-1)$$ $$(x)\cdot(-x)=(-x)\cdot x=-x^2$$ $$(-x^2)\cdot (-x)=(-x)\cdot(-x^2)=1$$ $$(-x^2)\cdot (x)=(x)\cdot(-x^2)=-1$$ $$1\cdot y=y\cdot 1=y\text{ for any }y\in A$$ $$\ldots$$

and so on. As you can see what actually is going on here is I simply write down the multiplication table. With that you can check associativity manually.

The "inheritance" you are refering to actually means that some subset can be treated as some other well known structure. And if that other structure is associative then so is the original one. This is also known as an isomorphism - an invertible function that preserves the binary operation. This automatically guarantess properties like associativity.

Anyway you are on the right track: if you know that $\{1,-1\}$ are numbers and $\cdot$ is the usual multiplication then yes, you can reuse the fact that the multiplication on those is associative.

freakish
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