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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!

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    $\begingroup$ 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... $\endgroup$ – Eric Wofsey Jul 23 '18 at 0:44
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    $\begingroup$ 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 $x^3 = 1$. $\endgroup$ – Qiaochu Yuan Jul 23 '18 at 0:45
  • $\begingroup$ Ok. The statement says nothing about the domain and codomain, you are right. Thanks! $\endgroup$ – manooooh Jul 23 '18 at 0:47
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    $\begingroup$ @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. $\endgroup$ – freakish Jul 23 '18 at 14:44
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    $\begingroup$ @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. :) $\endgroup$ – freakish Jul 23 '18 at 14:59
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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.

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