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There is this set $S=\{1,-1,\iota,-\iota\}$, where $\iota=\sqrt{-1}$, and I have to prove that ordinary multiplication "$\times$" is a binary operation on $S$, and that it is commutative and also associative on $S$. Now, I have proved that $\times$ is a binary operation on $S$ employing the Cayley table, and also that it is commutative by showing that the table is symmetric around the diagonal. But, to prove that $\times$ is associative, I found four cases for which $\times$ must be proved associative in order to prove the associativity of $\times$ on $S$. They are:

(1) Is $(1 \times \iota) \times -1 = 1 \times (\iota \times -1)$?

(2) Is $(1 \times -\iota)\times -1 =1 \times (-\iota \times-1)$?

(3) Is $(1 \times -\iota) \times \iota=1 \times (-\iota \times \iota)$?

(4) Is $(-1 \times -\iota) \times \iota=-1 \times (-\iota \times \iota)$?

The Question: If I prove, in all the four cases, that the answer to all the questions is positive, then will it suffice or must I also prove that $\times$ is associative for the cases in which all of the factors in the four cases above are differently arranged?

For example, for the first case, we can arrange the factors in the following different ways:

(1.1) $1 \times -1 \times \iota$ (1.2) $-1 \times 1 \times \iota$ (1.3) $-1 \times \iota \times 1$ (1.4) $\iota \times 1 \times -1$ (1.5) $\iota \times -1 \times 1$ and similarly for the other three cases we have different arrangements of the factors.

Related: How can we determine associativity of a binary structure from its Cayley table?

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Samama Fahim
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    If you already know that $\times$ is "ordinary multiplication", why not use the associativity inherited from $\mathbb C$? If you are for some reason "not allowed" to do that, you could still prove it in a general argument: $(a+bi)((c+di)(e+fi)) = ((a+bi)(c+di))(e+fi)$, expand both sides and see how it turns out. – Christoph Feb 19 '15 at 09:50
  • I can also prove commutativity by a similar general argument, can't I? – Samama Fahim Feb 19 '15 at 10:07
  • Yes, by keeping in mind that the definitions of addition and multiplication in $\mathbb C$ are $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)\cdot(c,d) = (ac-bd, ad+bd)$ and $i$ is defined to be $(0,1)$. Now you can prove all you need from those definitions. – Christoph Feb 19 '15 at 10:09
  • If we expand $(a+bi)((c+di)(e+fi))$, we will get $eac + eadi...$, and we still need to worry about $(ea)c$ being equal to $e(ac)$, if we are not allowed to use the associativity inherited from $\mathbb{C}$. – Samama Fahim Feb 19 '15 at 10:27
  • That's associativity in $\mathbb R$, maybe you can use that? – Christoph Feb 19 '15 at 10:28
  • No, I can't. I think I must prove it by solving for all of the arrangements. Should I? – Samama Fahim Feb 19 '15 at 10:29
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    If you really have to do it by checking all possible products of three elements, then you have $64$ things to check. – Gerry Myerson Feb 19 '15 at 10:50
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    @GerryMyerson Using commutativity and the fact we have only $3$ non-neutral elements we get this number down to 8. See my answer ;-) – Christoph Feb 19 '15 at 12:07

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Okay, so doing this from first principles, by using only the Cayley table as the definition of $\times$, this would be one way to go:

From the Cayley table we see that $\times$ is commutative and that $1\times x = x\times 1 = x$ for all $x\in S$, hence $1$ is a neutral element for $\times$. Thus, whenever one of $a$, $b$ or $c$ is $1$ we can immediately conclude $(a\times b)\times c = a\times(b\times c)$, simplifying your cases (1), (2) and (3). Now we are left with $a,b,c\in\{-1,i,-i\}$. If $a$, $b$ and $c$ are pairwise distinct, your statement (4) together with $(-1\times i)\times -i = -1\times(i\times -i)$ shows associativity, since you then have identities for each of $i$, $-i$ or $-1$ being outside of the parentheses. If $a=c$, you can use commutativity, $$ a\times(b\times a) = a\times(a\times b) = (a\times b)\times a. $$ Now you are left with the cases $a=b\neq c$ and $b=c\neq a$ for $a,b,c\in\{-1,i,-i\}$. For $a=b\neq c$, you have to prove the $6$ cases. Then, for $b=c\neq a$, we can conclude \begin{align*} a\times (b\times c) &= a\times(c\times c)\\ &= (c\times c)\times a\\ &= c\times(c\times a)\\ &= (c\times a)\times c\\ &= (a\times c)\times c\\ &= (a\times b)\times c. \end{align*}

Christoph
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  • What identities are you talking about? – Samama Fahim Feb 19 '15 at 11:57
  • As I said: Your (4), which is $(-1\times -i)\times i = -1\times(-i\times i)$ together with $(-1\times i)\times -i = -1\times(i\times -i)$ are enough to show associativity for $a,b,c$ being pairwise distinct and non of them equal to $1$. The reason being: Whenever you encounter $a\times(b\times c)$ for $a,b,c$ pairwise distinct, none of them equal to $1$, $a$ is on of $-1$, $i$ or $-i$ and the two equalities i mentioned, togehter with commutativity, cover all those cases. – Christoph Feb 19 '15 at 12:02