When I state $(a\cdot b) \cdot c = (a \cdot c) \cdot b$ , what properties am I using?

Question

When I state $(a\cdot b) \cdot c = (a \cdot c) \cdot b$ , am I using only the commutative property or both the commutative and associative property?

Answer 1

If you mean whether the statement you're posting relies upon commutativity and associativity, yes: You rely upon both. I am assuming $a, b, c \in \mathbb R$ and that by "$\cdot$" you mean "standard" multiplication:

$$\begin{align}(a\cdot b)\cdot c & = a\cdot (b\cdot c) \tag{by associativity}\\ \\ & = a\cdot (c \cdot b) \tag{by commutativity}\\ \\ & = (a \cdot c) \cdot b \tag{by associativity}\end{align}$$

Answer 2

The identity

$(a \cdot b) \cdot c = (a \cdot c) \cdot b$

can be proved using both commutativity and associativity.

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If your structure has an identity, then the statement implies both commutativity and associativity.

For commutativity. Let $b$ and $c$ be any element. Let $a$ be identity for $\cdot$, then

$ b \cdot c = (a \cdot b) \cdot c = (a \cdot c) \cdot b = c \cdot b$

Now to show associativity:

$(a \cdot b) \cdot c$

By commutativity shown above

$ = (b \cdot a) \cdot c$

By the statement

$= (b \cdot c) \cdot a$

By commutativity

$= a \cdot (b \cdot c)$

Hence associativity follows.

In conclusion if there is an identity for $\cdot$, then commutativity and associativity follows.

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In fact, if there is no identity, then you can not prove commutativity.

Consider the two element set $\{x,y\}$ with multiplication defined by $x \cdot x = x$, $y \cdot y = y$, $x \cdot y = x$ and $y \cdot x = y$. Note that the multiplication is not commutative and has no identity. However, the value of the product depends only on what the first element of the product is. Hence it easy to verify that the statement holds in this structure.

However associativity does hold in this structure. I feel that the statement should not be able to prove associativity. If I can come up with a counter example, I will post it later.

Answer 3

Another set of examples satisfying $(ab)c=(ac)b$ is given on any set $S$ by defining $xy=x$ for all $x,y \in S$. If $S$ has more than one element, this is not commutative (nor does it have a two sided identity element). It is however associative.

For a nonassociative example, take $x \cdot y=x-y$ on $\mathbb{R}$, and then the relation $(ab)c=(ac)b$ becomes $(a-b)-c=(a-c)-b.$

Just noticed: subtraction is not commutative either, so gives a single example where the relation holds, but it is neither commutative nor associative.

Answer 4

So far nobody has given counterexamples showing that associativity alone or commutativity alone are not sufficient to guarantee $(a\cdot b)\cdot c=(a\cdot c)\cdot b$.

Multiplication of $(n\times n)$-matrices is certainly associative, but not commutative when $n\geq2$. Choose any two non-commuting $(2\times2)$-matrices $b$ and $c$, and let $a$ be the $(2\times2)$ identity matrix. Then $(a\cdot b)\cdot c\ne(a\cdot c)\cdot b$.

For complex numbers $z$, $w$ define $$z\ *\ w:=\overline{z\cdot w}\ .$$ This product is certainly commutative, but $(1*1)*i\ne(1*i)*1$.