tl;dr: What properties does a set need for it to have the distributive law? Does it need to be an axiom?
Given a group (i.e. satisfies the closure, associative, inverse and identity properties/axioms), how do you show the distributive law? i.e is the distributive law a consequence of being a group or does it have to be an axiom? Does every group has as property the distributive law?
Say we have a group $(Z,+)$
How do you show:
$$
ab+ac=a(b+c)
$$
What threw me off a little was that the group was not defined under multiplication, so I was not sure what to try. A proof of the distributive property is what I am looking for. Also, I trying to gain intuition on why some operations do not distribute under some groups. Examples and reasons are welcome and helpful!
The following made me ask this question:
I have a book on number theory and group theory and it stated the following:
Theorem 6.4. Let G be an abelian group. Then for all $a, b \in G$ and $k, l \in Z$, we
have:
$$(i) \ k(la) = (kl)a = l(ka)$$
$$(ii) \ (k + l)a = ka + la$$
$$(iii) \ k(a + b) = ka + kb$$
And they leave the proof as an exercise, which I wanted to try and do. However, I was not sure how to even start because a group is defined under one operation but the expression provided show two operations. Also, I thought, that the distributive property was not bound only to abelian groups, thats why I didn't include the word abelian in my question.
Best Answer
Let $(G,\circ)$ be a group with neutral element $e$ and let $X$ be a set. We say that $G$ acts on the set $X$ if we are given a map $G\times X\to X$, $(g,x)\mapsto g\cdot x$ if the following natural requirements are fulfilled: $(g\circ h)\cdot x = g\cdot(h\cdot x)$ and $e\cdot x=x$ for all $g,h\in G$, $x\in X$. (The first requirement may allow one to be sloppy and use the same multiplication symbol for the group operation and the action and also drop parentheses).
If $X$ carries additional structure, say $(X,*)$ is also a group, we - naturally - demand that more requirements be respected before we say that the group $(G,\circ)$ acts on the group $(X,*)$, namely that $g\cdot(x*y)=(g\cdot x)*(g\cdot y)$. Note that this looks like the distributive law, but $g,x,y$ are (usually) not from the same set!
If $(X,*)$ is an abelian group with neutral element $e$, then there is always a standard way to define an action of the additive group $(\mathbb Z,+)$ on it: Define $0\cdot x=e$ first, then by the recursion $(n+1)\cdot x=(n\cdot x)*x$ define the action of all positive integers and finally by $(-n)\cdot x=(n\cdot x)^{-1}$ the action of negative integers. (Verify that this is an action, be careful with all those different operation symbols!). If we are a bit sloppy and use $+$ also for the group operation of $X$ (and negation for inverse), as is usual for abelian groups, then the fact that $(\mathbb Z,+)$ acts on the group $(X,+)$ precisely formalizes that "distributivity" holds.
As $(\mathbb Z,+)$ is also an abelian group, the above method can be used to define an action of $(\mathbb Z,+)$ on itself. We should be careful because we use $\cdot$ to denote this action and already have an intrinsic multiplication of integers that is also written with $\cdot$. Fortunately, the group action of $(\mathbb Z ,+)$ on itself is precisely the normal integer multiplication (check that without getting confused about the different operations).
As a bonus check that $(n\cdot m)\cdot x=n\cdot (m\cdot x)$ holds for all $n,m\in \mathbb Z$ and $x\in X$ with the group action of $(\mathbb Z,+)$ on the abelian group $(X,*)$ defined above, justifying again the use of the same symbol for both things (and allowing one to drop parentheses) even though they are strictly speaking not the same.