This is a follow up question on my earlier post (Updated): Showing that a set $M$ with two elements classifies as a field. I feel this post is necessary because I realize that what confuses me is how braces work in rigorous Mathematics. Consider the example below:
Example: Let $M= \lbrace u,g \rbrace $ be a set and the multiplication defined as follows: \begin{align} g \cdot g = u \cdot g = g \cdot u = g \tag{M1} \\ u \cdot u = u \tag{M2} \end{align}
I want to verify that the associative property for the above defined form of multiplication holds, assuming that I have already shown commutativity (which is clear by the definition)
So let $(x,y,z) \in M^3= \lbrace u ,g \rbrace ^2$, there are 8 cases I need to realize, I will only use two, one which is easy (for me) to show and one that I can only show with some 'magic'.
Case 1: Let $(x,y,z)=(u,g,u)$. Then I have: $$u\cdot (g \cdot u) \overset{M1}=u \cdot (g) \overset{com.}=(g) \cdot u \overset{M1}=(u \cdot g)\cdot u $$
Please note that during the whole process I kept my braces ( ) adamantly in place on purpose.
Case 2: Let $(x,y,z)=(u,u,g)$. Then I have: $$u\cdot (u \cdot g) \overset{M1}= u \cdot (g) \overset{com}= (g) \cdot u =(u \cdot g ) \cdot u\overset{M1}=(u \cdot u \cdot g) \cdot u \overset{*}=(u\cdot u) \cdot g \cdot u \\ \overset{M1}=(u \cdot u ) \cdot g$$
Where (*) clearly is some magic I cannot explain to myself mathematically, verbally all I did was moving the right ")" brace to the left and voila, but this way there would be no need to even verify the associative property.
Is there a way to show the associative property for Case 2? Are there any rules for placing/moving braces?
The easiest way for each case is to evaluate both sides and check they are equal.
$$u \cdot (u \cdot g) = u \cdot g = g$$ $$(u \cdot u) \cdot g = u \cdot g = g$$
So $u \cdot (u \cdot g) = (u \cdot u) \cdot g$.
Regarding your question in the comments, there is no difference between $u \cdot (g)$ and $u \cdot g$. Parentheses are only for grouping. They tell you to evaluate what's inside the parentheses first. In $u \cdot (g)$ there is nothing to evaluate in $(g)$ so it's the same as $u \cdot g$.