Let $G$ be the set $\mathbb R\setminus \{0\}$ and let $*$ denote a binary operation on the set, defined by:$$\forall a, b \in G,\;\;\;a*b= \dfrac{a\cdot b}2.$$ I need to show that
$[G,*]$ is an abelian group.
Does that mean I need to show that $G$ is a group, which is also abelian? I think I probably should show both, but I'm not sure where to start. Most of the groups I've studied up to now have had group operations of addition or multiplication, so I'm not sure how to work with this operator.
Best Answer
You need to show that for any two elements $a, b$ in $G$, $a*b = b*a$.
$$a*b = \dfrac{\color{blue}{a\cdot b}}2 = \dfrac{\color{blue}{b\cdot a}}2 = ba$$
Because "normal" multiplication on the real numbers is commutative, we know $\color{blue}{a\cdot b = b\cdot a}$. Hence, we have commutativity.
Now you need to establish that $G$ is in fact, a group:
Associative? Check whether, given any $3$ elements $a, b, c$ in $G$, is it true that $$a*(b*c) = (a*b)*c\;\;?$$
Existence of an Identity element $e$ such that for all $a \in G$, $a*e = e*a = a$? Let's check out whether $2$ meets this criteria: $$a*2 = \dfrac {a \cdot 2}2 =\color{blue} a = \dfrac{2\cdot a}2 = 2*a$$
Closure under inverses?: For every element $a\in G$, does there exist an element $a^{-1}$ such that $aa^{-1} = a^{-1}a = 2= e$? Let's check out whether $\frac 4a = a^{-1}$: $$a*\frac 4a= \frac {a\cdot \frac 4a} 2 = \frac 42 = \color{blue}2 = \frac{\frac 4a \cdot a}2 = \frac 4a*a$$
Once you confirm that associativity holds, we'll see that $[G, *]$ is a commutative group, i.e. and abelian group.