In a group $G$ with operation $\star$, can I apply $\star$ to both sides of an equation?
Solution 1:
You did just fine! No problems there.
For the nitpicker, you may want to start off, also, with "Since $G$ is a group,
- $G$ contains a unique identity, which we'll denote $e$, such that for all $a \in G,\;\; a\star e = e \star a = a.$
- And it also follows that every element $a \in G\,$ has a unique inverse, which we'll denote $a^{-1}$, such that $a\star a^{-1} = a^{-1}\star a = e$.
Then everything follows precisely as you argued. But as Gerry Myerson pointed out, you may also want to add justification as below for using associativity:$$a \star b = a\star c \tag{hypothesis}$$ $$\iff a^{-1}\star (a\star b)= a^{-1}\star (a\star c) \quad\quad\tag{left "multiplication" by $a^{-1}$}$$ $$\iff (a^{-1} \star a) \star b = (a^{-1}\star a) \star c \quad\tag{$G$ is associative}$$ $$ \iff e\star b = e\star c \tag{2}$$ $$\iff b = c \tag{1}$$