In a group $G$ with operation $\star$, can I apply $\star$ to both sides of an equation? **Theorem:** Let $G$ be a group with operation $\star$. For all $a,b,c\in G$, if $a\star b=a\star c$, then $b=c$. I've got a proof, but I'm not sure it is correct (I'm not sure that I can apply the operation $\star$ on both sides): _Proof:_ Every element in $G$ has an inverse, so let $a^{-1}$ be the inverse of $a$. Then we can say $$a^{-1}\star a\star b=a^{-1}\star a\star c.$$ But $a\star a^{-1}=e$, the identity element of $G$. Then $$e\star b=e\star c,$$ and since $e$ is the identity element, $b=c$.

You did just fine! No problems there.

For the nitpicker, you may want to start off, also, with "Since $G$ is a group,

1. $G$ contains a unique identity, which we'll denote $e$, such that for all $a \in G,\;\; a\star e = e \star a = a.$
2. 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:$$a \star b = a\star c \tag{hypothesis}$$ $$\iff a^{-1}\star (a\star b)= a^{-1}\star (a\star c) \quad\quad\tag{left "multiplication" by $a^{-1}$}$$ $$\iff (a^{-1} \star a) \star b = (a^{-1}\star a) \star c \quad\tag{$G$ is associative}$$ $$ \iff e\star b = e\star c \tag{2}$$ $$\iff b = c \tag{1}$$

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