Let $G$ be a group. Suppose that $G$ has a left action on a set $A$ denoted by $g\cdot a$. Denote the corresponding right action by $\sim'$.
So $a\sim b \Leftrightarrow a=gb for \ some \ g\in G$ And:
$a\sim^{'} b\Leftrightarrow a=bg \ for \ some\ g\in G$.
How to show then that there's equivalence bettween the two?
I need to show that $a\sim b \Leftrightarrow a\sim'b$
Clueless, really. I need to show that there exists $h\in G$ s.t $a=bh$, I thought of using $a\sim b$ which means there exist: $g\in G$ s.t $a=gb$, but how to continue?
I think you missed the text before the exercise. For a left action $g\cdot a$, the corresponding right action $a* g$ is defined by $a* g=g^{-1} \cdot a$. (I use $\cdot$ and $*$ for left and right action respectively to avoid confusion.)
Using this definition, we can show that if $a\sim b$, then $a=g\cdot b$ for some $g\in G$, which implies $b=g^{-1}\cdot a=a* g$. Therefore $a=b* g^{-1}$ and we conclude that $a\sim' b$. The converse follows in the similar way.