Let $G$ be a finite group. For any subset $S$ of $G$, and any group elements $g,h \in G$, I want to show that $|gSh|=|S|$, i.e. that the cardinality of $S$ is preserved when multiplying on the left and on the right.
My attempt:
$$ S = g^{-1}(gSh)h^{-1} \subseteq (gSh) \subseteq S $$
Am i right? If not, how can i prove the statement?
Note that the inclusion $gSh\subset S$ does not hold in general.
However, you can define a map $f_{g,h}:G\to G$ by $s\mapsto gsh$, and show that it is bijective. For that,it may be helpful to notice that this map is the composition of $s\mapsto gs$ and $s\mapsto sh$.