Let $H$ be a normal finite subgroup of $G$, and $g \in G$ has an order $n$. The only elements in $H$ that commutative to $g$ is only its identity element $e$. Show that a map $f: H \mapsto H$, $f$ is defined by $f(h) = g^{-1}h^{-1}gh$ is a bijection.
We must show that $f$ is injective and surjective. I don't know how to start honestly. We have stripped over paper but don't found exact solution. Do you have any idea?
It's enough for the map to be either injective or surjective, since it's a function from a finite set to itself.
Injection: Suppose $f(x) = f(y)$. Then $g^{-1} x^{-1} g x = g^{-1} y^{-1} g y$, so $x^{-1} g x = y^{-1} g y$, and hence $(xy^{-1})^{-1} g (xy^{-1}) = g$.
The step which gets you that final equality is one that it's well worth learning: it's a manipulation which turns up all the time.