Let G be a group and let $\displaystyle A \leq G$ be a subgroup. If $\displaystyle g \in G$, then $\displaystyle A^g \subseteq G$ is defined as

$\displaystyle A^g=\{a^g | a \in A \}$ where $\displaystyle a^g=g^{-1}ag \in G$.

Show that the sets $\displaystyle A$ and $\displaystyle A^g$ have the same cardinality.

Hint: sets $\displaystyle A$ and $\displaystyle A^g$ have the same cardinality if and only if there exists an injective and surjective map $\displaystyle f: A \to A^g$.

Attempt: So, in order to show that the mapping f is injective I have to show that

$\displaystyle f(a)=f(a^g)=f(g^{-1}ag) \iff a = a^g$ (for $\displaystyle a \in A$ and $\displaystyle a^g \in A^g$).

But how can I deduce that when I don't know what the function f is?