I have to prove that, if $G$ and $H$ are finite groups and $f: G \rightarrow H$ an epimorphism (an surjective homomorphism) then $|Ker(f)| = n \Rightarrow |f^{-1}[\{h\}]|=n$, $\forall h \in H$
Given an $ h \in H$ as f is surjective then $\exists g \in G $ such that $f(g)=h$ and my idea is to show that $gKer(f) = f^{-1}[\{h\}]$, is clear that if $a \in gKer(f) \Rightarrow a = gb $ for some $b \in ker(f)$ hence $f(a) = f(gb) = f(g)f(b) = h \Rightarrow a \in f^{-1}[\{h\}]$ since $b \in Ker(f) \Rightarrow f(b) = id$
$\therefore gKer(f) \subseteq f^{-1}[\{h\}]$ I need a hint to prove $f^{-1}[\{h\}] \subseteq gKer(f)$
Best Answer
Take $a \in f^{-1}[\{h\}]$. Then $f(a) = h = f(g)$ and so $f(g^{-1}a) = f(g^{-1})f(a) = f(g)^{-1}f(a) = 1$, that is, $g^{-1}a \in \ker f$, which means $a \in g \ker f$.