Show that the size of the pre-image $(\psi(x))$ is the same for all $x$. What is this size equal to?

Question

Let $\psi: G \rightarrow H$ be a homomorphism, and let $K=\operatorname{ker}(\psi)$. Define the canonical homomorphism $\varphi: G \rightarrow G/K$ by $\varphi(x) = xK$.

Show that the size of the pre-image $(\psi(x))$ is the same for all $x$. What is this size equal to? .

My thoughts:
First I showed that $\varphi^{-1}(x) =xK$ . and i know that logically this would be the size of the pre image of $\psi$ but I dont know how to show it.

Answer 1

We want to show that $\psi^{-1}(\psi(x))=xK$ for any $x \in G$, where $K=ker(\psi)$

Let $x\cdot k \in xK$, then $\psi(x\cdot k)=\psi(x)$ because $k\in K=ker(\psi)$, thus $xK\subseteq \psi^{-1}(\psi(x))$.

Now, let $g \in \psi^{-1}(\psi(x))$, then $\psi(g)=\psi(x)$ and $\psi(x \cdot g^{-1})=1_H$, meaning $x \cdot g^{-1}=k$ for some $k \in K$. Thus $g=x \cdot k^{-1} \in xK$, meaning $xK \subseteq \psi^{-1}(\psi(x))$

Now, from Lagrange's Theorem on subgroups, we know that all cosets of a subgroup are equal sized. In this case, we get $xK=x'K$ for any $x,x' \in G$. But we showed that $\psi^{-1}(\psi(x))=xK$ for any $x \in G$, then all pre-images of $\psi(x)$ all equal sized.

Answer 2

I assume $G$ is finite, or at least that $K$ is finite index? Anyway, in that case you're almost there--you need only show that the map [of sets] $K\to xK$ defined by $k\mapsto xk$ is a bijection. It's obviously surjective by definition of the coset $xK$, and if $xk=xk'$ we can left-multiply by $x^{-1}$ to yield $k=k'$. Now of course $|xK|=|x'K|$ for any $x,x'\in G$ since we just showed each of those cosets is the same size as $|K|$.