Prove the two preimage sets have the same size under a surjective group homomorphism (for a finite group)

Question

So I have a finite group $G$ and a surjective group homomorphism $\phi: G \to G'$. I was asked to show that for any $b,c\in G'$, $|\phi^{-1}(\{b\})|=|\phi^{-1}(\{c\})|$. As a hint, it told me to use the fact that $\phi(a)=\phi(b)$ iff $a^{-1}b\in\ker(\phi)$.

I first tried by choosing any $b_1,b_2\in\phi^{-1}(\{b\})\subseteq G$ and similarly for $c_1$ and $c_2$. Because the $b_i$'s are in the preimage of $b$, they map to the same element, i.e. $b$. Then I invoked the hint and I get $b_1^{-1}b_2\in\ker(\phi)$ and a similar result for the $c_i$'s. Now I am stuck and I have no clue how to proceed.

I thought of constructing a bijection between the preimage sets but I couldn't get anything out of it. Would using the left regular representation of $G$ help in anyway? I know it is supposed to feel something like a permutation of the group elements (and a permutation is a bijection). Would this idea prove the theorem if I apply it to the preimage sets?

Thanks in advance!

Answer 1

By the homomorphism theorem, $G/\ker(\phi)$ is isomorphic to $G'$, as $\phi$ is surjective.

The isomorphism is given by $gH\mapsto \phi(g)$, where $H=\ker(\phi)$.

The preimage of $g'\in G'$ is $gH$ (left coset), where $\phi(g)=g'$.

All left cosets have the same cardinality by the bijection $H\rightarrow gH:h\mapsto gh$.

Answer 2

We have $$\phi^{-1}(b)=\{kb\mid k\in\ker(\phi)\}\\\phi^{-1}(c)=\{kc\mid k\in\ker(\phi)\}$$ so both sets have the same size. To see this let $x,y\in\phi^{-1}(b)$, then $$\phi(x)=b\\\phi(y)=b$$ so we have $\phi(xy^{-1})=e$, that is, $xy^{-1}\in\ker(\phi)$. Therefore $x=(xy^{-1})y$, so all elements differ by an element in the kernel.

Answer 3

Pre-imaging establishes a partition of the domain for any map $f\colon X\to Y$, via the equivalence relation $x\sim x' \stackrel{(def.)}{\iff} f(x)=f(x')$. If $f$ is a group homomorphism, then the equivalence classes are the cosets of $\operatorname{ker}f$ (as hinted), which are all mutually equinumerous.