Group homomorphism question. T/F? There exist two groups $G$ and $H$ such that $|G|=12, |H|=8$, and there is a surjective homomorphism $\phi: G \to H$.
Answer says "False, because $|\phi [G]|$ must divide $|G|$."
Can someone explain this? Thank you.
 A: Let $G$ and $H$ be groups with $|G|=12$ and $|H|=8$. Seeking a contradiction, suppose there exists a surjective group homomorphism $\phi:G\rightarrow H$. Then the first isomorphism theorem ensures a group-isomorphism $H\simeq G/\mathrm{Ker}\,\phi$. It follows that $8=12/|\mathrm{Ker}\,\phi|$ so that $|\mathrm{Ker}\,\phi|=3/2$, a contradiction. Hence there exists no surjective group homomorphism $G\rightarrow H$.
A: Let's see why their claim is true, just from the basics.
Let $\phi:G\to H$ be any homomorphism, with $G$ a finite group, and consider the sets $$\phi^{-1}(h):=\{g\in G:\phi(g)=h\}.$$
Suppose $h\in H$ such that $\phi^{-1}(h)\ne\emptyset$, and fix any $g\in\phi^{-1}(h).$
For each $k'\in\phi^{-1}(1_H),$ we have that $$\phi(gk')=\phi(g)\phi(k')=h1_H=h,$$ so $gk'\in\phi^{-1}(h)$. Thus, $$|\phi^{-1}(h)|\ge|\phi^{-1}(1_H)|,\tag{1}$$ since $gk'=gk''$ if and only if $k'=k''$. On the other hand, for each $g'\in\phi^{-1}(h)$, we have $$\phi(g^{-1}g')=\phi(g^{-1})\phi(g')=\phi(g)^{-1}\phi(g')=h^{-1}h=1_H,$$ so $g^{-1}g'\in\phi^{-1}(h)$. Thus, $$|\phi^{-1}(h)|\le|\phi^{-1}(1_H)|,\tag{2}$$ since $g^{-1}g'=g^{-1}g''$ if and only if $g'=g''$.
By $(1)$ and $(2)$, we find that if $\phi^{-1}(h)\ne\emptyset,$ then $$|\phi^{-1}(h)|=|\phi^{-1}(1_H)|.$$
It should be clear (since $\phi$ is a function $G\to H$) that the sets $\phi^{-1}(h)$ are pairwise disjoint, and that their union is all of $G$. Thus, $$\begin{align}|G| &= \sum_{h\in H}|\phi^{-1}(h)|\\ &= \underset{\phi^{-1}(h)\ne\emptyset}{\sum_{h\in H}}|\phi^{-1}(h)|\\ &= \underset{\phi^{-1}(h)\ne\emptyset}{\sum_{h\in H}}|\phi^{-1}(1_H)|\\ &= |\phi^{-1}(1_H)|\cdot\bigl|\{h\in H:\phi^{-1}(h)\ne\emptyset\}\bigr|\\ &= |\phi^{-1}(1_H)|\cdot|\phi[G]|.\end{align}$$
Thus, $|\phi[G]|$ must divide $|G|$.
