# Fundamental Homomorphism Theorem

Let $$\phi$$ be a homomorphism from a group $$G$$ onto a group $$G'$$. Prove that if $$G$$ is finite, then $$G'$$ is also ﬁnite and $$|G'|$$ divides $$|G|$$.

I know i should be using the Fundamental Homomorphism Theorem.

I also know that the Fundamental Homomorphism Theorem relates the structure of two objects between which a homomorphism is given, and of the kernel and image of the homomorphism. Can you help me start this proof off?

What we do know by the homomorphism theorem, is that if $\phi : G \to G'$ homomorphism, then if $G$ is finite, we have that $\frac{G}{\ker \phi} \cong \operatorname{im} \phi$.

In your case, as $\phi$ is surjective, we have $\frac{G}{\ker \phi} \cong G'$. Now, since the above groups have the same size, $|G| = |G'| \times |\ker \phi|$. Hence, the result follows, as $|\ker \phi|$ is a natural number.

• how is this proving that if G is finite then G' is finite? – rattle Apr 27 '17 at 23:30
• @rochelle Because we have that $G' \cong \frac{G}{\ker \phi}$, so $G'$ is isomorphic to a finite group, hence is finite itself. Then we can take the sizes of the groups to get the conclusion. – астон вілла олоф мэллбэрг Apr 27 '17 at 23:32
• ok i understand and what does it mean to take the sizes of the group to get the conclusion? – rattle Apr 27 '17 at 23:35
• Each group is a set of objects, the cardinality of that set is referred to as the size of the group. So when two sets are bijective, don't we say their cardinality is the same? Here, an isomorphism is a bijection, so we can say that $G'$ and $\frac{G}{\ker \phi}$, as sets, have the same cardinality. This is what is meant by taking the size of both sides. – астон вілла олоф мэллбэрг Apr 27 '17 at 23:39

Onto means surjective. So $$\operatorname{im}(\phi) = G'$$ and by the fundamental theorem,

$$G/\ker \phi \cong \operatorname{im}(\phi) = G'.$$

Now you should have some result telling you that $$|G| = |\ker \phi||G'|$$.