Order of image of homomorphism divides order of G

[Mr Davis 97]

Homework Statement

Let $\phi : G \rightarrow G'$ be a group homomorphism. Show that if $| G |$ is finite, then $| \phi [G] |$ = $| I am (\phi)|$ is finite and a divisor of $| G |$

Homework Equations

The Attempt at a Solution

Assume that G is finite, with order n. Then im(G) is a subgroup of G' and has at most n elements, so must also be finite. Let im(G) have order m.

Next, we know that $\forall a \in G$, $a^n = e$. Also, we know that $\phi : G \rightarrow \phi [ G ]$ is a surjection, so $(\forall a' \in \phi [G]) (\exists a \in G)~ \phi (a) = a'$ Thus, $a^n = e \Rightarrow \phi (a^n) = \phi (e) \Rightarrow \phi (a) ^n = e \Rightarrow (a') ^n = e$, where a' is any element of the image of $\phi$. This means that n is a multiple of the order of a'. But m is also a multiple of the order of a'...

This is as far as I get. I can't seem how to conclude that n is a multiple of m from this information

 

[fresh_42]

The trick is to consider $G'/\Phi(G)$. This divides $G'$ into equivalence classes which are equally big.

 

[pasmith]

If you could show a bijection from the collection of left cosets of $\ker \phi$ to $\phi(G)$, then Lagrange's Theorem would complete the proof.

The trick is to consider $G'/\Phi(G)$. This divides $G'$ into equivalence classes which are equally big.

We are not given that $G'$ is finite.