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 [itex]\ker \phi[/itex] to [itex]\phi(G)[/itex], 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 [itex]G'[/itex] is finite.