Homomorphism between finite groups; showing the order of the image divides the orders of each group

I have been given a question:

Let G and H be finite groups and $f:G\rightarrow H$ a homomorphism. Show that $|f(G)|$ divides both $|G|$ and $|H|$.

So I try to proceed as follows:

let K = ker(f)

then |f(G)| = |G/K| = [G : K]

(Correct? I believe this is is a consequence of the first Iso Thm.)

Now by Lagrange, the index [G:K] divides |G|. But I am at a loss trying to show that |f(G)| divides |H|. One stumbling block for me is that f is not on to, so how to I account for the other members of H?

I think a little bit of direction here could quickly help me vastly improve my understanding of this material; I have only been introduced to it over the last week or so and it is not resonating yet. Thank you for any help!

I think the key here is to realize that $f(G)$ is a subgroup of $H$; thus $\vert f(G) \vert \mid \vert H \vert$ by Lagrange's theorem. Furthermore, $\ker f$ is a (normal) subgroup of $G$, with $G / \ker f$ isomorphic to $f(G)$. Thus $\vert G \vert = [G:\ker f] \vert \ker f \vert = \vert f(G) \vert \vert \ker f \vert$, showing $\vert f(G) \vert \mid \vert G \vert$. In fact, we have established the tidy relationship $\vert G \vert = \vert \ker f \vert \vert f(G) \vert = \vert \ker f \vert \vert \text{im} f \vert$. QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!