TITLE: Homomorphism between finite groups; showing the order of the image divides the orders of each group
QUESTION [12 upvotes]: 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!

REPLY [11 votes]: 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!!!