# Homomorphism between finite groups

I have to prove or disprove the following statement:

If $\phi:G \rightarrow H$ is a homomorphism between finite groups, with non-trivial image (i.e. $\phi(G)\neq\{e_H\}$), then $\#G$ and $\#H$ have a common divisor $>1$.

I think this statement is true because of the following reasoning:

Define $\ker(\phi)=K$, by the first isomorphism theorem we know that $K \triangleleft G$. Thus $\#K \mid \#G$ by Lagrange's theorem.

Now by the first isomorphism theorem we also know that $\psi: G/K \rightarrow img(\phi)$ is an isomorphism and $img(\phi)\leq H$. (I'm not sure if this statement is true, at first I thought that $img(\phi)=H$ but an homomorphism isn't always surjective. It seems obvious to me that $img(\phi)$ should be a subgroup of $H$ but I wouldn't know how to show this. I'm assuming that this is true though). Thus $\# img(\phi) | \# H$.

$\mid G/K\mid=\frac{\mid G\mid}{\mid K \mid}=\mid img(\phi) \mid$ rewriting this statement gives $\frac{|G|}{|img(\phi)|}=|K|$.

So my conclusion is that $img(\phi)$ divides both $\#G$ and $\#H$. And because $\phi$ has a non-trivial image this statement is true.

I would really like to know if this proof is correct and if so why the statement $img(\phi)\leq H$ is true.

• It is correct, go on and try to prove $\operatorname{im}(\varphi)$ is a subgroup of $H$ by the usual subgroup criterion. – Christoph Dec 21 '14 at 9:52
• It is correct. Nice solution. – Crostul Dec 21 '14 at 9:55
• thanks for checking my answer. With homomorphisms I keep forgetting that $\phi(xy)=\phi(x) \phi(y)$. This makes it easy to prove that the statement is true! – DeanTheMachine Dec 21 '14 at 9:56

$|G|=|Ker(\varphi)|\cdot|Im(\varphi)|$ and $|Im(\varphi)|\cdot |H:Im(\varphi)|=|H|$. Hence their common divisor is at least $|Im(\varphi)|$.