Let $G,H$ be finite groups where $|G|$ and $|H|$ are coprime. Prove that any homomorphism $\phi :G\rightarrow H$ must be trivial $($ie. $\phi (x)=e_H, $ the identity element of $H, \forall x\in G)$.
We know that $Ker(\phi )$ and $Im(\phi )$ are subgroups of $G$ and $H$, respectively. Then, the Lagrange Theorem asserts that $|Ker(\phi )|$ divides $|G|$ while $|Im(\phi )|$ divides $|H|$.
I am trying to show that $|Im(\phi )|=1 \implies |Ker(\phi )|=|G| \implies \phi$ is trivial.
I can show this last series of implications and the first part separately. How do I make the leap from where I left off to $|Im(\phi )|=1$?
Side note: I am using the Range-Kernel Theorem as well: $|Im(\phi )|\times |Ker(\phi )|=|G|$.
Best Answer
You have the right idea: $|\mathrm{Im}(\phi)|$ divides $|H|$, but on the other hand it is equal to $\frac{|G|}{|\ker(\phi)|}$, which is a divisor of $|G|$. Since $|H|$ and $|G|$ are coprime, this means that $|\mathrm{Im}(\phi)|=1$.