Clean and clever proofs to show every homomorphism of two groups with coprime orders is trivial? I'm practicing for a qualifying exam in algebra (the freebie attempt we get the week before my first semester of grad school). I don't have a lengthy or deep background with math so I'm especially interested in learning to write cleaner and more clever proofs or proofs that use different methods. Does anyone have feed back on my proof of the titular question or perhaps an alternative proof?


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*use the fact that $G$ and $H$ don’t have any subgroups of the same order besides the subgroup which is just the identity

*then use that the kernel of a homomorphism is a normal subgroup and the kernel defines the homormorphism

*But $G/\ker(f)$ is isomorphic to H. Since the order of $G/\ker(f)$ is a number composed of the product of primes that divide the order of G

*because the order of $G$ is coprime to the order of $H$ we know $|G/\ker(f)| = 1$ and that therefore $f$ maps all of $G$ into the identity of $H$ hence $f$ is a trivial homomorphism
 A: Say $|G|=n$ and $|K|=m$, with $\gcd(m,n)=1$. Find $r,s\in\mathbb{Z}$ such that $rm+sn=1$. Note that if $g\in G$ and $k\in K$, then $g^n=e$, $k^m=e$.
Let $f\colon G\to K$ be a morphism, and let $g\in G$. Then $g=g^{rm+sn} = g^{rm}(g^n)^s=g^{rm}$. So
$$f(g) = f(g^{rm}) = f(g)^{rm} = (f(g)^m)^r = e^r = e.$$
Thus, $f$ is the trivial map.

Feedback: your assertion in (3) that $G/\mathrm{ker}(f)$ is $H$ is incorrect. You can say that $G/\mathrm{ker}(f)$ is isomorphic to its image in $H$, but since you are not assuming $f$ is onto, you can’t assert it is isomorphic to $H$, just to a subgroup of $H$.
The rest of the argument is too convoluted: note that $|G/\mathrm{ker}(f)|$ divides both $|G|$ (since it is equal to $|G|/|\mathrm{ker}(f)|$, and $|H|$ (because it is isomorphic to a subgroup of $H$), hence divides their gcd, which is $1$.
A: The order of the image of $f$ divides the order of $G$ because $\operatorname{im} (f) \cong G/\ker(f)$.
The order of the image of $f$ divides the order of $H$ because $\operatorname{im} (f)$ is a subgroup of $H$.
Therefore, order of the image of $f$ divides $\gcd(|G|,|H|)=1$.
A: Basically the same as lhf's answer, but without the first homomorphism theorem. Say $f\colon G\to H$ such a homomorphism. Then $o(f(g))\mid o(g)$ for every $g\in G$, and hence $o(f(g))\mid |G|$. But $o(f(g))\mid |H|$ (Lagrange). Therefore, $o(f(g))$ divides both $|G|$ and $|H|$, which are coprime. So, $o(f(g))=1$ for every $g\in G$, and hence $f$ is the trivial homomorphism.
A: For any $g\in G$, by "homomorphismness", and Lagrange,  $\phi(g)$ has order dividing $\lvert G\rvert $.  For $$e=\phi(e)=\phi(g^{\lvert G\rvert})=\phi (g)^{\lvert G\rvert}$$.
And by Lagrange,  $\lvert \phi(g)\rvert \mid\lvert H\rvert $.
$\therefore \lvert \phi(g)\rvert =1$, that is $\phi(g)=e$, or $\phi$ is trivial.
