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
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then use that the kernel of a homomorphism is a normal subgroup and the kernel defines the homormorphism
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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
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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
Best Answer
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 $\gcd(|G|,|H|)=1$.