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Let $f: G\to H$ be a homomorphism and let $|G| = 10$ and $|H| = 4$

Prove that f is not onto.

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What have you tried? –  Tobias Kildetoft Dec 7 '12 at 14:44

3 Answers 3

Hint: Order of a subgroup always divides the order of the group, and then use the isomorphism theorems.

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+1 Asaf! (We haven't "crossed paths" as much recently; I hope you're not avoiding me!) –  amWhy Dec 7 '12 at 14:55
    
Oh, I am not avoiding you! I've seen your answers around, I just had nothing to add... :-) –  Asaf Karagila Dec 7 '12 at 15:05
    
@amWhy: He is always complete and brief in answers. –  Babak S. Dec 7 '12 at 15:26

By the first isomorphism theorem, $G/\ker f \cong \mathrm{im} f$. Also, the kernel is a subgroup of $G$ hence its order has to divide $10$ by Lagrange. Hence the possibilities for the order of the kernel of $f$ are $2$, $5$ or $1$. (If the order is $10$ then $f$ is the zero map and of course not surjective onto $H$.)

If the order of the kernel was $2$ (or $1$) then the order of the image would be $5$ (or $10$). Of course this is not possible since $H$ has only 4 elements.

If the order of the kernel is $5$ then the order of the image is $2$ so that $f$ is not onto surjective.

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$|\mbox{im} f| = (G:\ker f)$ divides |$G|=10$ and so $|\mbox{im} f|$ cannot be $4$.

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