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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.

3 Answers 3

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

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    @amWhy: He is always complete and brief in answers.2012-12-07
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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 $4$.