# Group homomorphism proof

Let $\phi : G \rightarrow G'$ be a group homomorphism. Show that if $|G|$ is finite, then $|\phi[G]|$ is finite and is a divisor of $|G|$.

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What have you tried? What do you know? Also please formulate it as a question. – Julian Kuelshammer Nov 17 '12 at 10:48
Here's my guess on why the question was downvoted: it looks like a homework question. If it is, then you should probably add the homework tag and follow these rules. – Dan Shved Nov 17 '12 at 10:48
Understood. Not a homework question, just reading through my book and wondering how to solve. Thanks. – achacttn Nov 17 '12 at 11:02

## 1 Answer

HINT: If $A$ is any finite set, and $f:A\to B$ is any surjection, then $B$ is finite, so the finiteness of $|\varphi[G]|$ is immediate. For the rest, use the fact that $\varphi[G]\cong G/\ker\varphi$, and $\ker\varphi$ is a subgroup of $G$.

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Is it a requirement for every group homomorphism to be surjective? – achacttn Nov 17 '12 at 10:58
@FortDover: No, and $\varphi$ may not map $G$ onto $G'$; but $\varphi$ definitely does map $G$ onto $\varphi[G]$, by definition. – Brian M. Scott Nov 17 '12 at 10:59