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 $|G|$.
Group homomorphism proof
-1
$\begingroup$
abstract-algebra
-
0Understood. Not a homework question, just reading through my book and wondering how to solve. Thanks. – 2012-11-17
1 Answers
6
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 $G$.
-
0@FortDover: No, and $\varphi$ may not map $G$ onto $G'$; but $\varphi$ definitely *does* map $G$ onto $\varphi[G]$, by definition. – 2012-11-17