Question:
Show that a homomorphism defined on a cyclic group is completely determined by its action on a generator of the group.
Let $G=\left \langle a \right \rangle$ where $a \in G$ be a cyclic group.
Let $\phi$ be a homomorphism.
$\phi\left ( G \right )=\phi\left ( \left \langle a \right \rangle \right )$ is also a cyclic group.
We want to show that $\phi\left ( \left \langle a \right \rangle \right )=\phi\left ( a \right )$.
Here's my attempt: It looks silly so I'd like some verification.
$\phi\left ( \left \langle a \right \rangle \right )=\phi\left \{ e,a,\cdot \cdot \cdot ,a^{n-1} \right \}$
=$\left \{ \phi\left ( e \right )=e,\phi\left ( a \right ),\cdot \cdot \cdot ,\phi\left ( a^{n-1} \right ) \right \}$
=$\left \{ e,\phi\left ( a \right ),\cdot \cdot \cdot ,\left [ \phi\left ( a \right ) \right ]^{n-1}\right \}$
=$\left \langle \phi\left ( a \right ) \right \rangle$
Thanks in advance.
Best Answer
It looks fine. You made a small mistake near the start: when you wrote “$\phi(\langle a\rangle)=\phi(a)$”, what you meant was “$\phi(\langle a\rangle)=\langle\phi(a)\rangle$”.