# Prove that if $\phi: G\to H$ is a homomorphism and $G_{1}\leq G$ is cyclic, then $\phi (G_{1})$ is cyclic.

Would someone tell me whether my proof is correct or is it missing anything important, please?

Prove that if $$\phi: G\to H$$ is a homomorphism and $$G_{1}\leq G$$ is cyclic, then $$\phi (G_{1})$$ is cyclic.

Let $$G_{1}$$ be cyclic. Then $$G_{1}=\left \langle g \right \rangle$$ for some $$g\in G_{1}$$. Since $$\phi(G_{1})=\left \{ \phi(g_{1}):g_{1}\in G_{1} \right \}$$ and $$g_{1}=g^{k}$$ for all $$g_{1}\in G_{1}$$ and $$k\in \mathbb{Z}$$, by definition of homomorphisms, $$\phi(g_{1})=\phi(g^{k})=(\phi(g))^{k}$$ for all $$\phi(g_{1})\in\phi(G_{1})$$. Then $$\phi(G_{1})=\left \langle \phi(g) \right \rangle$$ hence cyclic.

• Looks fine to me. Commented Mar 14, 2018 at 23:30
• @EthanBolker Thank you for taking the time to look through it :)
– user482939
Commented Mar 14, 2018 at 23:52
• Compare also with other posts here, e.g., this one. Commented Dec 21, 2019 at 15:11

The phrasing at the end, though, is a little off; try a new sentence reading "Hence $$\phi(G_1)$$ is cyclic." It's just a minor suggestion.