First isomorphism theorem $G=\mathbb{Z}/60\mathbb{Z}=<\overline{1}>$

Question

I have group G and this homomorphism,

$\begin {array}{rccl} \phi \colon & G & \longrightarrow & G \\ & \overline{a} &\longmapsto & \overline{a^3}\end{array}$

What is $Im\phi$??

Ker$\phi=\{\overline{a}:\overline{a^3}=\overline{1}\}$

Answer 1

From the OP we have $G=\Bbb Z_{60}^+$ ; $G$ is the additive group of the ring $\frac{\Bbb Z}{60\Bbb Z}$ ; $\phi:\bar a\mapsto\overline {3a}$. Since $G$ is cyclic, i.e. $G=\langle\bar 1\rangle$, then $\text{Im}(\phi)$ is cyclic with generator $\phi(\bar 1)$. $$\text{Im}(\phi)=\phi[G]=\phi[\{n\cdot\bar 1:n\in\Bbb Z\}]=\{n\cdot\phi(\bar 1):n\in\Bbb Z\}=\langle\phi(\bar 1)\rangle=\langle\overline{3}\rangle$$ $$\ker(\phi)=\{\bar x\in G:\overline{3x}=\bar 0\}=\{\bar 0,\overline{20},\overline{40}\}$$