Cyclicness is invariant under isomorphism

Dependencies:

1. Isomorphism on Groups
2. Mapping of power is power of mapping

If $\langle a \rangle \cong H$ via $\phi$, then $H = \langle \phi(a) \rangle$. This means that $H$ is cyclic.

Proof

\begin{align} & g \in \langle \phi(a) \rangle \\ &\iff \exists k, g = \phi(a)^k \\ &\iff \exists k, g = \phi(a^k) \\ &\iff g \in \phi(\langle a \rangle) = H \end{align}

Therefore, $\langle \phi(a) \rangle = H$.

Dependency for:

1. Zm × Zn is isomorphic to Zmn iff m and n are coprime

Info:

• Depth: 3
• Number of transitive dependencies: 4

Transitive dependencies:

1. Group
2. Homomorphism on groups
3. Mapping of power is power of mapping
4. Isomorphism on Groups