# If $\phi : G \rightarrow H$ is a group homomorphism and $G$ is cyclic, prove that $\phi(G)$ is also cyclic

Quoting " If $\phi : G \rightarrow H$ is a group homomorphism and $G$ is cyclic, prove that $\phi(G)$ is also cyclic."

Given that $\phi$ is a homomorphism, the group operation must be preserved: Given any $g_1,g_2 \in G$ and $h_1,h_2 \in H$ be the respective images of $g_1,g_2$.

$$\phi (g_1g_2)=\phi(g_1)\phi(g_2)=h_1h_2$$

Given G is a cyclic let $a$ be the generator: $G=<a>=\{a^n : n \in \Bbb Z\}$ Let's assume that H is a cyclic group such that $H=<b>=\{b^v, v \in \Bbb Z\}$ . Let $g_1=a^m$, $g_2=a^n$,$h_1=b^k$, and $h_2=b^r$. It follows that:

$$\phi(a^ma^n)=b^k b^r <=>\phi(a^{m+n}) =b^{k+r}$$

I believe, I need to show that through the homomorphism $\phi$, there exits an element $b$ that can generate the elements in $H$ that are the images of G .

As homomorphism sends identity to identity, using the same exponent operation as on the generator a, one can generate, using $b$, the identities: $$\phi(e_G=a^0=a^{m-m})=b^{m-m}=b^0=e_H, \space m \in \Bbb Z$$

As homomorphism sends inverse to inverse: $$\phi(a^{-m})=b^{-m}, \space m \in \Bbb Z$$

Therefore, as $b$ can generate all the elements of $\phi(G)$, $\phi(G)$ is also cyclic.

Any input is much appreciated.

• Take $\phi(a)$, then show that $\langle \phi(a) \rangle = \phi(G)$. Since $G$ is cyclic, if $f:G \rightarrow K$ is a group homomorphism, then $f$ is determined by its image on $a$. – Gilberto López Apr 19 '17 at 2:35

You can prove directly (without assuming that $H$ is cyclic) that the image of a generator for $G$ is a generator for the image of $G$. Here is the argument:
Let $a$ be a generator for $G$. For any $b\in \phi(G)$ there is $g\in G$ such that $\phi(g) = b$. Choose integer $k$ for which $g = a^k$. Then since $\phi$ is homomorphism, we have \begin{equation*} b = \phi(g) = \phi(a^k) = \phi(a)^k. \end{equation*}
If G is a cyclic group then $\exists g \in G$ such that $G=\{g^n : n \in \mathbb{N}\}$. Let $\phi: G \to H$ group homomorphism. Note that $\phi(g^k)=\phi(g)^k$ $\forall k \in \mathbb{N}$. This proves the desired result.