0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $a$. $\endgroup$ – Gilberto López Apr 19 '17 at 2:35
3
$\begingroup$

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*}

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.