# Homomorphisms and Calculating Elements

1. Nov 13, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
Suppose that $G$ is a cyclic group with generator $g$, that $H$ is some arbitrary group, and that $\phi : G \rightarrow H$ is a homomorphism. Show that knowing $\phi (g)$# lets you compute $\phi(g_1)$ $\forall g_1 \in G$

2. Relevant equations

$\phi(g^n) = \phi(\underbrace{g~\star_G~ g~ \star_G~ g ... \star_G~ g}_{n|}) = \underbrace{\phi (g) \star_H \phi(g) \star_H ... \star_H \phi(g)}_{|n|} = \phi(g)^n$

3. The attempt at a solution
Let $g_1 \in G$ be arbitrary. This implies that $\exists k \in \mathbb{Z}$ such that $g_1 = g^k$.

$\phi(g_1 \star_G g ) = \phi(g_1) \star_H \phi(g) \iff$

$\phi(g^k \star_G g) = \phi(g_1) \star_H \phi(g) \iff$

$\phi(g^{k+1}) = \phi(g_1) \star_H \phi(g) \iff$

$\phi(g)^{k+1} \star_H \phi(g)^{-1} = \phi(g_1) \iff$

$\phi(g)^k = \phi(g_1)$

This seems correct, but I am unsure. If this is correct, though, does this imply that $H$ is also cyclic, where $\phi(g)$ is the generator?

2. Nov 13, 2014

### Dick

Wouldn't it be much simpler just to say $\phi(g_1)=\phi(g^k)=\phi(g)^k$?? I don't know what all the extra stuff is for. And no, it doesn't mean H is cyclic. It does mean that the image of G under $\phi$ is a cyclic subgroup of H.

3. Nov 14, 2014

### Bashyboy

Hmm...yes, it would have been terribly simpler, had I done that...