I have two cyclic group $\langle G,*\rangle $ with order $7$ and a generator $a\in{G}$, $\langle H=\{\bar2,\bar4,\bar6,\bar8\},\cdot_{10}\rangle $.
I also have a homomorphism $\phi:G\rightarrow H , $where $ \phi(a)=\bar8$.
I have tried to see where each power of $a$ maps to and saw that :
$$\begin{align}
\phi(a)&=\bar8,\\
\phi(a^2)&=\bar4,\\
\phi(a^3)&=\bar2,\\
\phi(a^4)&=\bar6=e_H,\\
\phi(a^5)&=\bar8,\\
\phi(a^6)&=\bar4,\\
\phi(a^7)&=\bar2
\end{align}$$
I am a bit confused becauase I know that if $\phi$ is homomorphism then the identity element $a^7=e_G$ must map to the identity element of $H$. But I have $\phi(a^7)=\bar2$.
Best Answer
Note that if $\phi : G \to H$ is a homomorphism and for $g \in G$, $o(g) = n$, then $o(\phi(g))\mid n$.
So, if we suppose $\phi$ is a homomorphism in you example, then since $a$ is a generator of a cyclic group of order $7$, we have $o(a) = 7$. So, we must have $o(\phi(a)) \mid 7 \implies o(\phi(a)) = 1$ or $o(\phi(a)) = 7$. Latter case is not possible since $|H| = 4$. And for the former case, we must send $a$ to $\bar{6}$ (identity of $H$), which is not the case. So, our assumption of $\phi$ being a homomorphism is contradictory. From this, we can also conclude that in your example, if $\phi: G \to H$ is a homomorphism, then it must be trivial.