# Finding the kernel of homomorphism between two cyclic groups

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 $$\phi(a^7)=\bar2$$.

• The range of any homomorphism is a subgroup of the codomain isomorphic to a quotient of the domain, so its order must divide both. So there is no nontrivial homomorphism from a group of order 7 to a group of order 4. By the way, what is the identity element of $H$? – Berci Jan 4 at 10:03
• @Berci its $\bar6$ – user3133165 Jan 4 at 10:08

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.
• Yes I see now that $G$ needs to be of order 8 for this example to work – user3133165 Jan 4 at 10:33
• Yes, that is one of the possibilities for $|G|$. – ArsenBerk Jan 4 at 10:37
• Thank you for the clarification yes if it is of order 8 then it will work. $\phi(a^4)=\phi(a^8=e_G)=\bar6=e_H$ – user3133165 Jan 4 at 10:51