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

  • $\begingroup$ 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 $H$? $\endgroup$ – Berci Jan 4 at 10:03
  • $\begingroup$ @Berci its $\bar6$ $\endgroup$ – 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.

  • $\begingroup$ Yes I see now that $G$ needs to be of order 8 for this example to work $\endgroup$ – user3133165 Jan 4 at 10:33
  • $\begingroup$ Yes, that is one of the possibilities for $|G|$. $\endgroup$ – ArsenBerk Jan 4 at 10:37
  • 1
    $\begingroup$ Thank you for the clarification yes if it is of order 8 then it will work. $\phi(a^4)=\phi(a^8=e_G)=\bar6=e_H$ $\endgroup$ – user3133165 Jan 4 at 10:51

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