Epimorphism from $3\mathbb{Z}$ to $U_8$

Question

I'm trying to disprove/prove that an epimorphism exists from $3\mathbb{Z} $ to $U_8$

$U_8$ is acyclic, and is composed out of $\{1,3,5,7\}$. Therefore I can't map every $3k \in 3\mathbb{Z}$ to a cycle, so I'm thinking of creating a composition $f(x) = h(g(x))$ such that: $$g(x) = x \mod 12 \\ h(0) = 1, \: h(3) = 3,\: h(6) = 5,\: h(9) = 7 $$

However I'm stuck proving this is even an homomorphism.

Answer 1

Since all the elements of $U_8$ have order $2$, all the even elements of $\mathbb{Z}$ must map to $1$:$$f(2k)=f(k+k)=f(k)\cdot f(k)=1$$ Suppose that $f(3)=x$. Then $f(-3)=x^{-1}$ Now all the positive odd elements of $3\mathbb{Z}$ map to $x$, since $$f(3(2k+1))=f(6k)f(3)=x,$$and all the negative odd elements map to $x^{-1}=x$, so $f$ cannot be onto.

Answer 2

Consider these:

• $3 \mathbb Z$ is a cyclic additive group.
• A homomorphic image of a cyclic group is cyclic.
• $U_8 \cong C_2 \times C_2$ is not cyclic.