Why is this reasoning for proving that $\mathbb{Z}_4 \not\cong U(8)$ sound or unsound?

Question

Here is my proof that $\mathbb{Z}_4 \not\cong U(8)$.

1. We know that $\mathbb{Z}_4 \cong U(5)$ due to the fact that every element of a cyclic group of prime order is a generator of the group.

2. So what we really need to determine is whether or not $U(8) \cong U(5)$.

3. If $U(8)$ really were isomorphic to $U(5)$, then the projection epimorphism $\pi : U(8) \rightarrow U(5)$ st $x \mapsto y$ where $y$ is an element such that $x \equiv y \mod 5$ should preserve group structure.

4. But $\pi(5) = 0$ which obviously means the projection operator isn't even well-defined, let alone preserves group structure.

5. Therefore $\mathbb{Z}_4 \not\cong U(8)$.

I know that the conclusion of this argument is correct, because $U(8) \cong V_4 \not\cong \mathbb{Z}_4$. But I am not sure if the reasoning that reached this conclusion is sound. Why or why not?

Answer 1

Neither $Z_4$ nor $U(5)$ are groups of prime order, so I don't understand your reasoning for step (1). Why not just use the fact that since $3^2 (mod\;8) = 1, 5^2(mod\;8) = 1$, and $7^2(mod\;8) = 1$, then every non-identity element of $U(8)$ has order $2$, but in $Z_4$, $3$ is a non-identity element with order $4$, thus $Z_4$ cannot be isomorphic to $U(8)$.