Isomorphism of cyclic groups $U(p)$ and $\mathbb{Z}_{p-1}$

Question

For prime $p$ the group $U(p)$ is isomorphic to $\mathbb{Z}_{p-1}$.

Considering the case for $U(19)$ it is isomorphic to $\mathbb{Z}_{18}$.

Since isomorphism preserves order of elements so order of an element in $U(19)$ must be equal to order of the element in $\mathbb{Z}_{18}$.

But I calculated order of $8$ in $U(19)$ is $6$ while order of $8$ in $\mathbb{Z}_{18}$ is $9$. How is it possible?

Answer 1

You should find some generating group $U(19)$. After that construct an isomorphism. In this case $8\in U(19)$ is not mapped to $8\in \mathbb{Z}_{18}$ at all.

Addendum.

Since $\langle2\rangle=U(19)$, the mapping $k\to2^k$ is an isomorphism of the group $\mathbb{Z}_{18}$ to $U(19)$. Therefore, $8\to2^8$ or $3\to2^3=8$. But you can construct other isomorphisms of $\mathbb{Z}_{18}$ to $U(19)$.