1. ## isomorphism

prove:U(p)is isomorphic to Z(p-1),where p is prime.Z(p-1)is the additive group of modulo p-1.U(p) means all the positive intergers that is relatively prime to p.
Thanks a lot!!

2. ## More General result

In fact this is a consequence of a more general theorem.

A finite subgroup of the multiplicative group of a field is cyclic. In particular, if F is a finite field, then the multiplicative group $\displaystyle F^{\times}$ of nonzero elements of F is a cyclic group.

By The Fundamental Theorem, the finite subgroup can be written as a direct product of cyclic groups

$\displaystyle \mathbb{Z}/\mathbb{Z}_{n_1}\times \mathbb{Z}/\mathbb{Z}_{n_2} \times ... \times \mathbb{Z}/\mathbb{Z}_{n_k}$

Where $\displaystyle n_k|...|n_2|n_1$.

In general, if G is a cyclic group and $\displaystyle d||G|$, then G contains precisely d elements of order dividing d. Since $\displaystyle n_k$ divides the order of each of the cyclic groups in the direct product, it follows that each direct factor contains $\displaystyle n_k$ elements of order dividing $\displaystyle n_k$. If k were greater than 1, there would therefore be a total of more than $\displaystyle n_k$ such elements. But then there would be more than $\displaystyle n_k$ roots of the polynomial $\displaystyle p(x)=x^{n_k}-1$ in the field F, which contradicts the fact that a polynomial of degree n can have at most n roots. Thus k=1 and the group is cyclic.

Notice $\displaystyle \mathbb{Z}_p$ is a finite field, so your question is a trivial corollary of the above theorem.

3. Thank you for your solution! In your solution, there are some theorems that I still don't know. I am a newcomer of abstract algebra. So many things to be studied.....