Originally Posted by

**AlexP** True...the componentwise part doesn't make sense without the rest. Here's the whole proof. There's another part I don't understand so I'll have that part ready too."

"To see that the multiplicative group $\displaystyle \mbox{GF}(p^n)^\times$ of nonzero elements of $\displaystyle \mbox{GF}(p^n)$ is cyclic, we first note...that it is isomorphic to a direct product of the form $\displaystyle \mathbb{Z}_{n_1} \otimes \mathbb{Z}_{n_2} \otimes \cdots \otimes \mathbb{Z}_{n_k}$, where each $\displaystyle n_{i+1}$ divides $\displaystyle n_i$. So, for any element $\displaystyle a=(a_1, a_2, ..., a_k)$ in this product, we have $\displaystyle a^{n_1} = (n_1a_1, n_1a_2, ..., n_1a_k)$. (Remember, the operation is componentwise addition.)

Thus, the polynomial $\displaystyle x^{n_1}-1$ has $\displaystyle p^n-1$ zeros in $\displaystyle \mbox{GF}(p^n)$. Since the number of zeros of a polynomial over a field cannot exceed the degree of the polynomial, we know that $\displaystyle p^n-1 \le n_1$. On the other hand, since $\displaystyle \mbox{GF}(p^n)^\times$ has a subgroup isomorphic to $\displaystyle \mathbb{Z}_{n_1}$, we also have $\displaystyle n_1 \le p^n-1$. It follows, then, that $\displaystyle \mbox{GF}(p^n)^\times$ is isomorphic to $\displaystyle \mathbb{Z}_{p^n-1}$."