Primitive roots: An advanced approach

Proposition. Let $F$$F$ be a field, then any finite multiplicative subgroup of $(F^{\ast },\cdot )$$(F^*,\cdot)$ is a cyclic group.

Proof. Let $(G,\cdot )\subseteq (F^{\ast },\cdot )$$(G,\cdot)\subseteq (F^*,\cdot)$ be a finite subgroup. Since $G$$G$ is an abelian group, by the structure theorem of finitely generated abelian groups we have

$$\begin{array}{}\text{(1)}& G\cong C_{n_1}\times \cdots \times C_{n_k}\end{array}$$

Now we only need to prove that $\begin{array}{c}gcd(n_i,n_j)=\{\begin{array}{l}1,\text{ if }i\ne j,\\ n_i,\text{ if }i=j.\end{array}\end{array}$$\gcd(n_i,n_j)=\begin{cases}1,\text{ if }i\ne j,\\n_i,\text{ if } i=j.\end{cases}$ Then $G\cong C_{n_1\cdots n_k}.$$G\cong C_{n_1\cdots n_k}.$

Let $s=\mathrm{lcm}(n_1,\dots ,n_k)$$s=\mathrm{lcm}(n_1,\ldots,n_k)$, then for any $a\in G$$a\in G$, we have $a^s-1=0$$a^s - 1 = 0$, so all the elements of $G$$G$ are roots of $x^s-1=0$$x^s - 1 = 0$.

But in a field, $x^s-1$$x^s - 1$ has at most $s$$s$ roots. Hence we have $|G|=n_1\cdots n_k\le s=\mathrm{lcm}(n_1,\dots ,n_k).$$|G|=n_1\cdots n_k \leq s = \mathrm{lcm}(n_1,\ldots,n_k).$

Hence $\begin{array}{c}gcd(n_i,n_j)=\{\begin{array}{l}1,\text{ if }i\ne j,\\ n_i,\text{ if }i=j.\end{array}◻\end{array}$$\gcd(n_i,n_j)=\begin{cases}1,\text{ if }i\ne j,\\n_i,\text{ if } i=j.\end{cases}\quad \square$

Corollary. If $F$$F$ is a finite field, then $(F^{\ast },\cdot )$$(F^*,\cdot)$ is cyclic. In particular, $U(\mathbb{Z}/p\mathbb{Z})$$U(\mathbb{Z}/p\mathbb{Z})$ is cyclic.

Proposition. Let $p$$p$ be a prime number and $p\ne 2$$p\neq 2$, then the unit group of the local ring $\mathbb{Z}/p^n\mathbb{Z}$$\mathbb{Z}/p^n\mathbb{Z}$ is cyclic.

Proof. Since $\mathbb{Z}/p^n\mathbb{Z}$$\mathbb{Z}/p^n\mathbb{Z}$ is a local ring, the unit group is $\mathbb{Z}/p^n\mathbb{Z}-p\mathbb{Z}/p^n\mathbb{Z}.$$\mathbb{Z}/p^n\mathbb{Z} - p\mathbb{Z}/p^n\mathbb{Z}.$

We first claim that:

$$\begin{array}{}\text{(2)}& U(\mathbb{Z}/p^n\mathbb{Z})=\mathbb{Z}/p^n\mathbb{Z}-p\mathbb{Z}/p^n\mathbb{Z}\cong U(\mathbb{Z}/p\mathbb{Z})\oplus (1+p\mathbb{Z}/p^n\mathbb{Z})\end{array}$$

It is clear that

$$\begin{array}{}\text{(3)}& U(\mathbb{Z}/p\mathbb{Z})\cap (1+p\mathbb{Z}/p^n\mathbb{Z})=1.\end{array}$$

To see that

$$\begin{array}{}\text{(4)}& \mathbb{Z}/p^n\mathbb{Z}-p\mathbb{Z}/p^n\mathbb{Z}\cong U(\mathbb{Z}/p\mathbb{Z})\oplus (1+p\mathbb{Z}/p^n\mathbb{Z}),\end{array}$$

we only need to prove that they have the same cardinality.

It follows from

$$\begin{array}{}\text{(5)}& |U(\mathbb{Z}/p\mathbb{Z})\oplus (1+p\mathbb{Z}/p^n\mathbb{Z})|=(p-1)(p^{n-1})=p^n-p^{n-1}=|\mathbb{Z}/p^n\mathbb{Z}-p\mathbb{Z}/p^n\mathbb{Z}|.\end{array}$$

Now we know that

$$\begin{array}{}\text{(6)}& U(\mathbb{Z}/p\mathbb{Z})\cong C_{p-1},2\mid (p-1).\end{array}$$

If we can prove that

$$\begin{array}{}\text{(7)}& 1+p\mathbb{Z}/p^n\mathbb{Z}\cong C_{p^{n-1}},\end{array}$$

then we finish the proof.

Lemma.

(1) Let $p$$p$ be a prime number and $k\ge 1$$k\geq 1$, if $a\equiv b\mathrm{mod}{p}^k$$a\equiv b\mod{p^k}$, then $a^p\equiv b^p\mathrm{mod}{p}^{k+1}.$$a^p\equiv b^p\mod{p^{k+1}}.$

(2) If $p\ne 2$$p\neq 2$ and $k\ge 2$$k\geq 2$, then $(1+cp{)}^{p^{k-2}}\equiv 1+c{p}^{k-1}\mathrm{mod}{p}^k.$$(1 + cp)^{p^{k-2}}\equiv 1 + cp^{k-1}\mod{p^k}.$

Proof.

(1) Let $a=b+c{p}^k$$a = b + cp^k$, then

$$\begin{array}{}\text{(8)}& a^p=b^p+(\genfrac{}{}{0ex}{}{p}{1})b^{p-1}c{p}^k+\sum _{i=2}^p{\lambda }_i{p}^{ki}\equiv b^p\mathrm{mod}{p}^k.\end{array}$$

(2) Consider induction on $k$$k$. It is obviously true when $k=2$$k = 2$. Assume it is true for $k\ge 2$$k \geq 2$:

$$\begin{array}{}\text{(9)}& (1+cp{)}^{p^{k-2}}\equiv 1+c{p}^{k-1}\mathrm{mod}{p}^k.\end{array}$$

From (1), we know that

$$\begin{array}{}\text{(10)}& (1+cp{)}^{p^{k-1}}\equiv (1+c{p}^{k-1}{)}^p\mathrm{mod}{p}^{k+1}.\end{array}$$

Expanding the right-hand side, we get

$$\begin{array}{}\text{(11)}& (1+cp{)}^{p^{k-1}}\equiv 1+(\genfrac{}{}{0ex}{}{p}{1})c{p}^{k-1}+(\genfrac{}{}{0ex}{}{p}{2})c^2{p}^{2(k-1)}+\cdots +c^p{p}^{p(k-1)}\mathrm{mod}{p}^{k+1}.\end{array}$$

Since $\mathrm{\forall }n>1,p^{1+n(k-1)}\mid (\genfrac{}{}{0ex}{}{p}{n})c^n{p}^{n(k-1)}$$\forall n > 1, p^{1 + n(k-1)} \mid \binom{p}{n} c^n p^{n(k-1)}$ and $1+n(k-1)\ge k+1$$1 + n(k-1) \geq k+1$, we conclude:

$$\begin{array}{}\text{(12)}& (1+cp{)}^{p^{k-1}}\equiv 1+c{p}^k\mathrm{mod}{p}^{k+1}.◻\end{array}$$

Proposition. Let $1+cp\in 1+p\mathbb{Z}/p^n$$1 + cp \in 1 + p\mathbb{Z}/p^n$ and $c\notin p\mathbb{Z}/p^n\mathbb{Z}$$c \notin p\mathbb{Z}/p^n\mathbb{Z}$, then $⟨1+cp⟩\cong (\mathbb{Z}/p^{n-1}\mathbb{Z},+)$$\langle 1 + cp \rangle \cong (\mathbb{Z}/p^{n-1}\mathbb{Z},+)$.

Proof. The order of $1+cp$$1 + cp$ has to divide $p^{n-1}$$p^{n-1}$.

By Lemma (2), we know that $(1+cp{)}^{p^{n-2}}\equiv 1+c{p}^{n-1}\mathrm{mod}{p}^n$$(1 + cp)^{p^{n-2}}\equiv 1 + cp^{n-1}\mod{p^n}$, and if $p$$p$ does not divide $c$$c$, then $1+c{p}^{n-1}\ne 1$$1 + cp^{n-1} \neq 1$ in $\mathbb{Z}/p^n\mathbb{Z}.◻$$\mathbb{Z}/p^n\mathbb{Z}. \square$

Hence $U(\mathbb{Z}/p^n\mathbb{Z})\cong U(\mathbb{Z}/p^n\mathbb{Z})\cong C_{p-1}\times C_{p^{n-1}}\cong C_{(p-1)p^{n-1}}=C_{\phi (p^n)}.$$U(\mathbb{Z}/p^n\mathbb{Z}) \cong U(\mathbb{Z}/p^n\mathbb{Z}) \cong C_{p-1}\times C_{p^{n-1}} \cong C_{(p-1)p^{n-1}} = C_{\varphi(p^n)}.$