Cyclicity of the multiplicative group of the integers modulo a prime

Question

(NB: I'm aware that there's plenty of proofs for the Theorem: here I'm rather interested in whether the guessed argument can be brought to an end.)

Trying to generalize the argument in the Example, I'm looking for a proof of:

• the Lemma 2, based on the Lemma 1 and its Corollary;
• the Theorem, based on the Lemma 2.

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Example. Let $p=7$. In this case, the general notation below specializes into: \begin{alignat}{1} &s_1=1 \\ &s_2=2+4+1 \\ &s_3=3+2+6+4+5+1 \\ &s_4=4+2+1 \\ &s_5=5+4+6+2+3+1 \\ &s_6=6+1 \\ \end{alignat} and, by "completing cyclically the sums and summing over each column": \begin{alignat}{1} &\sigma_1&=1+1+1+1+1+1 \\ &\sigma_2&=2+4+1+2+4+1 \\ &\sigma_3&=3+2+6+4+5+1 \\ &\sigma_4&=4+2+1+4+2+1 \\ &\sigma_5&=5+4+6+2+3+1 \\ &\sigma_6&=6+1+6+1+6+1 \\ \\ &\hline\\ &c_i &= 0\space\space\space\space\space0\space\space\space\space\space0\space\space\space\space\space0\space\space\space\space\space0\space\space\space\space6\space(=p-1) \\ \end{alignat} Now, if $3$ and $5$ had multiplicative order -say- $3$, then we'd have: \begin{alignat}{1} &\sigma_1&=1+1+1+1+1+1 \\ &\sigma_2&=2+4+1+2+4+1 \\ &\sigma_3&=3+2+\color\red1\space... \\ &\sigma_4&=4+2+1+4+2+1 \\ &\sigma_5&=5+4+\color\red1\space... \\ &\sigma_6&=6+1+6+1+6+1 \\ \\ &\hline\\ &c_i &= 0\space\space\space\space\space0\space\space\space\space\color\red4\space\space\space etc. \\ \end{alignat} a "contradiction", etc.

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Now the attempt of the general argument. Let $p$ be an odd prime. Hereafter, all the (non)equalites are modulo $p$.

Lemma 1. For every $k\in(\mathbb Z/p\mathbb Z)^\times$, $k\ne1$: $$s_k:=\sum_{i=1}^{m_k}k^i=0\tag1$$ where $m_k$ is the multiplicative order of $k$.

Proof. By contradiction, suppose $s_k\in(\mathbb Z/p\mathbb Z)^\times$; then, being $ks_k=s_k$, from the cancellation law in $(\mathbb Z/p\mathbb Z)^\times$ follows $k=1$, a contradiction. Therefore, $s_k=0$. $\space\Box$

Corollary. For every $k\in(\mathbb Z/p\mathbb Z)^\times$, $k\ne1$: $$\sigma_k:=\sum_{i=1}^{p-1}k^i=0\tag2$$ Proof. By the Lemma 1, the sum splits into $\frac{p-1}{m_k}$ partial sums, each congruent to $0$ modulo $p$. $\space\Box$

Lemma 2. For every $i=1,\dots,p-2$: $$c_i:=\sum_{k=1}^{p-1}k^i=0\tag3$$ Proof. ...

Theorem. $(\mathbb Z/p\mathbb Z)^\times$ is cyclic.

Proof. By contradiction, suppose that $\mathfrak m:=\max\{m_k\mid k\in(\mathbb Z/p\mathbb Z)^\times\}<p-1$. We will prove then that $c_{i}\ne0$ for some $i\le\mathfrak m$, thus contradicting $(3)$. ...

Answer 1

I believe your lemmas $1$ and $2$ can be combined in the following way :

Lemma : let $G$ be a finite group and $f : G \to (\mathbb{Z}/p\mathbb{Z})^*$ a group homomorphism. If you define $S_f = \sum_{g \in G} f(g)$, then either $S_f = 0$ or $f = 1$.

Proof : take any $g_0 \in G$, we can write :

$$f(g_0) S_f = \sum_{g \in G} f(g_0 g) \underset{\color{blue}{g' = g_0 g}}{=} \sum_{g' \in G} f(g') = S_f$$

So $(1-f(g_0)) \, S_f = 0$. If $f$ is non trivial, we can pick $g_0$ so that $1-f(g_0) \ne 0$, which gives the result. Note that when $f = 1$, then $S_f = |G|$ (modulo $p$).

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From the lemma above, we deduce lemma $1$ or it's corollary by considering either $f_k : i \in \mathbb{Z}/m_k\mathbb{Z} \mapsto k^i \in (\mathbb{Z}/p\mathbb{Z})^*$ for lemma $1$ or $g_k : i \in \mathbb{Z}/(p-1)\mathbb{Z} \mapsto k^i \in (\mathbb{Z}/p\mathbb{Z})^*$ for the corollary. And lemma $2$ follows by considering the homomorphism $\varphi_i : k \in (\mathbb{Z}/p\mathbb{Z})^* \mapsto k^i \in (\mathbb{Z}/p\mathbb{Z})^*$.

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Now to prove your final result, we need to show that the maximum of the $m_k$ is also the lowest common multiple of the $m_k$ :

$$m = \max\{m_k | k \in (\mathbb{Z}/p\mathbb{Z})^*\} = \operatorname{lcm} \{m_k | k \in (\mathbb{Z}/p\mathbb{Z})^*\}$$

To do that, denote $M = \{m_k | k \in (\mathbb{Z}/p\mathbb{Z})^*\}$ the set of all orders. We'll show the set is stable under the lowest common multiple operation (so its maximum is also its lowest common multiple):

First notice that if $x$ is of order $m$, then the order of $x^a$ divide $m$ (namely it's $\frac{m}{\gcd(a,m)}$).

Now let $m_{1}, m_{2} \in M$ be the orders of $x_1$ and $x_2$. We aim to show that $\operatorname{lcm}(m_1, m_2) \in M$ by constructing an element of such order.

If $m_1$ and $m_2$ are coprime, we can check $x_1 x_2$ fits the bill : because $(x_1 x_2)^{m_1 m_2} = 1$, the order divides $m_1 m_2$. And conversely if $(x_1 x_2)^m = 1$ then $x_1^m = (x_2^{-1})^m$. The order of the left hand side divides $m_1$, and the order of the right hand side divides $m_2$, so it can only be $1$. Which means $1 = x_1^m = (x_2^{-1})^m$, which implies $m$ divides both $m_1$ and $m_2$ so it divides the product.

Now $m_1$ and $m_2$ aren't coprime, we simply $d = \gcd(m_1,m_2)$ and notice $k'_2 = k_2^d$ is of order $m'_2 = \frac{m_2}{d}$. So $x_1 x'_2$ is of order $m_1 m'_2 = \operatorname{lcm}(m_1, m_2)$ using the previous case.

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Once you know this, you can complete the proof. Define

$$m := \max\{m_k | k \in (\mathbb{Z}/p\mathbb{Z})^*\} \underset{\text{prop}}{=} \operatorname{lcm} \{m_k | k \in (\mathbb{Z}/p\mathbb{Z})^*\}$$

The sums $c_i$ are non zero exactly when $i$ is a multiple of $m$ :

$$c_i \ne 0 \iff \forall k \in (\mathbb{Z}/p\mathbb{Z})^*, k^i = 1 \iff \forall k \in (\mathbb{Z}/p\mathbb{Z})^*, i | m_k \iff i| m$$

and when that sum is non zero, it's always -1. So we get :

$$\begin{align*} -1 &= \sigma_1\\ &= \sum_{k \in (\mathbb{Z}/p\mathbb{Z})^*} \sigma_k \\ &= \sum_{i = 1}^{p-1} \sum_{k \in (\mathbb{Z}/p\mathbb{Z})^*} k^i \\ &= \sum_{i = 1}^{p-1} c_i \\ &= \sum_{i | m} -1 \\ &= -1 \times \frac{p-1}{m} \end{align*}$$

So finally $\frac{p-1}{m} = 1$ (the equality is modulo $p$, but since $m < p$, it implies $m = p-1$)