Congruent zeta functions. No.2

Yoshifumi Tsuchimoto

In this lecture we define and observe some properties of conguent zeta functions.

Proof of Lemma 1.3 (5). We prove the following more general result

LEMMA 2.1   Let $K$ be a field. Let $G$ be a finite subgroup of $K^\times$ (=multiplicative group of $K$ ). Then $G$ is cyclic.

PROOF.. We first prove the lemma when $\vert G\vert=\ell^k$ for some prime number $\ell$ . In such a case Euler-Lagrange theorem implies that any element $g$ of $G$ has an order $\ell^s$ for some $s\in \mathbb{N}$ , $s \leq k$ . Let $g_0\in G$ be an element which has the largest order $m$ . Then we see that any element of $G$ satisfies the equation

$$x^m=1 .$$

Since $K$ is a field, there is at most $m$ solutions to the equation. Thus $\vert G\vert\leq m$ . So we conclude that the order $m$ of $g_0$ is equal to $\vert G\vert$ and that $G$ is generated by $g_0$ .

Let us proceed now to the general case. Let us factorize the order $\vert G\vert$ .

$$\vert G\vert=\ell_1^{k_1} \ell_2 ^{k_2} \dots \ell_t ^{k_t} \qquad(\ell_1,\ell_2,\dots, \ell_t:$$ prime number,$$k_1,k_2,\dots,k_t \in \mathbb{Z}_{>0}).$$

Then $G$ may be decomposed into product of $p$ -subgroups

$$G=G_1\times G_2 \times \dots \times G_t \qquad (\vert G_j\vert=\ell_j^{k_j} (j=1,2,3,\dots,t)).$$

By using the first step of this proof we see that each $G_j$ is cyclic. Thus we conclude that $G$ is also a cyclic group. $\qedsymbol$

EXERCISE 2.1   Let $G$ be a finite abelian group. Assume we have a decomposition $\vert G\vert=m_1 m_2$ of the order of $G$ such that $m_1$ and $m_2$ are coprime. Then show the following:

1. Let us put
   $$H_j=\{g\in G; g^{m_j}=e_G\} \qquad(j=1,2)$$
   Then $H_1,H_2$ are subgroups of $G$ .
2. $\vert H_j\vert=m_j$ ($j=1,2$ ).
3. We have
   $$G=H_1 H_2.$$

EXERCISE 2.2   Let $G_1,G_2$ be finite cyclic groups. Assume $\vert G_1\vert$ and $\vert G_2\vert$ are coprime. Show that $G_1\times G_2$ is also cyclic.