----- > [!proposition] Proposition. ([[finite subgroup of multiplicative field of a group is cyclic]]) > Let $\mathbb{F}$ be a [[field]] and let $G \leq G$ be a finite [[subgroup]] with $|G|=n$. Then > 1. For any $d | n$, le[](field.md)md)denote the number of elements of $G$ of [[order of an element in a group|order]] [[divides|dividing]] $d$, and let $b(d)$ denote the number of elements of order exactly $d$. Then $d=a(d)=\sum_{d' | d}^{}b(d').$ > 2. $b(d)=\phi(d)$ for all $d | n$. ^42baab > [!proposition] Corollary. ($(\mathbb{Z} / p\mathbb{Z})^{\times}$ is cyclic of order $p-1$) > $(\mathbb{Z} / p\mathbb{Z})^{\times}$ is a multiplication [[subgroup]] of the [[prime field]] $\mathbb{F}_{p}=\mathbb{Z} / p\mathbb{Z}$. The result shows us that the number of elements of order $n=|(\mathbb{Z} / p\mathbb{Z})^{\times}|$ is $\phi(n) \geq 1$, meaning that the [[group]] is [[cyclic group|cyclic]] with $\phi(n)$ different [[generating set of a group|generators]]. ^8e785d > [!proof]- Proof. ([[finite subgroup of multiplicative field of a group is cyclic]]) > **Part 1.** First, it is clear that $a(d)=\sum_{d' | d}^{}b(d')$. So we'll just show $d=a(d)$. $x \in G$ has order that [[divides]] $d$ if and only if $x^{d}=1$, i.e., if and only $x^{d}-1=0.$ There are at most $d$ roots in $\mathbb{F}$ of this degree-$d$ [[polynomial]], from which we conclude $a(d) \leq d$. It remain to be shown that $a(d) \geq d$. I cannot figure out how to show this. \ **Part 2.** '' We know that: \ $|G|=\sum_{d | n}^{} b(d)$ (part $(1)$ of the proposition) \ $|G|=\sum_{d | n}^{} \phi(d)$ (see the proof of [[number of elements of each order in a cyclic group]]; note that it this part does not depend on $G$ being cyclic) \ We want to show that these sums are term-by-term equal. Since each term in each sum is nonnegative, it suffices to show for any $d | n$ that $b(d) \leq \phi(d)$. If $b(d)=0$ then we're done; assume $b(d)>0$. Obtain an element $h_{}$ in $G$ of order $d$ and define $H:=\langle h \rangle$. By [[Lagrange's Theorem]], any $x \in H$ satisfies $x^{d}-1=0. \ \ (*)$ We know that $(*)$ has at most $d$ solutions; since $|H|=d$, this implies that $H$ contains all the elements in $G$ of order $d$. So, to find all elements in $G$ of order $d$, it suffices to find all [[generating set of a group|generators]] of $H$. By [[number of elements of each order in a cyclic group]], we see that the number of elements in $H$ of order $d$ equals $\phi(d)$. Thus $b(d) \leq \phi(d)$. ^6c9289 ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```