# Finite subgroups of the multiplicative group of a field are cyclic

In Grove's book Algebra, Proposition 3.7 at page 94 is the following

If $G$ is a finite subgroup of the multiplicative group $F^*$ of a field $F$, then $G$ is cyclic.

He starts the proof by saying "Since $G$ is the direct product of its Sylow subgroups ...". But this is only true if the Sylow subgroups of $G$ are all normal. How do we know this?

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Multiplication is commutative. So $G$ is abelian and every subgroup is normal. –  jspecter Aug 26 '11 at 14:03
For a finite group, $G$ is nilpotent if and only if it is the direct product of its Sylow subgroups. –  user1729 Aug 26 '11 at 14:31
A slight generalization of the lemma/theorem you are wondering about is topic of this question (in the moment there is no answer, but a good comment by Geoff). –  Someone Aug 26 '11 at 14:59

There's a simple proof which doesn't use Sylow's theory.

Lemma. Let $G$ a finite group with $n$ elements. If for every $d \vert n$ we have $\# \{x \in G \mid x^d = 1 \} \leq d$, then $G$ is cyclic.

If $G$ is a finite subgroup of the multiplicative group of a field, then $G$ satisfy the hypothesis because the polynomial $x^d - 1$ has $d$ roots at most.

Proof. Fix $d \vert n$ and consider the set $G_d$ made up of elements of $G$ with order $d$. Suppose that $G_d \neq \emptyset$, so there exists $y \in G_d$; it is clear that $\langle y \rangle \subseteq \{ x \in G \mid x^d = 1 \}$. But the subgroup $\langle y \rangle$ has cardinality $d$, so from the hypothesis we have that $\langle y \rangle = \{ x \in G \mid x^d = 1 \}$. Therefore $G_d$ is the set of generators of the cyclic group $\langle y \rangle$ of order $d$, so $\# G_d = \phi(d)$.

We have proved that $G_d$ is empty or has cardinality $\phi(d)$, for every $d \vert n$. So we have: $$n = \# G = \sum_{d \vert n} \# G_d \leq \sum_{d \vert n} \phi(d) = n,$$ Therefore $\# G_d = \phi(d)$ for every $d \vert n$. In particular $G_n \neq \emptyset$. This proves that $G$ is cyclic. QED

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Very nice proof. It may be noted that the last equality $\sum_{d|n}\phi(d)=n$ is derived from the very same argument applied when $G$ is the cyclic group of order$~n$, using the additional knowledge that in this case elements of every order $d|n$ do exist. In other words no knowledge at all about the values $\phi(d)$, apart from the fact that the are well defined, is used. –  Marc van Leeuwen Apr 28 '13 at 5:24
@MarcvanLeeuwen, what are the $\phi(n)$? –  Juan Pablo Jul 12 '13 at 21:27
@JuanPablo: That's Euler's totient function, the number of non-negative integers${}<n$ that are relatively prime to$~n$. –  Marc van Leeuwen Jul 12 '13 at 22:06

Note that this result is not true if $F$ is a skew field (division ring), as is illustrated by the quaternion group $Q_8$ inside the quaternions. So one must use commutativity somewhere, and this usually happens implictly by using that the polynomial $X^d-1$ can have at most $d$ roots in $F$; this is for instance the case in the answer by Andrea, where the proof of the lemma does not use commutativity. Here is a somewhat different approach that exploits commutativity a second time.

Lemma. The set of orders of elements in a finite Abelian group is closed under taking least common multiples.

Proof. The set of orders (in any group) is certainly closed under taking divisors: if $x$ has order $n$ and $d\mid n$ then $x^{n/d}$ has order $d$. Now if $a,b$ are orders of elements in an Abelian group and $\def\lcm{\operatorname{lcm}}m=\lcm(a,b)$, then there are relatively prime $a',b'$ with $a'\mid a$, $b'\mid b$, and $a'b'=m$: it suffices to retain in $a'$ those and only those prime factors of $a$ whose multiplicity in $a$ is at least as great as in $b$, and to retain in $b'$ all other prime factors of $b$ (those whose multiplicity exceeds those in $a$). Now if $x$ has order $a'$ and $y$ has order $b'$, then these orders are relatively prime, whence $\langle x\rangle\cap\langle y\rangle=\{e\}$, and their product is$~m$ so that $$x^iy^i =e\iff x^i=e=y^i\iff \lcm(a',b')=a'b'=m\mid i,$$ and therefore $xy$ has order $m$. QED

Now to prove the proposition, let $n=\#G$, and let $m$ be the least common multiple of all the orders of elements of $G$. By Lagrange's theorem one has $m\mid n$, and, and since all $x\in G$ are roots of $X^m-1$ one has $n\leq m$. Therefore $n=m$, and by the lemma (using that $G$ is commutative since $F$ is so) $G$ has an element $g$ of order $m=\#G$, so that $G=\langle g\rangle$ is cyclic.

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We know that if $G$ is a finite abelian group, $G$ is isomorphic to a direct product $\mathbb{Z}_{(p_1)^{n_1}} \times \mathbb{Z}_{(p_2)^{n_2}} \times \cdots \times \mathbb{Z}_{(p_r)^{n_r}}$ where $p_i$'s are prime not necessarily distinct. Consider each of the $\mathbb{Z}_{(p_i)^{n_i}}$ as a cyclic group of oreder $p_i^{n_i}$ in multiplicative notation. Let $m$ be the $lcm$ of all the $p_i^{n_i}$ for $i=1,2,\ldots,r.$ Clearly $m\leq {p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}.$ If $a_i \in \mathbb{Z}_{(p_i)^{n_i}}$ then $(a_i)^{({p_i}^{n_i})}=1$ and hence $a_i^m=1.$ Therefore for all $\alpha \in G,$ whe have $\alpha^m=1$ that is every element of $G$ is a root of $x^m=1.$ However $G$ has ${p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}$ elements, while the polynomial $x^m-1$ can have at most $m$ roots in $F.$ So, we deduce that $m={p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}.$ Therefore $p_i$'s are distinct primes, and the group $G$ is isomorphic to the cyclic group $\mathbb{Z}_m.$

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