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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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    $\begingroup$ Multiplication is commutative. So $G$ is abelian and every subgroup is normal. $\endgroup$
    – jspecter
    Commented Aug 26, 2011 at 14:03
  • $\begingroup$ For a finite group, $G$ is nilpotent if and only if it is the direct product of its Sylow subgroups. $\endgroup$
    – user1729
    Commented Aug 26, 2011 at 14:31
  • $\begingroup$ 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). $\endgroup$
    – Someone
    Commented Aug 26, 2011 at 14:59
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    $\begingroup$ See also : mathoverflow.net/questions/54735 $\endgroup$
    – Watson
    Commented Nov 11, 2017 at 13:05
  • $\begingroup$ See also Stroppel, Locally compact groups, Theorem 6.32 and corollary 6.33. $\endgroup$
    – Watson
    Commented Nov 11, 2017 at 14:23

7 Answers 7

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

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

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

Proof. Fix $d \mid n$ and consider the set $G_d$ made up of elements of $G$ with order $d$. Suppose that $G_d \neq \varnothing$, 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 $d \mid n$. So we have:

$$\begin{align} n &= \# G\\ & = \sum_{d \mid n} \# G_d \\ &\leq \sum_{d \mid n} \phi(d) \\ &= n. \end{align}$$

Therefore $\# G_d = \phi(d)$ for every $d \vert n$. In particular $G_n \neq \varnothing$. This proves that $G$ is cyclic. QED

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    $\begingroup$ 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. $\endgroup$ Commented Apr 28, 2013 at 5:24
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    $\begingroup$ @JuanPablo: That's Euler's totient function, the number of non-negative integers${}<n$ that are relatively prime to$~n$. $\endgroup$ Commented Jul 12, 2013 at 22:06
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    $\begingroup$ @abeliangrape - That is obvious. Either $\#G_d=0$ or $\#G_d=\phi(d)$; in either case, $\#G_d\leq\phi(d)$. $\endgroup$
    – mr_e_man
    Commented Jan 21, 2022 at 4:41
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    $\begingroup$ @Koro by easy consequences of the definition of order, since $y$ has order $d$, $1, y, y^2, ..., y^{d-1}$ are all distinct elements in $G$, and they are the elements of the group generated by $y$, $\langle y\rangle$ $\endgroup$
    – Fra
    Commented Feb 28, 2023 at 13:11
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    $\begingroup$ @Fra: Yes. I had misunderstood the definition of $G_d$. Since y is in $G_d$, it is of order $d$, hence the cardinality of $\langle y\rangle$ is $d$. Thanks. $\endgroup$
    – Koro
    Commented Feb 28, 2023 at 15:07
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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 order $p_i^{n_i}$ 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 $\alpha \in G,$ we 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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  • $\begingroup$ In the second paragraph third sentence: It is not true that if $a_i \in \mathbb{Z}_{(p_i)^{n_i}}$, then $(a_i)^{({p_i}^{n_i})}=1$. Take $2 \in \mathbb{Z}_{3^2}=\mathbb{Z}_9$. We have $2^9 =512 \equiv 8 \pmod 9$. $\endgroup$
    – user5826
    Commented Nov 19, 2018 at 16:12
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    $\begingroup$ @AlJebr I don't know if I understand this correctly, but $2^9$ should be supposed to mean $2+2+2+\dots+2$ for 9 times ? So the result is clearly divisible by 9. $\endgroup$
    – hephaes
    Commented Dec 7, 2018 at 8:12
  • $\begingroup$ As, @user5826 pointed out , there is a glitch in the proof . But that can be rectified . Observe that , order of every element in G , devides m , so they are roots of the equation $ x^m-1$ Now how many roots can , $x^m-1$ , can have ? Because F is a field , it will have exactly m roots and number of distinct roots $ \le m$ . Number of elements in G is , $p_{1}^{a_{1}}...p_{k}^{a_{k}}$ . So , $p_{1}^{a_{1}}....p_{k}^{a_{k}})\le m$ . We previously saw the reverse inequality, thus , $p_{1}^{a_{1}}...p_{k}^{a_{k}}= m$ . m being l.c.m , no two $p_{i}$'s can be equal . So , gcd of any pair is 1. $\endgroup$ Commented Feb 13 at 5:27
  • $\begingroup$ Excuse , the glitches above ( the 5-min editing time frame passed before noticing much !!!) . $\endgroup$ Commented Feb 13 at 5:35
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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 implicitly 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.

(Edit: This happens to be the subject of another math.SE question. It may seem quite hard, unless one realises that in Abelian torsion groups, different prime factors can be considered independently due to a canonical direct sum decomposition, after which the question becomes trivial. Here I'll leave my original proof below, which follows another answer to that question.)

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 the order of every element divides$~n$, whence $m\mid n$ by the property of least common multiples. But one also has $n\leq m$ since all $n$ elements of $G$ are roots of the polynomial $X^m-1$ in the field$~F$. Therefore $n=m$, and by the lemma (using that $G$ is commutative since $F$ is so) $G$ has an element $g$ of order $m=n=\#G$, so that $G=\langle g\rangle$ is cyclic.

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  • $\begingroup$ For another proof done the same way see this May 2011 post. Note that Marc had earlier critiqued this method of proof in this Sep 2012 post so it's a bit puzzling to see him repeat it years later. $\endgroup$ Commented Mar 21, 2022 at 18:56
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Let's have a number-theoretic proof .

Let $o(G) = n$ .
Then , $d|n$ $\Rightarrow$ $x^{d}-1|x^{n}-1$ .
Or, $x^{n}-1=g(x).(x^{d}-1)$ , where , $g(x)$ is a polynomial of degree , $n-d$ .
Now, if $x^{d}-1$ has less than d distinct roots , F being a field,$g(x)$ can't have more than $n-d$ roots resulting less than $n$ distinct roots for $x^{n}-1$ , which is absurd as $o(G)=n$ .

Thus , $G'=(\alpha\in G : \alpha^{d}=1)$ , is a subgroup of $G$ .
Let, $\psi(d)$ be the number of elements of order d in $G'$ , then , $\sum_{c|d} \psi(d)= d$ .

By , Möbius Inversion formula ,
$\psi(d)=\sum_{c|d}\mu(c).\frac{d}{c}$.....(1) , where $\mu$ is Möbius function.

We know , $\sum_{c|d}\phi(c)=d$ . Applying , Möbius Inversion formula to it we observe , $\sum_{c|d}\mu(c).\frac{d}{c}$....(2).

Using (1) and (2) , we may conclude, $\psi(d)=\phi(d)\geq1$ .

Putting , n in place of d , the theorem is proven.

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This is another elementary proof that uses only the fact that a polynomial of degree $n$ has at most $n$ roots in a field.

So, let $F$ be a finite field let $g$ be an element of maximal possible order $n$ in the multiplicative group $F^*=F\setminus\{0\}$ of $F$. Let $H:=\{g^k:0\le k<n\}$ be the cyclic subgroup of $F^*$, generated by the element $g$. If $H=F^*$, then the group $F^*$ is cyclic and we are done. So, assume that $H\ne F^*$. Observe that every element $x\in H$ satisfies the equation $x^n-1=0$, which has at most $n$ roots in the field $F$. Therefore, $H=\{x\in F:x^n=1\}$. Let $p$ be the smallest positive number for which there exists an element $y\in F^*\setminus H$ such that $y^p\in H$. The minimality of $p$ implies that $p$ is a prime number. Let $k\in\{0,\dots,n-1\}$ be the smallest possible number such that $y^p=g^k$ for some element $y\in F^*\setminus H$. We claim that $k<p$. Assuming that $k\ge p$, we can observe that the element $z=yg^{-1}\in F^*\setminus H$ has $z^p=(yg^{-1})^p=g^{k-p}$, which contradicts the minimality of $k$. If $k=0$, then $y^p=g^k=1$ and $y\notin H$ imply that $p$ does not divide $n$. In this case the element $yg\in F^*$ has order $pn>n$, which contradicts the maximality of $n$. This contradiction shows that $k>0$. It follows from $0<k<p$ that the element $g^k$ has order $d>\frac{n}{p}$. Then the element $y$ has order $pd>n$, which contradicts the maximality of $n$. This is a final contradiction showing that the group $F^*=H$ is cyclic.

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Here is an alternative argument, essentially the 8th proof presented by Keith Conrad here, but organized a little differently and trying to be clear about the flow of ideas, and what is and is not really necessary. So, let $G$ be a finite subgroup of order $n$ of the multiplicative group $K^{\times}$ of a field. Construct the monic polynomial

$$f(x) = \prod_{g \in G} (x - g).$$

Since $G$ is a subgroup, it is closed under multiplication by every $g \in G$, so $f(gx) = f(x)$ for every $g \in G$, since the two polynomials have the same roots and the same nonzero constant term. Comparing the leading terms gives $g^n = 1$ for every $g \in G$. Then $f(x)$ and $x^n - 1$ are two monic polynomials of degree $n$ with the same roots, namely the elements of $G$, so

$$f(x) = x^n - 1.$$

(This is a lightly disguised version of the easy proof of Lagrange's theorem for finite abelian groups where we observe that $\prod_{a \in G} a = \prod_{a \in G} ga$. We do not need Lagrange's theorem for nonabelian groups.)

This tells us that $G$ is exactly the group of $n^{th}$ roots of unity in $K$. Next we need some basic facts about the cyclotomic polynomials.

Proposition 1: Let $\Phi_n(x) = \prod_{\gcd(k, n) = 1} \left( x - e^{ \frac{ 2\pi i k}{n}} \right) \in \mathbb{C}[x]$ be the monic polynomial over $\mathbb{C}$ whose roots are the primitive $n^{th}$ roots of unity. Then $\Phi_n(x) \in \mathbb{Z}[x]$ has integer coefficients, and $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$.

Proof. The identity just says that every $n^{th}$ root of unity is a primitive $d^{th}$ root of unity for a unique divisor $d \mid n$, so the interesting statement here is that the cyclotomic polynomials have integer coefficients. This follows from rewriting the factorization identity as

$$\Phi_n(x) = \frac{x^n - 1}{\prod_{d \mid n} \Phi_d(x)}$$

and then applying strong induction on $n$, starting from the base case $\Phi_1(x) = x - 1$. That is, we know the denominator is a monic polynomial, and by the inductive hypothesis it has integer coefficients. We also know the quotient is a polynomial, which therefore also has integer coefficients. $\Box$

The significance of the cyclotomic polynomials having integer coefficients is that they can be interpreted as polynomials over any field, and moreover the identity $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$ holds over any field, so we get that

$$f(x) = \prod_{d \mid n} \Phi_d(x) \in K[x].$$

Now we can prove the desired result.

Proposition 2: If $K$ is a field with all $n^{th}$ roots of unity (meaning $x^n - 1$ splits completely and has distinct roots) then the roots of $\Phi_n(x)$ over $K$ are exactly the primitive $n^{th}$ roots of unity. Hence the number of primitive $n^{th}$ roots of unity over $K$ is the same as the number of primitive $n^{th}$ roots of unity over $\mathbb{C}$, and in particular there is at least one, so the group of $n^{th}$ roots of unity is cyclic, generated by any primitive $n^{th}$ root of unity.

Here, for clarity, a primitive $n^{th}$ root of unity in a field $K$ is an element $\zeta \in K$ satisfying $\zeta^n = 1$ but $\zeta^d \neq 1$ for any proper divisor $d \mid n$.

Proof. A root of $\Phi_n(x)$ is an $n^{th}$ root of unity (since $\Phi_n(x) \mid x^n - 1$) but not a $d^{th}$ root of unity for any proper divisor $d \mid n$ (since the roots of $x^n - 1$ are distinct, meaning $\Phi_n(x)$ is relatively prime to $\Phi_d(x)$ where $d$ is a proper divisor). Conversely, if $\zeta$ is a primitive $n^{th}$ root of unity then it is a root of $x^n - 1$ but not a root of $x^d - 1$ for any proper divisor $d$, so can only be a root of $\Phi_n(x)$.

It follows that the number of primitive $n^{th}$ roots of unity over $K$ is $\deg \Phi_n(x)$, which by definition is the number of primitive $n^{th}$ roots of unity over $\mathbb{C}$. There exists at least one, namely $e^{ \frac{2 \pi i }{n} }$, so the same is true over $K$.

Now let $\zeta \in K$ be any primitive $n^{th}$ root of unity. Then the elements $\{ 1, \zeta, \dots \zeta^{n-1} \}$ are distinct (otherwise $\zeta^k = 1$ for some $k < n$ gives $\zeta^{\gcd(k, n)} = 1$) and are all $n^{th}$ roots of unity, so must be the roots of $x^n - 1$. Hence $\zeta$ generates the group of $n^{th}$ roots of unity. $\Box$

(This last argument again avoids Lagrange's theorem. Of course it's not hard to show that $\deg \Phi_n(x) = \varphi(n)$ but we don't need this. It is possible for $x^n - 1$ to have repeated roots over a field of characteristic $p \mid n$ which is why it was necessary to explicitly state that condition. Otherwise the cyclotomic polynomials are not necessarily relatively prime. For example in characteristic $p$ we have $x^p - 1 = (x - 1)^p$ so $\Phi_p(x) \equiv (x - 1)^{p-1} \bmod p$ is not relatively prime to $\Phi_1(x) = x - 1$. In this case there are no primitive $p^{th}$ roots of unity.)


To my mind the significance of this argument is the following.

  1. This is really a fact about fields, not a fact about finite abelian groups. Almost no group theory is required in this argument and the hypothesis on a finite abelian group that there are at most $d$ elements of order dividing $d$ only comes up in this case and nowhere else that I know of. The usual statement of this fact as "every finite subgroup of..." is arguably misleading because in fact the first step is to show that the only such finite subgroups are given by the $n^{th}$ roots of unity for some $n$. So this is really a fact about how the roots of unity behave in any field.

  2. We deduce the statement over arbitrary fields from the statement over $\mathbb{C}$, by using the cyclotomic polynomials to "transfer" information about the roots of unity over $\mathbb{C}$ to arbitrary fields. This is a nice and understandable special case of a general strategy with many applications, and our arguments can be understood abstractly in terms of the group scheme $\mu_n$ of $n^{th}$ roots of unity, although this is of course not necessary.

Let me say a bit more about why this "transfer" is surprising. Suppose $K$ is a field of characteristic $p$ (which might be $0$) containing all $n^{th}$ roots of unity (this requires that $p \nmid n$) and let $\mathbb{F}_p \subset K$ be its prime subfield (the subfield generated by $1 \in K$), where if $p = 0$ then $\mathbb{F}_0 = \mathbb{Q}$. Let $K_n$ be the subfield generated by the $n^{th}$ roots of unity. This is a splitting field of $x^n - 1$ over $\mathbb{F}_p$, so using the uniqueness of splitting fields it follows that this splitting field in $K$ is isomorphic to the same splitting field in another field $K'$ for any other field also containing all $n^{th}$ roots of unity of the same characteristic. It follows that the group of $n^{th}$ roots of unity in $K$ and in $K'$ are isomorphic.

This can be used to deduce the desired statement for $K$ a field of characteristic zero from the statement over $\mathbb{C}$, since they have the same characteristic. The surprising thing about the "transfer" argument is that it does not depend on the characteristic, so can transfer the desired result from characteristic $0$ to positive characteristic, using the key fact that the cyclotomic polynomials are defined over $\mathbb{Z}$ and so make sense over every field.

The 9th proof in Keith Conrad's blurb makes this transfer from characteristic $0$ to positive characteristic more explicit using the $p$-adics which is quite nice.

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Let $G$ be a finite group with $n$ elements of a field with the field's multiplication operation. Let $d \: | \: n$ and consider the set $G_d$ containing elements of $G$ with order $d$. Suppose $G_d \neq \varnothing$, so there is $y \in G_d$.

Let's generate a cyclic group $\langle y \rangle$. From group theory, generators of $\langle y \rangle$ are elements of $\langle y \rangle$ that are relatively prime to $|\langle y \rangle| = d$. And because any other elements having orders different than $d$ could not generate $\langle y \rangle$, $G_d$ contains all elements that generate $\langle y \rangle$, so $\#G_d$ = $\phi(d)$.

($\#G_d$ is number of elements of $G_d$; $\phi$ is Euler's totient function.)

Let's partition $G$ into sets of elements with the same order. For example if $G = Z_{10}$, then it would be partitioned into sets of elements: $\{0\}$, $\{5\}$, $\{2, 4, 6, 8\}$ and $\{1, 3, 7, 9\}$ of order $1$, $2$, $5$, and $10$ respectively. By Lagrange's theorem, all elements of a group must have their orders divide the order of the group, so by partitioning G into all $d \: | \: n$, we would have the whole $G$, and having a formula as following:

$$n = |G| = \sum_{d|n} \#G_d = \sum_{d|n} \phi(d)$$

Borrowing Gauss' divisor sum we also have the following formula:

$$\sum_{d|n} \phi(d) = n$$

To recapitulate, we have shown that $\#G_d = 0$ or $\#G_d = \phi(d)$, but if $\#G_d$ could be equal to $0$, then we would have:

$$n = |G| = \sum_{d|n} \#G_d \leq \sum_{d|n} \phi(d) = n$$

which is false because we would have $|G|$ less than $n$. So $\#G_d = \phi(d)$, could not be $0$, and in particular for $d = n$ and $y_n \in G_n \neq \varnothing$, we are guaranteed to have $y_n$, so is $\langle y_n \rangle$ as well.

Then note that $\langle y_n \rangle = \{x \in G \: | \: x^n = 1\}$, and as $|\langle y_n \rangle| = n$, that is all elements of $G$ (because $G$ is already defined as a finite group with $n$ elements). This proves that $G \cong \langle y_n \rangle$ hence cyclic.

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