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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 Aug 26 '11 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 Aug 26 '11 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 Aug 26 '11 at 14:59
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    $\begingroup$ See also : mathoverflow.net/questions/54735 $\endgroup$ – Watson Nov 11 '17 at 13:05
  • $\begingroup$ See also Stroppel, Locally compact groups, Theorem 6.32 and corollary 6.33. $\endgroup$ – Watson Nov 11 '17 at 14:23
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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: $$ n = \# G = \sum_{d \mid n} \# G_d \leq \sum_{d \mid n} \phi(d) = n, $$ 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. $\endgroup$ – Marc van Leeuwen Apr 28 '13 at 5:24
  • $\begingroup$ @MarcvanLeeuwen, what are the $\phi(n)$? $\endgroup$ – Juan Pablo Jul 12 '13 at 21:27
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    $\begingroup$ @JuanPablo: That's Euler's totient function, the number of non-negative integers${}<n$ that are relatively prime to$~n$. $\endgroup$ – Marc van Leeuwen Jul 12 '13 at 22:06
  • $\begingroup$ @Andrea or whoever can answer, can you explain what the # notation means, please? $\endgroup$ – ALannister Sep 27 '18 at 1:47
  • $\begingroup$ @ALannister It simply means the cardinality of the following finite set (the # notation is typically only used when talking about finite sets, so "cardinality" here means "number of elements"). $\endgroup$ – Nubok Oct 28 '18 at 0:04
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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 $2^9 =512 \equiv 8 \pmod 9$. $\endgroup$ – Al Jebr Nov 19 '18 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 Dec 7 '18 at 8:12
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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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