MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can we name some examples of theorems in group theory which imply (in a relatively straight-forward way) interesting theorems or phenomena in number theory?

Here are two examples I thought of:

The existence of Golod-Shafarevich towers of Hilbert class fields follows from an inequality on the dimensions of the first two cohomology groups of the ground field.

Iwasawa's theorem on the size of the $p$ part of the class groups in $\mathbb{Z}_p$-extensions follows from studying the structure of $\mathbb{Z}_p[\![T]\!]$-modules.

Can you name some others?

share|cite|improve this question
I'm not sure that your second example is a straight-forward result coming from group theory, I'd rather think that it is more a result from commutative algebra. Although you could argue that it is in fact a pro-p group representation theoretic result. – Guillermo Mantilla Dec 3 '09 at 23:20

10 Answers 10

Brauer's theorem implies meromorphic continuation of Artin L-functions (indeed, I believe that was Brauer's motivation).

share|cite|improve this answer

The fact (from class field theory) that ideals become principal in the Hilbert class field follows from the fact that the Verlagerung $V:G^{\text{ab}}\rightarrow H^{\text{ab}}$ is zero if $G$ is any finite group and $H$ is its commutator subgroup.

share|cite|improve this answer

I'm sure you omitted this just because it's too classic: big part of group theory was invented to prove that most algebraic numbers cannot be constructed by radical extensions.

It's still the best, most direct connection between [nt.number-theory] and [] I know of.

For a more "advanced" version of this, do computations of group cohomology count?

share|cite|improve this answer
Even more classic: no-one seems to have mentioned the insolubility of the quintic by radicals follows from the simplicity of the group A_5. Whether that is straightforward depends on what you already know. – Simon Wadsley Dec 4 '09 at 8:54
This is exactly what I am saying (I guess I disguised this too well): "big part" = "solvable groups", "cyclic extensions" = solvable by radicals. – Ilya Nikokoshev Dec 4 '09 at 19:52
I don't get it. Roots of unity are in a cyclic extension. Do you mean roots of integer polynomials? – Dror Speiser Oct 28 '10 at 20:48
Edited to make the italizicized statement correct. Previous version said "roots of unity" were not in radical extensions, but $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is a radical extension. – David Speyer Nov 2 '15 at 14:13

I'd say that classification of subgroups of GL$_2(F_p)$ plays a big part in Serre's result about the almost surjectivity of $\ell$-adic Galois representations of CM Elliptic curves.

Propriétés galoisiennes des points d'ordre fini des courbes elliptiques. (French) Invent. Math. 15 (1972), no. 4, 259-331.

share|cite|improve this answer

The fundamental theorem of arithmetic (uniqueness of factorization of integers into primes) is an immediate consequence of the Jordan-Holder theorem on uniqueness of composition factors of finite groups.

share|cite|improve this answer

In Conway's book The Sensual (Quadratic) Form he covers Zolotarev's proof of quadratic reciprocity:

The legendre symbol (a|m) is defined as the sign of the permutation "multiplication by a mod m". This happens to match up with the usual definition. (Note the Cayley type replacement of 'a' with the function 'multiplication by a').

Then quadratic reciprocity is proved just using group theory, and as Conway points out this has no mention of square number or even prime numbers!

share|cite|improve this answer
You need to clarify on what set you are acting by multiplication by $a \bmod m$. Multiplication by $a \bmod m$ on $\mathbf Z/m$ and on $(\mathbf Z/m)^\times$ need not have the same sign. – KConrad Nov 2 '15 at 1:47

The notion of arithmetically equivalent number fields is a good example of a connection between group theory and number theory, see for example:

a couple of specific applications:

Lemma: Let $G$ be a finite $p$-group. Any two subgroups of index $p$ are quasi-conjugated.

Corollary: Two number fields $K$, $L$ of degree $p$ prime are arithmetically equivalent if and only if $[KL:Q] \neq p^2$ See "A remark about zeta functions of number fields of prime degree" by R. Perlis.

Also by doing some basic group theory one can prove that any two arithmetically equivalent number fields of degree less than $7$ must be isomorphic.(This is also proven in a paper by Perlis but I don't remember what paper.)

Another result that comes to my mind with this question (totally unrelated to arithmetical equivalence) is that every group of odd order can be realized as a Galois group over Q(odd order theorem plus Shafarevich).

share|cite|improve this answer

Galois classified the transitive solvable groups of prime degree $p$ (subgroups of the symmetric group ${\frak S}_p$ which are solvable and act transitively on the $p$ letters) . This is a crucial ingredient in the classification of all separable degree-$p$ extensions of local fields of residual charactertistic $p$. As an application, one gets an elementary proof of Serre's mass formula in prime degree.

See Serre's "formule de masse" in prime degree arXiv:1005.2016 [math.NT]

See also Monatshefte 166 (2012) 1, 73--92.

share|cite|improve this answer

Zagier's famous one-sentence proof of Fermat's Theorem ( that every prime $p \equiv 1$ (mod $4$) is the sum of two integer squares) relies on the very elementary group-theoretic fact that if two involutions act on a finite set $S$ and one of them fixes an odd number of points, so does the other.

share|cite|improve this answer

In an abelian group, if $x$ has order $m$, and $y$ has order $n$, and $\gcd(m,n)=1$, then the order of $xy$ is $mn$. This group-theoretical fact has the number-theoretical consequence that if $p$ is a prime, then there is a primitive root modulo $p$.

[Edited in response to comment by Emanuele Tron]

share|cite|improve this answer
What do you mean by this? Take $y=x^{-1}$... – Emanuele Tron Nov 2 '15 at 10:58
The way I remember it is largely based on this group-theoretic fact, but is stated a little differently: if the exponent of a finite abelian group equals its order, then the group is cyclic. A consequence being that a finite multiplicative subgroup of a field is cyclic. – Todd Trimble Nov 2 '15 at 13:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.