Given two positive integers $n,m$, which positive integers $d$ appear as the degree of $\mathbb{Q}(a,b)$ for two algebraic numbers $a$ and $b$ of degrees $n$ resp. $m$?

Two necessary conditions are $\mathrm{lcm}(n,m) \mid d$ and $d \leq nm$. In particular, if $n,m$ are coprime, only $ d=nm$ is possible. Are they sufficient?

I have chosen $\mathbb{Q}$ just to fix ideas. Maybe the same analysis works for any field, or at least any field of characteristic zero. Answers treating more general cases are appreciated as well.

  • $\begingroup$ I have asked this before at math.stackexchange.com/questions/2359474. The comment there shows that $d \mid nm$ is not necessary. $\endgroup$
    – HeinrichD
    Oct 31, 2017 at 13:34
  • 4
    $\begingroup$ Using Galois theory, it suffices to prove the following purely group-theoretic statement: there exists a finite group $G$ together with subgroups $H,K $ of index $n,m $ such that $H \cap K $ has index $d $. Of course, the group $G $ should also be realizable as a Galois group over $\mathbf{Q} $ (or over the base field). I suggest adding the group-theory tag. $\endgroup$ Nov 1, 2017 at 19:17
  • $\begingroup$ This is a good point. The question is not about algebraic numbers, but about algebraic extensions. It asks for which $d$ there are finite extensions $E,F \subseteq \mathbb{C}$ of $\mathbb{Q}$ of degrees $n$ resp. $m$ whose compositum $EF \subseteq \mathbb{C}$ has degree $d$. $\endgroup$
    – HeinrichD
    Nov 3, 2017 at 23:12
  • $\begingroup$ This really depends on the field. If $p,q$ are distinct prime numbers, and $a,b$ are primitive $p$-th and $q$-th roots of unity, then the degrees of ${\mathbb Q}(a), {\mathbb Q}(b)$ are $p-1,q-1$ but their compositum has degree $(p-1)(q-1)$; this has to do with the fact that there are no unramified extensions of $\mathbb Q$. $\endgroup$ Nov 4, 2017 at 2:59

1 Answer 1


The necessary condition is not sufficient. In particular, it is impossible for $|\mathbb{Q}(a) : \mathbb{Q}| = |\mathbb{Q}(b) : \mathbb{Q}| = 5$ and $|\mathbb{Q}(a,b) : \mathbb{Q}| = 15$. This is because there is no group $G$ with subgroups $H$ and $K$ with $|G : H| = |G : K| = 5$ and $|G : H \cap K| = 15$. (So the argument works for arbitrary separable extensions.) I'm not happy with the proof I have of this, and I welcome cleaner arguments.

We may assume without loss of generality that $G$ is finite. Consider first the case in which $H$ and $K$ are conjugate in $G$. The action of $G$ on the cosets of $H$ gives a homomorphism from $G$ into $S_{5}$ with kernel contained in $H \cap K$, so we can assume that $G$ is a transitive subgroup of $S_{5}$. The only such groups of order a multiple of $15$ are $A_{5}$ and $S_{5}$ which are both doubly transitive and so the intersection of two point stabilizers has index $20$ in $G$, not $15$.

Now, suppose that $H$ and $K$ are not conjugate and consider the partition of $G$ into double cosets $HxK$ for $x \in G$. The double coset $HK$ has size $|H| |K|/|H \cap K| = \frac{3}{5} |G|$. The size of the double coset $|HxK| = |K| |H : H \cap xKx^{-1}| = |H| |K : K \cap x^{-1} Hx|$. Since $H \ne xKx^{-1}$, then $|HxK| = |K| |H : H \cap xKx^{-1}| \geq 2|K| = \frac{2}{5} |G|$. Thus, there are exactly two double cosets in the partition: $HK$ and $HxK$, with sizes $\frac{3}{5} |G|$ and $\frac{2}{5} |G|$.

It follows that $|K : K \cap x^{-1} Hx| = |H : H \cap xKx^{-1}| = 2$ and so $M_{1} = K \cap x^{-1} Hx$ is normal in both $K$ and $x^{-1} Hx$. Likewise $M_{2} = H \cap xKx^{-1}$ is normal in both $H$ and $xKx^{-1}$. Since $H$ and $K$ are maximal, it follows that $M_{1}$ and $M_{2}$ are both normal in $G$. Thus, $xM_{1}x^{-1} = x(K \cap x^{-1}Hx)x^{-1} = xKx^{-1} \cap H = M_{2}$, but since $M_{1}$ is normal in $G$, $M_{1} = M_{2}$. However, $M_{1} \subseteq H$ and $M_{2} \subseteq K$ and thus, $M_{1} = M_{2} = H \cap K$ actually has index $2$ in $H$ and $K$, which contradicts the assumption that $|G : H \cap K| = 15$.

EDIT: Here's a more general claim taking into account YCor's approach.

Claim: Suppose that $m = n$ is prime. Then $d = kn$ where

$\bullet$ $k = n$ or

$\bullet$ $k | n-1$ or

$\bullet$ if $n = \frac{p^{q}-1}{p-1}$ for primes $p$ and $q$, then $k = p^{q-1}$ or

$\bullet$ $n = 11$ and $k = 6$.

Proof: Letting $G$ act on the cosets of $H$ we get a homomorphism $\phi : G \to S_{n}$. Let $N = \ker \phi$. Then $NK$ is a subgroup of $G$ containing $K$. If $NK = G$ then $|G : N \cap K| = n^{2}$ but $N \subseteq H$ and this implies that $|G : H \cap K| = n$ or $n^{2}$.

In the case that $NK = K$, then we have that $N \subseteq K$. Replacing $G$ with $G/N$ we have that $G$ is a transitive subgroup of $S_{n}$. A theorem of Burnside from 1911 implies that either $G$ is solvable (and hence contained in the normalizer of a Sylow $p$-subgroup) or $G$ is doubly-transitive. If $G$ is solvable, then $G \cong (\mathbb{Z}/n\mathbb{Z}) \rtimes (\mathbb{Z}/k\mathbb{Z})$ where $k | n-1$ and $H$ and $K$ are conjugate in $G$ (since they are Hall $\pi$-subgroups). The orbits of a point stabilizer in such a group have size $1$ and $k$ and so $|H \cap K| = 1$ which has index $nk$.

In the case that $G$ is doubly-transitive, then for any minimal normal subgroup $N$ of $G$, $N$ is transitive and this implies that $C_{G}(N) = 1$. It follows that $G \subseteq {\rm Aut}(N)$. It therefore suffices to find the finite simple groups that are doubly-transitive. This determination follows from the classification of finite simple groups (and is given by Peter Cameron in "Finite Permutation Groups and Finite Simple Groups" published in the Bulletin of the London Mathematical Society in 1981; see the table on page 8). The only such groups that can have prime degree are $A_{n}$, ${\rm PSL}_{q}(\mathbb{F}_{p})$ of degree $\frac{p^{q}-1}{p-1}$, ${\rm PSL}_{2}(\mathbb{F}_{11})$, $M_{11}$ and $M_{23}$. Moreover, if the subgroups $H$ and $K$ are conjugate the double transitivity implies that $|G : H \cap K| = n(n-1)$. Therefore, examples where this doesn't occur must arise from simple groups with more than one conjugacy class of subgroups of index $n$. This occurs only for ${\rm PSL}_{q}(\mathbb{F}_{p})$ with $q > 2$ and ${\rm PSL}_{2}(\mathbb{F}_{11})$.

In the case of ${\rm PSL}_{q}(\mathbb{F}_{p})$, one class of subgroups comes from the stabilizers of one-dimensional subspaces of $\mathbb{F}_{q}^{d}$ and the other class comes from stabilizers of codimension one subspaces. The stabilizer of a codimension one subspace has two orbits on the one-dimensional subspaces: one of size $\frac{p^{q-1} - 1}{p-1}$, and the other of size $p^{q-1}$. The former is a divisor of $n-1$.

Finally, in the case of ${\rm PSL}_{2}(\mathbb{F}_{11})$ there are two classes of subgroups of index $11$. If $H$ comes from the first class, the orbits of $K$ have sizes $5$ and $6$, and this gives rise to the possibility $m=n=11$ and $k = 6$. QED

  • $\begingroup$ The non-conjugate case can be discarded as follows: in $S_5$, no transitive subgroup has a proper subgroup of index 3 (as it would have cardinal divisible by 15, hence equals $A_5$ or $S_5$ which have no such subgroup). So, if $K$ were transitive on $G/H$, the image of $K$ in $Sym(G/H)$ would be the same as the image of subgroup of index 3 $H\cap K$, which is not transitive, contradiction. Hence $K$ is not transitive on $G/H$, which means that $K$ is contained in a conjugate of $H$, hence they are conjugate. $\endgroup$
    – YCor
    Nov 4, 2017 at 9:32
  • $\begingroup$ It seems you jump from "$K$ is not transitive on $G/H$" to "$K$ has an orbit of size $1$ on $G/H$". The main case I have to rule out is a situation where $K$ is not transitive, but has an orbit of size $2$ and an orbit of size $3$ acting on $G/H$. $\endgroup$ Nov 4, 2017 at 12:03
  • $\begingroup$ Thanks, you're right. This can be fixed as follows: indeed write $p:G\to Sym(G/H)$. Then since $H,K$ are not transitive on $G/H$, $p(K)$ are both proper subgroups, hence have index 5, and now assuming $K$ has no fixed point, $p(H)\neq p(K)$. Now a non-transitive subgroup on 5 elements with no fixed point has order 6 or 12, and the only transitive subgroups containing it are $A_5$ and $S_5$. So $p(H)$ and $p(K)$ are conjugate, and this boils down to the case when $H,K$ are conjugate. $\endgroup$
    – YCor
    Nov 4, 2017 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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