Extra Facts

Statements and proofs of some important extra facts.

Cyclic Groups

It is easy to prove that if a finite group is cyclic then it has at most one subgrup of each order (dividing its order, by Lagrange, in which case this subgrup exists). The converse is also true, and useful for studying fields.

This is an alternative, somewhat simpler, to my taste, proof of Corollary 8.10. The book’s proof by way of Lemma 8.11 is still quite interesting, though – one can use 8.11(b) as a starting point for proving the fundamental theorem of finite abelian groups. Compare also Theorem 9.57.

Note that field theory is what suggests an induction using an element of minimal order here, rather than maximal order, as in the book’s proof.

Theorem 1 Let \(G\) be a finite group of order \(n\). Suppose that \(G\) has at most \(1\) subgroup of each order \(d\leq n\). Then \(G\) is cyclic.

Proof. We proceed by induction on the order \(n\); if \(n\) is prime then we are done. Note that the hypothesis on \(G\) is inherited by subgroups. Therefore, we may assume that all proper subgroups of \(G\) are cyclic. It is also useful to observe that the assumption of the theorem implies every subgroup of \(G\) is normal: the conjugate of a subgroup has the same order, and hence coincides with the original.

Let \(g\) be a non-identity element of minimal order, so its order is a prime \(p\). By induction, \(G/\langle g \rangle\) is cyclic and therefore generated by the image of some \(h\) in \(G\), from which it follows that \(G\) is generated by \(g\) and \(h\). Let \(d\) be the order of \(h\).

If \(p\) divides the order of \(h\) then \(h^{d/p}\) has order \(p\), and so \(\langle g\rangle\) and \(\langle h^{d/p}\rangle\) are subgroups of \(G\) of the same order, which implies \(g\) is in \(\langle h \rangle\) and therefore \[G = \langle g,h \rangle = \langle h \rangle\] is cyclic.

Otherwise, \(p\) does not divide \(d\). By Lagrange, then, the intersection of \(\langle h\rangle\) and \(\langle g \rangle\) is trivial. Since the two subgroups are normal, this implies \(g\) and \(h\) commute and moreover that \[G = \langle g,h\rangle \cong \langle g \rangle \times \langle h \rangle.\] But the product of cyclic groups of coprime order is cyclic, so again \(G\) is cyclic!

Corollary 1 The multiplicative group \(\mathbb F_q^*\) of a finite field is cyclic. In particular, it has an element of order \(d\) if and only if \(d\) divides \(q-1\).

Exercise: adapt the argument to show that any finite multipicative subgroup of a field is cyclic.

Proof. Let \(H\) be a subgroup of \(\mathbb F_q^*\) of order \(d\). By Lagrange’s theorem, every element of \(H\) is a root of \(x^d - 1\). That polynomial has at most \(d\) roots, and so its roots are precisely the subgroup \(H\) – so \(H\) is determined entirely by the integer \(d\), and hence \(\mathbb F_q^*\) has at most one subgroup of each order \(d\), according to whether or not \(x^d - 1\) splits completely.

Therefore, Theorem 1 implies the group \(\mathbb F_q^*\) is cyclic.

Corollary 2 The Galois group of \(\mathbb F_{q^s}\) over \(\mathbb F_q\) is cyclic of order \(s\).

Proof. We have established that there is at most one finite field of each order, hence at most one of each degree, contained in \(\mathbb F_{q^s}\), hence at most one extension of \(\mathbb F_q\) contained in \(\mathbb F_{q^s}\). Therefore, the Galois correspondence tells us that there is at most one subgroup of this Galois group of each order and so Theorem 1 tells us that the Galois group is cyclic!

Linear Algebra

We’ll use the following handy lemma a few times.

Lemma 1 Let \(V\) be a vector space over an infinite field \(K\). Then \(V\) is not the union of a finite number of proper subspaces.

Proof. Suppose, for contradiction, that \(V\) can be written as such a union, and let \(W_1,..., W_n\) be a minimal collection of proper subspaces of \(V\) whose union is \(V\). Enlarging \(W_1\) if necessary, we may further assume that \(W_1\) has codimension \(1\).

By minimality, we can take some \(w\in W_1\) which is not in \(W_j\) for any \(j\neq i\). By properness of \(W_1\), we can also find some \(v\) not in \(W_1\). Consider sums \[v + aw\] for \(a\in K\). Since \(K\) is infinite, there are infinitely many such sums, and because \(v\) is not in \(W_1\), none of them are in \(W_1\), so some \(W_j\) contains \(v+aw\) and \(v+bw\) for some \(a\neq b\)… but then \((a-b)w\) is in \(W_j\), hence \(w\) is in \(W_j\), a contradiction!

Corollary 3 If an extension \(K/k\) has only finitely many sub-extensions, then \(K/k\) is a primitive extension.

Conversely, if \(K/k\) is primitive, then it has finitely many subfields.

Proof. There is nothing to prove if \(k\) is finite, so assume it is infinite.

A proper subextension field is also a proper subspace of \(K\) as a vector space over \(k\). If there are only finitely many proper subfields, then their union cannot cover \(K\). By Lemma 1, there is some \(\alpha\) in \(K\) not contained in any of those proper subextensions, and so \(k(\alpha)\) must be all of \(K\).

In the other direction, suppose \(K = k(\alpha)\) with minimal polynomial \(f(x)\). Let \(L\) be an intermediate extension and let \(g(x)\) be the minimal polynomial of \(\alpha\) over \(L\). Since \(K/k\) must be finite, so too is \(K/L\). Observe that the extension \(L'\) obtained by adjoining the coefficients of \(g\) to \(k\) is a subfield of \(L\), but the degree of \(\alpha\) over \(L'\) is at most that of \(\alpha\) over the larger field \(L\), so \(L=L'\). But there are only finitely many such \(L'\) because \(f(x)\) has finitely many divisors.