solvable groups

basics about solvable groups

A couple of years ago I remember watching a 3Blue1Brown video about the unsolvability of the quintic, and being extremely excited by it but also a bit anguished by the fact that I thought I probably wouldn’t ever truly understand it. At the time I was in high school and, although I liked math, I thought that pure math just wasn’t for me. Fortunately that perception changed quite a bit during my undergraduate journey, and now I just finished a course in Group Theory and Galois Theory that proved the unsolvability of the quintic at the very end of the course. I don’t claim to ‘‘truly’’ understand this result – whatever that means – but I do think I have a far better understanding of it than my past self would’ve ever dreamed of, and I’m quite happy about it.

For these and other reasons, I decided to make a series of posts that will eventually lead up to the Fundamental Theorem of Galois Theory and the unsolvability of the quintic. I will start with some Group Theory basics – in this post namely solvable groups – and then go through the basics of Galois extensions and the whole bunch.

Derived and Characteristic Subgroups

We start with a group \(G\), and we define the commutator of two elements \(a,b\) in \(G\) as

\[[a,b] := a^{-1}b^{-1}ab\]

hence \([a,b] = 1\) iff \(a\) and \(b\) commute. We can then define the derived subgroup \(G'\) of \(G\) as the subgroup generated by all commutators, that is,

\[G' := \langle [a,b]:a,b \in G\rangle\]

We say that a subgroup \(K \leq G\) is a characteristic subgroup if it is invariant under automorphisms of \(G\), in other words, if \(\varphi \in \text{Aut}(G)\), then \(\varphi(K) = K\). Since conjugation by any element of \(G\) is an automorphism, it follows that any characteristic subgroup is normal, and in particular, we can see that

\[\varphi([a,b]) = [\varphi(a),\varphi(b)]\]

hence \(G'\) is also a characteristic subgroup. We now prove some basic facts about these subgroups.

Proposition 1: If \(G\) is a group, \(K,H \leq G\) are subgroups where \(K \subseteq H\), then the following hold:

• If \(K\) is a characteristic subgroup of \(H\), and \(H\) is a characteristic subgroup of \(G\), then \(K\) is also a characteristic subgroup of \(G\);
• If \(K\) is a characteristic subgroup of \(H\), and \(H\) is a normal subgroup of \(G\), then \(K\) is also a normal subgroup of \(G\);
• The quotient \(G/G'\) is an abelian group, and if \(N \trianglelefteq G\) is a normal subgroup such that \(G/N\) is abelian, then \(G' \leq N\).

Proof:

• Let \(\varphi \in \text{Aut}(G)\) be a group automorphism, and note that since \(H\) is characteristic w.r.t. \(G\), the restriction of \(\varphi\) onto \(H\) is also an automorphism of \(H\), and since \(K\) is contained in \(H\), this implies that \(K\) is invariant w.r.t. \(\varphi\), and thus it is also characteristic w.r.t. \(G\).
• Since any conjugation by an element in \(G\) is an automorphism of \(G\), the result follows by an argument similar to the previous item.
• Let \(aG',bG' \in G/G'\) be two elements of the quotient group, and note that \([aG',bG'] = [a,b]G' = 1_{G'}\), hence \(aG',bG'\) commute, implying that the quotient is indeed abelian. Now if \(G/N\) is abelian for some normal subgroup \(N\), then for any \(aN,bN \in G/N\), we have \([aN,bN] = 1\), thus \(N\) contains all elements of the form \([a,b]\), and hence it contains \(G'\), as desired.

\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]

The quotient \(G/G'\) is often called the abelianization of \(G\).

Normal, subnormal and derived series

If \(G\) is a group, and if

\[G = G_0 \geq G_1 \geq \ldots \geq G_k = 1\]

is a finite chain of subgroups, we say call this chain a normal series if each \(G_i\) is a normal subgroup of \(G\). We say that a series is subnormal if for each \(i\) we have \(G_{i+1} \trianglelefteq G_i\), and of course any normal series is also by definition subnormal. We say that a group \(G\) is solvable if there exists a normal series of subgroups

\[G = G_0 \geq G_1 \geq \ldots \geq G_k = 1\]

such that each quotient \(G_{i}/G_{i+1}\) is abelian. Naturally, any abelian group is solvable, and it can also be shown that any \(p\)-group is also solvable.

Solvable groups are intimately connected to Galois Theory, and we will ultimately see that any solvable group gives rise to a polynomial with roots that can be expressed with the usual arithmetic expressions – addition, subtraction, multiplication and division – and a finite number of root operations – that is, taking the \(n\)-th root for some natural \(n\). A key result in this connection is the fact that, for values of \(n\) greater than or equal to \(5\), the symmetric group on \(n\) elements \(S_n\) is not solvable. There a multiple ways of proving this fact, but in the next post we’ll see that this actually follows from the fact that the altenating group \(A_n\) is simple for \(n \geq 5\) – that is, contains no proper nontrivial normal subgroups.

Given any group \(G\), we can always consider the following series:

\[G = G^{(0)} \geq G^{(1)} \geq \ldots \geq G^{(k)} \geq \ldots\]

where \(G^{(1)} = G'\), and \(G^{(i+1)}\) is the derived subgroup of \(G^{(i)}\). From the previous proposition, it follows that this series is normal: \(G^{(i)}\) is both characteristic and normal w.r.t. \(G\) by induction, hence \(G^{(i+1)}\) also is. It also follows that each quotient \(G^{(i)}/G^{(i+1)}\) is abelian, and thus we have a pretty natural way of checking if a given group is solvable: if we manage to find some \(k\) for which \(G^{(k)} = 1\), we are done. What we’ll see now is that this is in fact a necessary and sufficient condition.

Characterization and properties of solvability

We can now characterize solvable groups in terms of the derived series.

Theorem 1: If \(G\) is a group, the following are equivalent:

1. The group \(G\) is solvable;
2. There exists a subnormal series of \(G\) in which each of the quotients is abelian;
3. There exists some \(k\) for which \(G^{(k)} = 1\).

Proof:

(1 \(\Rightarrow\) 2) This simply follows from the fact that any normal series is also subnormal.

(2 \(\Rightarrow\) 3) Let

\[G = G_0 \trianglerighteq G_1 \trianglerighteq \ldots \trianglerighteq G_k = 1\]

be a subnormal series in which each quotient \(G_i/G_{i+1}\) is abelian. We’ll show that \(G^{(k)} \leq G_k\) and conclude the result, and in order for us to do this we proceed via induction on \(k\). For the base step, we note that since \(G/G_1\) is abelian, it follows that \(G' \leq G_1\). Now for the general case, we know that \(G_{k-1}/G_k\) is abelian, hence \(G_{k-1}' \leq G_k\), but since by the inductive hypothesis we have that \(G^{(k-1)} \leq G_{k-1}\), this implies that \(G^{(k)} \leq G_k\). Since \(G_k = 1\), this shows that \(G^{(k)} = 1\), as desired.

(3 \(\Rightarrow\) 1) The derived series is a normal series that satisfies the desired properties.

\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]

I’ll end this post with some properties of solvable groups that will be useful for future posts.

Proposition 2: If \(G\) is a group, then the following hold:

• Any subgroup of a solvable group is solvable;
• If \(\varphi\) is a homomorphism between two groups \(G,H\), then \(\varphi(G^{(i)}) = \varphi(G)^{(i)}\) for any \(i\). In particular, solvability is a property that is preserved under isomorphisms;
• If \(G\) is a solvable group and \(N\) is a normal subgroup, then \(G/N\) is solvable;
• If \(N\) is a normal subgroup of \(G\) such that both \(N\) and \(G/N\) are solvable, then \(G\) is solvable.

Proof:

• Let \(G \geq G^{(1)} \geq \ldots \geq G^{(k)} = 1\) be the derived series of \(G\). We’ll show by induction on \(k\) that \(H^{(k)} \leq G^{(k)}\) and conclude the result. For the base case, since \(G/G'\) is abelian, it follows that \(HG'/G' \leq G/G'\) is also abelian, but since \(HG'/G'\) is isomorphic to \(H/H\cap G'\), it follows that \(H' \leq H\cap G' \leq G'\). Now for the general case, let \(G^{(k-1)}/G^{(k)}\) be abelian, and note then that \(H^{(k-1)}G^{(k)}/G^{(k)}\) must also be abelian, and again since the latter is isomorphic to \(H^{(k-1)}/H^{(k-1)}\cap G^{(k)}\), it follows that \(H^{(k)} \leq G^{(k)} = 1\), as desired.
• Note that if \(G,H\) are groups and \(\varphi\) is an homomorphism, then \(\varphi([a,b]) = [\varphi(a),\varphi(b)]\) for any \(a,b \in G\), hence \(\varphi(G') = \varphi(G)'\), which by induction immediately implies that \(\varphi(G^{(i)}) = \varphi(G)^{(i)}\) for any \(i\).
• Let \(G \geq G^{(1)} \geq \ldots \geq G^{(k)} = 1\) be the derived series of \(G\), and let \(\pi\) be the canonical projection of \(G\) onto \(G/N\). Note then that by the previous item we have \(\pi(G^{(i)}) = (G/N)^{(i)}\) since \(\pi\) is surjective, hence \(1 = \pi(G^{(k)}) = (G/N)^{(k)}\), implying that \(G/N\) is solvable.
• Let \(k,l\) be such that \((G/N)^{(k)} = 1,N^l = 1\), and consider the subgroup \(G^{(k+l)}\) of \(G\). We note that \(\pi(G^{(k)}) = (G/N)^{(k)} = 1\), hence \(G^{(k)} \leq N\), and thus \(G^{(k+l)} \leq N^{(l)} = 1\), implying that \(G^{(k+l)} = 1\), and so we conclude that \(G\) is solvable.

\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]