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Eric Allender is one of the world’s experts on computational complexity. He is especially a master at the care and feeding of the complexity classes that are at the weaker end of the spectrum. These include classes like ACC 0 or TC 0 among many others. Just because these classes seem to be weak does not mean they give out their secrets easily. Proving almost anything new about them seems well beyond all the techniques that are available to us.

We will talk about a pretty result of Eric’s, and how it connects to Chapter 8’s discussion on partial results.

I have never had the honor of working on a project with Eric, even though he was at Rutgers and I was nearby at Princeton for many years. We did however have an interesting situation arise where Eric and his colleagues wiped out Anastasios Viglas and me.

It was back in 1999, and for some reason the question of what is the computational cost of deciding whether a given number is prime occurred to both groups, one at Rutgers and one at Princeton. Proving lower bounds on general Turing machines is too hard, so both groups asked: could it be proved that testing whether a number is prime or not at least cannot be too easy. In particular, could we prove that primality did not lie in some low-level complexity class.

The answer from Rutgers was “yes”; the answer from Princeton was “maybe.” Eric, along with Michael Saks and Igor Shparlinski, proved that primality could not be in ACC 0[p] for any prime p. Viglas and I proved a slightly weaker result, and what was worse, our result needed an unproved assumption about the structure of primes. It was one of those assumptions that must be right, but is probably hard to prove.

Unfortunately for us both papers were submitted to exactly the same conference, the 14th Annual IEEE Conference on Computational Complexity. Theirs got in, while ours did not. It was clearly the right decision, but I have wondered what would have happened if we had submitted alone. Oh well.

10.1 Allender on the Permanent

Now I will talk about another particularly pretty result, an even more notable theorem of Eric’s from the late 1990s. The actual results proved in his paper are stronger, but for our purposes the following simple statement is sufficient:

Theorem 10.1

The permanent is not in uniform TC 0.

Recall that TC 0 is a constant-depth circuit model that allows arbitrary threshold gates. In the theorem the circuits must be uniform: this means roughly that there is a simpler-than-logspace computation that tells which wires connect to which gates. There is some care needed in getting the model just right, so please look at Eric’s paper for all the details.

The proof uses a number of facts from complexity theory, but shows that a contradiction would follow if the permanent were in the threshold class. In all it is a very pretty and clever argument. I think the result is well known, but I do not think Eric gets as much credit as he should for it. It is definitely one of the strong results about the complexity of permanent, and also one of the Mason–Dixon lines in the frontiers of lower bounds.

10.2 A Result to Dream of

I decided that after the previous discussion on partial results I would put out a partial result. The goal is to state a conjecture—actually a family of conjectures—so that if one is proved, then we can prove the following statement:

Conjecture 10.1

The complexity class NC 1 is properly contained in the class \(\mathsf{\#P}\).

The high-level issue is whether there is a loophole in the work by David Barrington and others that an efficient group-algebra simulation of Boolean gates requires a non-solvable group. The issue involves the so-called commutator subgroup [H,H] of a subgroup H of a group G. This is the group generated by the commutators [a,b]=aba −1 b −1 where a,b∈H. The commutators themselves need not form a group. The question is, what group I do they generate, and is it all of H?

Well, if H is an Abelian subgroup then I={e}, the trivial identity subgroup, and we’ve figuratively “hit bottom.” If I is not {e} and not all of H, then one can iterate and ask the same question about [I,I], and so on. A group G is solvable provided you always hit bottom. Put another way, G is non-solvable provided it has a subgroup H≠{e} such that [H,H]=H. Barrington’s proof employs such a subgroup H, indeed where H is simple.

Of course, most of us probably believe that solvable groups cannot replace the simple groups in Barrington’s theorem, but that is an open problem. Suppose they can. Then the later work by Barrington and others on the solvable-group case would make NC 1 equal to (a subclass of) TC 0. Allender’s theorem would then make NC 1 properly contained in #P (which is equivalent for our purposes to the language class PP). The partial work has thus been to establish a connection between a family \({\mathcal{SOLVE}}\) of existential questions about solvable groups and results that would then fit together to prove an open separation theorem.

Moreover, if you don’t believe NC 1=TC 0, i.e. if you believe Barrington’s characterization is tight, then you can read the connection the other way: results in complexity theory imply the impossibility of some pure statements in group theory. Maybe you can refute the particular statements we’ll propose by direct algebra, but if they stay open then we gain a meaningful dialogue between complexity and algebra.

10.3 \({\mathcal{SOLVE}}\)

I will state one of the questions in the family \(\mathcal{SOLVE}\). They all imply as I stated that

$$\mathsf{NC^1} \subsetneq\mathsf{\#P}. $$

The following is the question. Some group theory notation is needed. If x,y are in a group, then

$$yxy^{-1} $$

is the conjugate of x by y. Also 〈a,b〉 denotes the group generated by the elements a and b. Finally, the order of an element x in G is denoted by o(x).

  • Suppose that G is a solvable group. Are there four elements a,b,c,d in the group so that the following are true?

    1. (1)

      Let K be the subgroup 〈a,b〉. Then we require that some conjugate of c and some conjugate of d lie in K.

    2. (2)

      Let L be the subgroup 〈c,d〉. Then we require that some conjugate of c and some conjugate of d lie in L.

    3. (3)

      And the following two numbers are relatively prime: o(a)o(b) and o(c)o(d).

The last clause may force that H=〈a,b,c,d〉 must equal its commutator subgroup, which would violate G being a solvable group. That would shoot down this member of \({\mathcal{SOLVE}}\), though we have others that might survive.

10.4 Open Problems

Prove \({\mathcal{SOLVE}}\). Or disprove it. If you do that then I will put out some even weaker conditions. Let me know.

10.5 Notes and Links

Original post:

Eric Allender, Michael Saks, and Igor Shparlinski, “A lower bound for primality,” Journal of Computer and Systems Sciences 62(2), 2001, 356–366. Also available at:

In comments, as we tell in Chap. 11, Colin Reid turned aside the idea. The similarly-titled Chap. 43 gives more technical background, while we moved this and the next chapter up to highlight the social process.