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Finding an isomorphism requires to show that for 2 groups $G$ and $H$, there exists a bijective map $\phi : G\to H$ such that $$\phi(ab)=\phi(a)\phi(b)$$ For all $a,b \in G$. This is (probably naively) pretty straight forward, and there are plenty of theorems that allow us to show that groups of specific orders must be isomorphic to some specific set of groups. So, my question (which I hope isn’t too loaded) is

What intrinsically makes finding if 2 groups are isomorphic so hard? Is it showing that they are bijective, is it showing that they are operation preserving, or is it something entirely different?

Ty Jensen
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    I guess you mean the 'group isomorphism problem' based on (finite) group representations by generators and relations. – Berci May 15 '20 at 21:24
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    See this. – Shaun May 15 '20 at 21:24
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    @Berci, you mean presentations, right? – Shaun May 15 '20 at 21:25
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    Yes, group presentation – Berci May 15 '20 at 21:26
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    Several factors make this difficult. (1) Given a presentation, there's no obvious way to tell whether the group is finite. (2) Even if you know that the group is finite, there's no obvious way to determine its order. (3) Even if you know its order and have a complete list of all groups of that order (up to isomorphism), that list can be way longer than the order of the group. (4) Even if you think your group is isomorphic to a particular known group, there are lots ($n!$ for groups of order $n$) of bijections to search through for an isomorphism. – Andreas Blass May 16 '20 at 03:22
  • There is no algorithm that, given two group presentations, decides whether they present isomorphic groups. There is no algorithm that, given a group presentation decides whether it presents a finite group. There is (if I remember correctly) no algorithm that, given a presentation of a finite group, computes its order. – Andreas Blass May 16 '20 at 03:25
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    Given whole multiplication tables of two finite groups (which is much more information then just presentations), there is an algorithm that tells whether the groups are isomorphic; just check all bijections to see whether they preserve multiplication. Because there are so many bijections, this algorithm won't run in polynomial time. More efficient algorithms exist, but none are known (and it's widely believed that none exist) that work in polynomial time. – Andreas Blass May 16 '20 at 03:29
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    @AndreasBlass Why didn’t you write that as an answer? – Jonas Linssen May 16 '20 at 08:59

1 Answers1

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There are two variants of Group Isomorphism (GpI):

  • CayleyGpI: The groups are given by their multiplication tables.
  • SuccinctGpI: This includes permutation groups, matrix groups, and black-box groups.

It is known that (under polynomial-time computable, and even $\textsf{AC}^{0}$-computable reductions) that:

  • CayleyGpI reduces to Graph Isomorphism (GI), and under $\textsf{AC}^{0}$-reductions there is no reduction in the other direction.
  • GI reduces to SuccinctGpI.

While GI is known to be in $\textsf{NP} \cap \textsf{coAM}$, the best upper bound we have for SuccinctGpI is $\textsf{Promise}\Sigma_{2}^{p}$. In the setting of black-box groups, even verifying the group axioms is a $\Pi_{2}^{p}$-problem- so we need a promise that the inputs are groups. If you are working with permutation or matrix groups, then you get a better upper bound of $\Sigma_{2}^{p}$. In the multiplication table setting, we can check the group axioms in $\textsf{AC}^{0}$.

In the CayleyGpI setting, we have the generator enumeration algorithm. Every finite group has a generating set of size at most $\log n$. So compute a generating set $S$ for $G$, and check all ways of embedding $S$ into $H$. Do any of those extend to an isomorphism? This check takes $n^{\log n + O(1)}$ time. (Note that if a group has a bounded number of generators, this check takes polynomial time). Rosenbaum and Wagner improved this to $n^{(1/2) \log n + O(1)}$ using composition-series enumeration. In the case of solvable groups, they get $n^{(1/4) \log n + O(1)}$ using a technique called bidirectional collision.

The Computational Group Theory community has done very impressive work on SuccinctGpI. Even their techniques, however, do not beat $|G|^{\Theta(\log |G|)}$ in the worst case (where here, I stress $|G|$ to be the order of the group, and not the size of the succinct input which is $O(\log |G|)$).

For some intuition on why the model is important, consider the case of groups of cube-free order. These groups have generating sets of size $3$. So for CayleyGpI, this problem is solvable in polynomial-time. On the other hand, placing this problem into $\textsf{P}$ in the permutation group setting took work. See this recent result of Dietrich & Wilson (https://arxiv.org/abs/1810.03467).

Another piece of intuition is that constructive recognition of finite simple groups is hard in the succinct setting. See for instance:

ml0105
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