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Suppose we have a word in $t$ letters, and suppose we have a finite group $G$. We can view the word as a map $G^t \to G$- by plugging into our letters and then doing the multiplication in the group.

I am interested in the existence of an efficiently algorithm to determine if this map is trivial (i.e gives the trivial element).

For example a word with one letter is $a\cdot a.... \cdot a$, (repeated $|G|$ times), will always give the trivial element.

A related question is that it's open if the trivial random tries works (I am also interested in random algorithms of course)- see •Can Erdős-Turán $\frac{5}{8}$ theorem be generalised that way? .

Some clarifications -

  1. Assume we have the group multiplication table, i.e if $G$ is of size $n$ and the word of size $w$, I want a $Poly(n,w)$ algorithm.

  2. There is a decent method that works for very specific types of word- if you can find a set of $s$ letters, so that their removal gives you disjoint words no pair of which share a letter, then you can recurse easily (by asking the more general algorithm of determining if a word gives you a specific element)- this gives you $n^s + recursion$ work.

  3. I feel that taking the representations of the group should be helpful.

  4. Obvoiusly this may be NP hard, but I don't know many examples of NP hard problems to try and reduce it to.

user135743
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  • How do you represent the word? If it is a string of generators, you'll need $t-1$ group operations and clearly cannot do better in the general case. – ahulpke Jun 14 '19 at 01:12
  • @ahulpke The word isn't an element of the group, it's a function, you can plug each group element to each letter. I would be happy to answer my question in t-1 time, the naive is $G^t$ – user135743 Jun 14 '19 at 13:47
  • OK, so you want to check whether the group is in the variety generated by the word. – ahulpke Jun 14 '19 at 14:23
  • @ahulpke No, I am trying to see if the word is useless, i.e it can only give the trivial element. – user135743 Jun 14 '19 at 16:11

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