LetS be a semigroup, letAbe a finite set and letψ :A∗ →Sbe a surjective homomorphism. For u, v ∈ A∗ we write u =S v if ψ(u) = ψ(v). We say S
has solvable word problem if there exists an algorithm that can decide for any two words u, v ∈A+ whether u=S v.
Theorem 5.36. [37, Theorem 5.1] Let S be a finitely generated semigroup and let T be a subsemigroup of finite Rees index. Then S has solvable word problem if and only if T has solvable word problem.
Theorem 5.37. [1, Theorem 5.4] A Rees matrix semigroup S = [G;I,Λ;P] has solvable word problem if and only if G has solvable word problem.
Theorem 5.38. (A) If S is a finitely generated semigroup with one end and has no infinite R-class then S has solvable word problem.
(B) LetS be a finitely generated semigroup with one end, an infiniteR-class and a right group G×E as a subsemigroup of finite Rees index. Then S has solvable word problem if and only if Ghas solvable word problem.
Proof. (A) Let A={a, u1, u2, . . . , un, i1, i2, . . . , im} be a generating set forS
as in Theorem 5.27. The normal forms for all but finitely many elements of S are given in Corollary 5.28. Thus if there exists an algorithm to decide if a word lies in this finite set X the semigroup S has solvable word problem.
From Corollary 5.25 and Lemma 5.7 we have elements of I are of the form apu
bik. We retain the notation u0 for a possibly external identity element.
Lets =a1a2. . . an be a word in A+. We now show that we can decide if
s ∈Xor not. Recall from Corollary 5.25 elements ofX come in two flavours, those that lie in S \aNS and those of the form apu
bij where µ((b, j),(b, j))
exists. We can rewrite s to be in the form apu
b or apubij using Corollary
5.25. If p≥N then s is certainly in aNS.
Ifp < N ands =apu
b then we apply rules of the formaπ(b,c)ub =aπ(c,b)uc
where possible to get different representatives of the form apiu for different elementsu. We need only apply at most n rules as at mostnelements of the form apu can be equal. Otherwise apu = aqu for some p < q which cannot
happen. If there is a pi ≥N then s is in aNS otherwise s∈X.
If p < N and s = apubij then we first check if µ((b, j),(b, j)) exists,
if this is the case then s ∈ X. If not then we apply rules of the form aµ((b,j),(c,k))u
bij = aν((b,j),(c,k))ucik where possible to get various elements of
the form apiu
i. If we apply these rules more than (n+ 1)m times and get at
and q1 ≤q2 such that aq1ucik=aq2ucik. It follows that apubij =aq1ucik =aq2u cik =aq2−ν((b,j),(c,k))+µ((b,j),(c,k))u bij
and hence as q2 > q1 the rule µ((b, j),(b, j)) exists. It follows the rules only need to be applied at most (n+ 1)m times. If any pi ≥ N then s ∈ aNS
otherwise s ∈X.
We are now able to decide if a word s ∈ A+ is in X and if not we can apply rewrite rules from Corollary 5.28 to get it in unique normal form. Let s ∈A∗ represent an element of X.
If s is of the form apu
b then there are only at most n words of the form
aqu
c such that apub = aquc. We can list these using the rules of the form
aπ(b,c)ub =aπ(c,b)uc.
If s is of the form apu
bij where µ((b, j),(b, j)) does not exist then there
are only at most (n+ 1)mwords of the formaqu
cik such thatapubij =aqucik.
We can list these using the rules of the formaµ((b,j),(c,k))ubij =aν((b,j),(c,k))ucik.
If s is of the form apu
bij where µ((b, j),(b, j)) does exist then either
p < µ((b, j),(b, j)), in which case we apply the argument above, or p ≥
µ((b, j),(b, j)). If p ≥ µ((b, j),(b, j)) then we apply the rule µ((b, j),(b, j))
to get s in the form aµ((b,j),(b,j))+ru
bij where 0 ≤ r < ν((b, j),(b, j)) −
We can list these using the rules of the formaµ((b,j),(c,k))ubij =aν((b,j),(c,k))ucik
for (b, j)6= (c, k).
By inspection one can see that this is solvable in linear time.
(B) By Theorem 5.36 S has solvable word problem if and only if G×E has solvable word problem and from Theorem 5.37 G×E has solvable word problem if and only if G has solvable word problem. Thus S has solvable word problem if and only if G has solvable word problem.
As in the case for finite presentability we can use an ideal to make a semigroup with one end and an infiniteR-class with unsolvable word problem from a group G even if G has solvable word problem.
A subset S of Zn is said to be Diophantine if there exists a polyno-
mial P(x1, x2, . . . , xn, y1, y2, . . . ym) such that (s1, . . . , sn) ∈ S if and only if
P(s1, s2, . . . , sn, y1, y2, . . . ym) has an integer root.
A subset S of Zn is said to be recursive if both S and Zn\S are Dio- phantine.
Example 5.39. Let Σ be a Diophantine non-recursive set of positive integers. The group H = Gpha, b, c, d | a−ibai = c−idci(i ∈ Σ)i has unsolvable word
problem, see [27, Theorem 7.7]. The group G =Gpha, b, c, d | i ×Z has one end by Lemma 2.45 and has solvable word problem. However, the semigroup constructed by taking G and adding an element z with the property that
z2 = z, z · g = z for all g ∈ G and g−1a−ibaigd = g−1c−idcigd for all i∈Σ and all g ∈G gives a semigroup with one end and an unsolvable word problem.