The Term Equivalence Problem in M2(Z3)

The Term Equivalence Problem in M2(Z3)


Chris Eagle and Collin Roberts

2009-03-03


Abstract

Abstract

The term equivalence problem for a ring R asks whether a givenequation is always true in R.

In this talk, we shall prove that the term equivalence problem iscoNP-complete for the ring M2(Z3). This result was proved bySzabo and Vertesi in their paper ‘The complexity of checkingidentities for finite matrix rings’, Algebra univers. 51 (2004)439-445.


The Semigroup M2(Z3)


The semigroup M2(Z3) contains 81 elements.

Of these, 48 elements are invertible and under matrix multiplicationthese form a group which we denote M2(Z3)∗.

The remaining 33 elements are singular and under multiplicationthese form a semigroup which we denote M2(Z3)#.

We can characterize these matrices by their actions on the1-dimensional subspaces of Z3 × Z3.




There are 4 1-dimensional subspaces of Z3 × Z3, which are spanned bythe following vectors:

v1 =

[10

]= u2

v2 =

[01

]= u1

v3 =

[11

]= u4

v4 =

[12

]= u3




Define the following relation on M2(Z3).

A ∼ B ⇐⇒ (A = B or A = 2B)

Since in Z3, 2 · 2 = 1, it is easy to see that this is an equivalencerelation.

Since for any A, B, C and D, A ∼ B and C ∼ D imply thatAC ∼ BD, this equivalence relation is compatible with the operationof matrix multiplication. Therefore ∼ is a congruence on thesemigroup M2(Z3).

It is also clear that A ∼ B if and only if A and B induce the sametransformation on the 1-dimensional subspaces of Z3 × Z3.

Last, definePM2(Z3) = M2(Z3)/ ∼




As it is easier to work with, we first prove a result for PM2(Z3) which issimilar to the main result of the talk.

Theorem 2.1

TERMEQ is coNP-complete for the semigroup PM2(Z3).




We will need the following result to characterize the group PM2(Z3)∗.

Proposition 2.2

If n > 1, then the symmetric group Sn has a presentation with generatorsx1, . . . , xn−1 and relations

1 = x2i = (xjxj+1)3 = (xkxl)

2

for 1 ≤ i ≤ n − 1, 1 ≤ j ≤ n − 2, 1 < l ≤ k − 1 < n − 1.

Proof: Theorem 2.2.2 in [2]. �

Proposition 2.3

Considering the action of invertible matrices on the 1-dimensionalsubspaces of Z3 × Z3, the group of invertible matrices, PM2(Z3)∗, isisomorphic to the 4-letter symmetric group S4.




Proof: By 2.2, for S4, we need to exhibit x1, x2, x3 ∈ M2(Z3)∗ such that(x1)2 = 1, (x2)2 = 1, (x3)2 = 1, (x1x2)3 = 1, (x2x3)3 = 1, (x3x1)2 = 1.Taking the following matrices, it is routine to verify that the requiredrelations hold.

x1 =

[0 11 0

], x2 =

[1 20 2

], x3 =

[1 00 2

]

So the subgroup

⟨[0 11 0

],

[1 20 2

],

[1 00 2

]⟩≤ PM2(Z3)∗ is

isomorphic to a quotient of S4. We need to show that this quotientcontains more than 12 distinct matrix classes. Then the quotient canonly we the whole of S4. And then since |PM2(Z3)∗| = |S4| = 24, theonly possibility is that PM2(Z3)∗ ∼= S4, as required. Here are the 13distinct classes with their factorizations in terms of the generators.




1 x1 =

[0 11 0

]2 x2 =

[1 20 2

]3 x3 =

[1 00 2

]4 x1x2 =

[0 21 2

]5 x1x2x1x2 =

[2 12 0

]6 x2x3 =

[1 10 1

]7 x2x3x2x3 =

[1 20 1

]8 x1x3 =

[0 12 0

]9 x1x2x3 =

[0 11 1

]10 x3x2x1 =

[2 11 0

]11 x1x2x3x1x2x3 =

[1 11 2

]12 x1x3x2 =

[0 22 1

]13 x2x3x1 =

[1 11 0

]�




Note that in all the computations that follow, we composepermutations from left to right (as if the permutations are actingfrom the right). This choice is completely arbitrary. We could justas easily have made the permutations act from the left andcomposed from right to left.

From our earlier definitions of the vi s and ui s, we have that each ui

represents the unique line orthogonal to vi . Thus uTt vs = 0 if and

only if t = s.




Proposition 2.4

Every non-zero singular matrix in PM2(Z3)# has a unique representationof the form vsuT

t .

Proof: It is routine to verify this. �Denote this class of vsuT

t by 〈s, t〉.




As every class of matrices induces a natural right and left action on therepresentatives of the lines, we can consider these matrices aspermutations on the set {1, 2, 3, 4}.

Proposition 2.5

The semigroup PM2(Z3) is isomorphic to

S = S4 ∪ {〈s, t〉 : 1 ≤ s, t ≤ 4} ∪ {0}

where multiplication is defined by

0 · a = a · 0 = 0 for all a ∈ PM2(Z3)

multiplication in S4 is the usual composition

〈s, t〉π = 〈s, π(t)〉 for all π ∈ S4

π〈s, t〉 = 〈π−1(s), t〉 for all π ∈ S4

〈s1, t1〉〈s2, t2〉 =

{0 if t1 = s2

〈s1, t2〉 if t1 6= s2




Proof: It is easy to check that the given multiplication operation isassociative.Define

ϕ : PM2(Z3)# −→ S#[0 00 0

]7→ 0

vsuTt 7→ 〈s, t〉

By (2.4), this is a bijection. Now we have to prove it is also a semigrouphomomorphism. Observe that

(vs1 uTt1

)(vs2 uTt2

)

= vs1 (uTt1

vs2 )uTt2

=

[

0 00 0

]if uT

t1vs2 = 0⇐⇒ t1 = s2

vs1 uTt2

if uTt1

vs2 6= 0⇐⇒ t1 6= s2




and therefore

ϕ((vs1 uTt1

)(vs2 uTt2

))

=

{0 if uT

t1vs2 = 0⇐⇒ t1 = s2

〈s1, t2〉 if uTt1

vs2 6= 0⇐⇒ t1 6= s2

while

ϕ(vs1 uTt1

)ϕ(vs2 uTt2

)

= 〈s1, t1〉〈s2, t2〉

=

{0 if t1 = s2

〈s1, t2〉 if t1 6= s2

So the definitions agree and we have an isomorphismϕ : PM2(Z3)# → S#.




By 2.3, we also have an isomorphism Ψ : PM2(Z3)∗ → S∗ ∼= S4. Nowdefine

Φ : PM2(Z3) −→ Sa 7→ Ψ(a), ∀a ∈ PM2(Z3)∗

b 7→ ϕ(b), ∀b ∈ PM2(Z3)#

Then Φ is a bijection, since Ψ and ϕ are. If a, b both come from eitherthe group of invertible matrices or the semigroup of singular matrices,then Φ commutes with the group / semigroup multiplication, by earlierresults. We still need to check the following cases.




a ∈ PM2(Z3)∗, b = vsuTt ∈ PM2(Z3)#: Then

a(b)

= a(vsuTt )

= (avs)uTt

= (vΨ(a)−1(s))uTt

=⇒ Φ(ab) = 〈Ψ(a)−1(s), t〉= Ψ(a)〈s, t〉= Ψ(a)ϕ(b)

= Φ(a)Φ(b)




a ∈ PM2(Z3)∗, b = vsuTt ∈ PM2(Z3)#: Then

(b)a

= (vsuTt )a

= vs(uTt a)

= vsuTΨ(a)(t)

=⇒ Φ(ba) = 〈s, Ψ(a)(t)〉= 〈s, t〉Ψ(a)

= ϕ(b)Ψ(a)

= Φ(b)Φ(a)

So we have exhibited a semigroup isomorphism Φ : PM2(Z3)→ S ,completing the proof. �




For longer products, we will need the following result.

Lemma 2.6

Let aj = 〈sj , tj〉, for 1 ≤ j ≤ n and π1, . . . , πn+1 ∈ S4. Then

1 a2j = a3

j

2 a1π2a2 = 0 if and only π2(t1) = s2

3 π1a1π2a2 · · ·πnanπn+1 = 0 if and only there exists 1 < k ≤ n suchthat ak−1πkak = 0 (i.e. πk(tk−1) = sk)

4 If π1a1π2a2 · · ·πnanπn+1 6= 0 then

π1a1π2a2 · · ·πnanπn+1 = 〈π−11 (s1), πn+1(tn)〉




Proof: (1) aj is a 2× 2 matrix of rank 1. Since det(aj) = 0, one of itseigenvalues must be zero. Therefore its characteristic polynomial has theform x2 + ax , for some a ∈ Z3. By the Cayley-Hamilton Theorem, aj

must satisfy its characteristic polynomial. There are three cases toconsider.

1 (a = 0) The characteristic polynomial is x2, which says that

(aj)2 = 0. Then (aj)

3 = 0 = (aj)2.

2 (a = 1) The characteristic polynomial is x2 + x , which says that

(aj)2 + aj = 0⇐⇒ (aj)

2 = −aj . Then multiplying both sides by aj

gives (aj)3 = −(aj)

2, implying (aj)3 ∼ (aj)

2.

3 (a = 2) The characteristic polynomial is x2 + 2x , which says that

(aj)2 + 2aj = 0⇐⇒ (aj)

2 = aj . Then multiplying both sides by aj

gives (aj)3 = (aj)

2.




(2) (⇒) Assume that a1π2a2 = 0. Then

0 = a1π2a2

= 〈s1, t1〉π2〈s2, t2〉= 〈s1, π2(t1)〉〈s2, t2〉

=

{0 if π2(t1) = s2

〈s1, t2〉 if π2(t1) 6= s2

And since every 〈s1, t2〉 is non-zero, we must have that π2(t1) = s2, asrequired.(⇐) Assume that π2(t1) = s2. Then

a1π2a2 = 〈s1, t1〉π2〈s2, t2〉= 〈s1, π2(t1)〉〈s2, t2〉= 〈s1, s2〉〈s2, t2〉= 0




(3) (⇒) The proof is by induction on n.Notice that by definition, a1 is never zero. Thus for anyπ1, π2 ∈ S4, π1a1π2 is never zero. So for the base case (n = 1), assumethat π1a1π2a2 = 0. This implies a1π2a2 = 0, and therefore by (2), wehave that π2(t1) = s2.Now assume the result holds up to n = k . Then for n = k + 1, supposethat π1a1π2a2 · · ·πk+1ak+1πk+2 = 0. If πk+1ak+1πk+2 = 0, then by (2),we have that πk+1(tk) = sk+1. Otherwise, it must be thatπ1a1π2a2 · · ·πkakπk+1 = 0, and the result follows by the inductionhypothesis.(⇐) Assume that for some k , ak−1πkak = 0. Then it is clear from themultiplication law that π1a1π2a2 · · ·πnanπn+1 = 0, as required.




(4) The proof is by induction on n. For the base case (n = 1), we have

π1a1π2 = π1〈s1, t1〉π2

= 〈π−11 (s1), π2(t1)〉

as required.Now assume the result holds up to n = k . For n = k + 1, we compute,using the induction hypothesis for the first part

π1a1π2a2 · · ·πk+1ak+1πk+2

= 〈π−11 (s1), πk+1(tk)〉〈sk+1, tk+1〉πk+2

= 〈π−11 (s1), πk+1(tk)〉〈sk+1, πk+2(tk+1)〉

=

{0 if πk+1(tk) = sk+1

〈π−11 (s1), πk+2(tk+1)〉 if πk+1(tk) 6= sk+1

By hypothesis, the product is not zero, and we get the desired result.�




Proposition 2.7

1 S4/V ∼= S3, where V is the Klein 4-group

2 The commutator subgroup of S4 is A4

Proof:1 The normal subgroups of S4 are {1}, V , A4 and S4. So the quotient

S4/V makes sense. |S4/V | = 244 = 6. Therefore S4/V ∼= C6 or S3.

Since S4 contains no element of order 6, neither does S4/V .Therefore S4/V cannot be isomorphic to C6. Therefore S4/V ∼= S3.

2 The commutator subgroup, S ′4, of S4 is always a normal subgroup.S4 is not abelian, therefore S ′4 6= {1}. By Proposition 7 on p169 of[1], we also know that the quotient S4/S ′4 is abelian, and is thelargest abelian quotient of S4. It is clear that S4/A4

∼= C2, since itsorder is 2. This is abelian, therefore S ′4 6= S4. By (1) S4/V ∼= S3,which is not abelian. The only possibility that remains is thatS ′4 = A4.

�




We can then write the cosets of S4/V ∼= S3 as follows, using thepresentation S3 = 〈a, b : a2 = b3 = 1, aba = b2〉

1 : V = { 1, (12)(34), (13)(24), (14)(23) }a : (12)V = { (12), (34), (1423), (1324) }

ab : (13)V = { (13), (1432), (24), (1234) }ab2 : (14)V = { (14), (1342), (1243), (23) }

b : (123)V = { (123), (243), (142), (134) }b2 : (124)V = { (124), (234), (143), (132) }

Now define

D = { (12)(34), (13)(24), (14)(23) }T = { (123), (132), (124), (142), (134), (143), (234), (243) }

We can then write the commutator subgroup of S4, A4 as

A4 = {id} ∪ D ∪ T

We list a few useful properties of S4.




Lemma 2.8

Let u, v , y , a ∈ S4 and put w = [y , uv−1] = y−1(uv−1)−1y(uv−1) fory ∈ S4. Then the following hold:

1 a12 = id

2 w ∈ A4, the commutator subgroup of S4.

3 w 6 = id

4 w 3 = id if and only if w ∈ {id} ∪ T .

5 For u, v ∈ S4, there exists an e ∈ S4 such that [e, uv−1] ∈ D if andonly if uv−1 6= id, that is, if and only if u 6= v.

6 If w ∈ D, then the subgroup generated by { u, v , y } is transitiveon { 1, 2, 3, 4 }.

7 If w ∈ D, then the set { abc : a, b, c ∈ {u, v , y} } is transitiveon { 1, 2, 3, 4 }.




Proof: (1) The exponent of S4 is LCM{ 1, 2, 3, 4 } = 12.(2) This is clear since w is a commutator.(3) w lies in A4, which contains elements of orders 1, 2 and 3. Thereforethe exponent of A4 is 6.(4) Since w lies in A4 = {id} ∪ D ∪ T , and D contains all the elements

of order 2, it is easy to see that w 3 = 1 if and only if w ∈ {id} ∪ T .(5) (⇒) Assume that we have such an e. For a contradiction, assume

u = v ⇔ uv−1 = id . Then

[e, uv−1] = [e, id ]

= id

/∈ D

giving us the required contradiction.




(⇐) It suffices to exhibit a choice of e ∈ S4 that has the requiredproperty for each non-identity choice of uv−1.

Coset of S4/V uv−1 e [e, uv−1]V

(12)(34) (13) (13)(12)(34)(13)(12)(34) = (13)(24)(13)(24) (12) (12)(13)(24)(12)(13)(24) = (12)(34)(14)(23) (12) (12)(14)(23)(12)(14)(23) = (12)(34)

(12)V(12) (1423) (1324)(12)(1423)(12) = (12)(34)(34) (1423) (1324)(34)(1423)(34) = (12)(34)(1423) (12) (12)(1324)(12)(1423) = (12)(34)(1324) (12) (12)(1423)(12)(1324) = (12)(34)




(13)V(13) (1234) (1432)(13)(1234)(13) = (13)(24)(24) (1234) (1432)(24)(1234)(24) = (13)(24)(1234) (13) (13)(1432)(13)(1234) = (13)(24)(1432) (13) (13)(1234)(13)(1432) = (13)(24)

(14)V(14) (1243) (1342)(14)(1243)(14) = (14)(23)(23) (1243) (1342)(23)(1243)(23) = (14)(23)(1342) (14) (14)(1243)(14)(1342) = (14)(23)(1243) (14) (14)(1342)(14)(1243) = (14)(23)




(123)V(123) (234) (234)(132)(234)(123) = (14)(23)(243) (143) (134)(234)(143)(243) = (12)(34)(142) (132) (123)(124)(132)(142) = (12)(34)(134) (124) (142)(143)(124)(134) = (14)(23)

(124)V(124) (123) (132)(142)(123)(124) = (12)(34)(234) (123) (132)(243)(123)(234) = (14)(23)(143) (142) (124)(134)(142)(143) = (14)(23)(132) (243) (234)(123)(243)(132) = (14)(23)




(6) The authors do not give a proof of the last 2 parts of this lemma. Ihave not found an elegant proof yet, so I am left with brute force. Noticealso that (7) implies (6), so proving (7) suffices for both parts. The workto carry out the brute force proof can be cut down by observing thefollowing facts. We use the cosets of S4/V ∼= S3 and the presentation ofS3 which we identified earlier.Case 1 (uv−1 ∈ D) uv−1 is a product of 2 disjoint cycles. Then y comes

from a coset other than V (if not, then the commutator [y , uv−1] willbe trivial since V is abelian, which is a contradiction).




Case 2 (uv−1 ∈ (12)V or (13)V or (14)V ) We claim that y must come

from the same coset as uv−1 or from D. We prove this in the case whenuv−1 ∈ (12)V . All other proofs can be done in exactly the same way. Weidentify (12)V with a ∈ S3. We need our commutator to lie in D ⊂ Vwhich we identify with 1 ∈ S3.

1 y ∈ ab =⇒ (ab)−1a−1(ab)a = b−1a−1a−1aba = b2b2 = b 6= 1

2 y ∈ ab2 =⇒ (ab2)−1a−1(ab2)a = b−2a−1a−1ab2a = b2 6= 1

3 y ∈ b =⇒ (b)−1a−1(b)a = b−1a−1ba = b2b2 = b 6= 1

4 y ∈ b2 =⇒ (b2)−1a−1(b2)a = b−2a−1b2a = b2 6= 1




Case 3 (uv−1 ∈ (123)V or (124)V ) Then uv−1 is a 3-cycle. Then ymust lie in either V , (123)V or (124)V (so y is a 3-cycle). Also y is notuv−1 or the inverse of uv−1 (if not, then the commutator [y , uv−1] istrivial, a contradiction). We prove this in the case when uv−1 ∈ (123)V .The other proof can be done in exactly the same way. We identify(123)V with b ∈ S3. We need our commutator to lie in D ⊂ V which weidentify with 1 ∈ S3.

1 y ∈ a =⇒ (a)−1b−1ab = ab2ab = bb = b2 6= 1

2 y ∈ ab =⇒ (ab)−1b−1(ab)b = b−1a−1b−1ab2 = b2ab2ab2 =b2bb2 = b2 6= 1

3 y ∈ ab2 =⇒ (ab2)−1b−1(ab2)b = b−2a−1b−1ab2b = bab2a = bb =b2 6= 1

Having done all this, we can reduce the number of commutatorpossibilities we need to check, as shown in the attached table. Here aresome examples to demonstrate how all the resulting cases would thenneed to be checked for the transitivity of the set of 3-fold products.




(7) u = (12), v = (34), y = (34).

v 3 : 1 7→ 1 y 3 : 3 7→ 1u3 : 1 7→ 2 uyu : 3 7→ 2y 3 : 1 7→ 3 u3 : 3 7→ 3

yvu : 1 7→ 4 v 3 : 3 7→ 4u3 : 2 7→ 1 uvy : 4 7→ 1y 3 : 2 7→ 2 vyu : 4 7→ 2

uyu : 2 7→ 3 v 3 : 4 7→ 3uyv : 2 7→ 4 u3 : 4 7→ 4




u = (12), v = id , y = (1432).

uy 2 : 1 7→ 1 u2y : 3 7→ 1u3 : 1 7→ 2 y 3 : 3 7→ 2y 3 : 1 7→ 3 u3 : 3 7→ 3

u2y : 1 7→ 4 uy 2 : 3 7→ 4u3 : 2 7→ 1 y 3 : 4 7→ 1

uy 2 : 2 7→ 2 u2y : 4 7→ 2u2y : 2 7→ 3 uy 2 : 4 7→ 3y 3 : 2 7→ 4 u3 : 4 7→ 4




u = (12), v = (13), y = (243).

y 3 : 1 7→ 1 v 3 : 3 7→ 1u3 : 1 7→ 2 vy 2 : 3 7→ 2v 3 : 1 7→ 3 u3 : 3 7→ 3

uvy : 1 7→ 4 uy 2 : 3 7→ 4u3 : 2 7→ 1 y 2u : 4 7→ 1v 3 : 2 7→ 2 y 2v : 4 7→ 2

y 2u : 2 7→ 3 u2y : 4 7→ 3v 2y : 2 7→ 4 u3 : 4 7→ 4

�




Lemma 2.9

Let u, v , y ∈ S4 and w as in (2.8), and let b be a non-invertible elementof PM2(Z3). If w 3 6= id, then (

∏cr∈{y ,u,v}(bc1c2c3))b = 0.

Proof: Let b = 〈s, t〉. By (4) of (2.8), w ∈ D. By (7) of (2.8), there is aproduct of the form c1c2c3 taking t to s. Hence, by (3) of (2.6), theproduct is equal to 0. �


Break

Break

Please take a few minutes now to stretch your legs before we prove themain result. . .


The Main Result

The Main Result

We reduce graph 24-colourability, which is known to becoNP-complete, into TERMEQ for PM2(Z3).

Let Γ = (V ,E ) be a finite simple graph, with V = {v1, . . . , vn} andE = {e1, . . . , em}.For each element of V and E we introduce a variable of the samename.

For each i ∈ {1, . . . ,m}, with ei = {vk , vl}, we letwi = e11

i vlv11k eivkv 11

l .


The Main Result

The Main Result

Observe that, if ei , vl , vk ∈ S4 then wi = [ei , vkv−1l ].

We also introduce the following, where ei = {vk , vl}:

P =∏ei∈E

(xw 3i )12

Q =∏ei∈E

(xw 6i )12

Hi,j =∏

cr∈{ei ,vl ,vk}

(wjc1c2c3wj)12

H =∏

ei ,ej∈E

Hi,j

p = PPPxH

q = PQPxH

We show that there exists a substitution for the variables such that p 6= qif and only if Γ is 24-colourable.


The Main Result

The Main Result

Case 4.1

All variables are in S4.

Proof: Since the exponent of S4 is 12, we see that

P =∏ei∈E

(xw 3i )12 =

∏ei∈E

id = id

andQ =

∏ei∈E

(xw 6i )12 =

∏ei∈E

id = id

andHi,j =

∏cr∈{ei ,vk ,vl}

(wjc1c2c3wj)12 =


id = id


The Main Result

The Main Result

SoH =

∏ei ,ej∈E

Hi,j =∏

ei ,ej∈E

id = id

and hencep = PPPxH = x

q = PQPxH = x

So p = q in this case. �


The Main Result

The Main Result

Case 4.2

There exists wj = 〈s, t〉 /∈ S4 and there exists i such that wi ∈ D.

Proof: We observe that

Hi,j =∏

cr∈{ei ,vk ,vl}

(wjc1c2c3wj)12

=∏

cr∈{ei ,vk ,vl}

(〈s, t〉c1c2c3〈s, t〉)12


The Main Result

The Main Result

Next,

〈s, t〉c1c2c3〈s, t〉 = 〈s, c1(t)〉c2c3〈s, t〉= 〈s, c2(c1(t))〉c3〈s, t〉= 〈s, c3(c2(c1(t)))〉〈s, t〉

=

{0 if c3(c2(c1(t))) = s

〈s, t〉 otherwise

=

{0 if c3(c2(c1(t))) = s

wj otherwise


The Main Result

The Main Result

So if c3(c2(c1(t))) = s, or if s = t, then

Hi,j =∏

cr∈{ei ,vk ,vl}

(wjc1c2c3wj)12 = 0

So H = 0.

Otherwise,

Hi,j =∏

cr∈{ei ,vk ,vl}

(wjc1c2c3wj)12

=∏

cr∈{ei ,vk ,vl}

(wj)12

= w 12·3!j

= wj


The Main Result

The Main Result

Since c3(c2(c1(t))) 6= s for all choices of c1, c2, c3, we have, for somec ′1, c

′2, c′3 ∈ {ei , vk , vl},∏

cr∈{ei ,vk ,vl}

(wjc1c2c3) =∏

cr∈{ei ,vk ,vl}

〈s, c3(c2(c1(t)))〉

= 〈s, c ′3(c ′2(c ′1(t)))〉

So

Hi,j = wj

= 〈s, t〉= 〈s, c ′3(c ′2(c ′1(t)))〉〈s, t〉= (


(wjc1c2c3))wj


The Main Result

The Main Result

Now by (2.9),

(∏

cr∈{ei ,vk ,vl}

(wjc1c2c3))wj = 0

and hence Hi,j = 0 in this case as well.Thus we have shown that in this case we always have Hi,j = 0, and soH = 0 and p = q = 0. �


The Main Result

The Main Result

Case 4.3

There exists wj /∈ S4, and for all other ei ∈ E we have wi ∈ PM2(Z3)\D.

Proof: Since A4 = T ∪ D ∪ {id} and, for any k , wk ∈ A4 or wk /∈ S4, wehave that wi ∈ T ∪ {id} or wi /∈ S4.If wi ∈ T ∪ {id} then, since every element of T is a 3-cycle, w 3

i = id .If wi /∈ S4 then, by (2.6), we have w 2

i = w 3i , so w 6

i = w 3i .

Thus in either case w 3i = w 6

i , and so

P =∏ei∈E

(xw 3i )12

=∏ei∈E

(xw 6i )12

= Q

Thusp = PPPxH = PQPxH = q

�


The Main Result

The Main Result

Case 4.4

For all i , wi ∈ S4, and x = 〈s, t〉 with s 6= t.

Proof: Since s 6= t, 〈s, t〉〈s, t〉 = 〈s, t〉, so by induction, for any n ≥ 2,〈s, t〉n = 〈s, t〉. By (2.8), w 6

i = id . Thus

Q =∏ei∈E

(xw 6i )12

=∏ei∈E

x12

= x12|E |

= 〈s, t〉12|E |

= 〈s, t〉= x

So q = PQPxH = PxPxH.


The Main Result

The Main Result

Next we examine P.Since wi ∈ S4, so actually wi ∈ A4, and hence w 3

i ∈ A4. For each i , letzi = w 3

i . Then P =∏

ei∈E (xzi )12.

Now xzi = 〈s, t〉zi = 〈s, zi (t)〉.


The Main Result

The Main Result

There are two cases to consider.

1 First, if there exists zi such that zi (t) = s then, for that i , we havexzi = 〈s, s〉, and so (xzi )

12 = ((xzi )2)6 = (〈s, s〉〈s, s〉)6 = 06 = 0.

Thus in this case P = 0, so p = q = 0.

2 Otherwise, if there is no i with zi (t) = s, then

P =∏ei∈E

(〈s, t〉zi )12

=∏ei∈E

〈s, zi (t)〉12

=∏ei∈E

〈s, zi (t)〉

= 〈s, z|E |(t)〉


The Main Result

The Main Result

So, since z|E |(t) 6= s,

p = PPPxH

= 〈s, z|E |(t)〉3〈s, t〉H= 〈s, z|E |(t)〉〈s, t〉H= 〈s, t〉H= xH

While

q = PQPxH

= PxPxH

= 〈s, z|E |(t)〉〈s, t〉〈s, z|E |(t)〉〈s, t〉H= 〈s, t〉2H

= 〈s, t〉H= xH

So p = q throughout this case. �


The Main Result

The Main Result

Case 4.5

For all i , wi ∈ S4, and x = 〈s, s〉.

Proof: We have wi ∈ A4, so by (2.8) w 6i = id . Also, 〈s, s〉2 = 0, so

Q =∏ei∈E

(xw 6i )12

=∏ei∈E

x12

=∏ei∈E

0

= 0


The Main Result

The Main Result

So q = 0. Thus p 6= q ⇐⇒ there is a substitution such that p 6= 0.Now with wj ∈ S4 we have wjc1c2c3wj ∈ S4, so since the exponent of S4

is 12, we have

Hi,j =∏

cr∈{ei ,vk ,vl}

(wjc1c2c3wj)12 =


id = id

So H = id as well, and p = PPPx .Next, if there is an i such that zi (s) = s then for that i ,

(xzi )12 = (〈s, s〉zi )

12 = 〈s, zi (s)〉12 = 〈s, s〉12 = 0


The Main Result

The Main Result

Otherwise,

(xzi )12 = (〈s, s〉zi )

12 = 〈s, zi (s)〉12 = 〈s, zi (s)〉 = xzi

Thus

P =∏ei∈E

(xw 3i )12

=∏ei∈E

(xzi )12

=

{0 if there exists i such that zi (s) = s∏

ei∈E (xzi ) otherwise

=

{0 if there exists i such that zi (s) = s

〈s, z|E |(s)〉 otherwise


The Main Result

The Main Result

Now, if zi (s) 6= s, then

xz1 . . . xz|E | = 〈s, s〉z1〈s, s〉z2 . . . 〈s, s〉z|E |= 〈s, z1(s)〉〈s, s〉z2 . . . 〈s, s〉z|E |= 〈s, s〉z2 . . . 〈s, s〉z|E |= . . .

= 〈s, s〉z|E |= 〈s, z|E |(s)〉

So, as 〈s, z|E |(s)〉 6= 0, we have

P3 =


〈s, z|E |(s)〉 otherwise

So

P3x =


x otherwise


The Main Result

The Main Result

So there exists a substitution such that p 6= q ⇐⇒ there is no i suchthat zi (s) = s. That is, if and only if no w 3

i stabilizes s.Now by (2.8), w 3

i (s) = s ⇐⇒ wi ∈ T ∪ {id}.So there is a substitution such that p 6= q ⇐⇒ for all i , wi /∈ T ∪ {id}.Now in the case we are considering wi is a commutator, sowi ∈ A4 = D ∪ T ∪ {id}. We thus have, writing wi = [ei , vkv−1

j ], soei = {vk , vj},

There is a substitution p 6= q

⇐⇒ ∀i ,wi /∈ T ∪ {id}⇐⇒ ∀i ,wi ∈ D

⇐⇒ ∀i , vk 6= vj(by (2.8))

⇐⇒ for all ei = {vk , vj} ∈ E , vk 6= vj

⇐⇒ Adjacent vertices are assigned different elements of S4

⇐⇒ Γ is 24-colourable

�


The Main Result

The Main Result

We have thus shown that p 6= q ⇐⇒ Γ is 24-colourable, as required.By two additional results proved by Professor Lawrence, we get thedesired result for M2(Z3).


The Main Result

The Main Result

Theorem (Lawrence) 4.6

Let F be any field. Let n be a positive integer greater than or equal to 2.If A, B ∈ Mn(F ) and B 6= ±A, then there exists C ∈ Mn(F ) such thatACA 6= BCB.

Proof: For a contradiction, assume that for all C ∈ Mn(F ),

ACA = BCB (1)

Write A = (aij), B = (bij). Let C = Eji , the matrix that has a 1 inposition (j , i) and zeros elsewhere. By (1), AEjiA = AEjiA. Consider the(i , j) entry of each side. Equality then gives (aij)

2 = (bij)2. There are two

cases to consider.Case 1 (F has characteristic 2) If F has characteristic 2, then square

roots in F are unique, so (aij)2 = (bij)

2, implies bij = aij . Since (i , j) wasarbitrary, this says that (aij) = (bij), i.e. A = B. This contradicts ourhypothesis, and it completes this case.


The Main Result

The Main Result

Case 2 (F has characteristic different from 2) If F has characteristic

different from then 2, (aij)2 = (bij)

2, implies bij = ±aij .Now suppose there is some (s, t) and (u, v) such that bst = ast 6= 0 andbuv = −auv . Let C = Etu. Checking the (s, v) entry of both sides of (1),one gets astauv = bstbuv . But from the above facts, −astauv = bstbuv .This is the required contradiction. �


The Main Result

The Main Result

Theorem (Lawrence) 4.7

W1 ≈W2 is an identity in PM2(Z3) if and only if W1YW1 ≈W2YW2 isan identity in M2(Z3), where Y does not occur in W1 or W2.

Proof: (⇒) If W1 ≈W2 in PM2(Z3), then W2 = ±W1 for anysubstitution. Then for all Y , W1YW1 = W2YW2 in M2(Z3).(⇐) If W1YW1 ≈W2YW2 in M2(Z3), then by (4.6), W2 = ±W1 (forany substitution), i.e. W2 ≈W1 in PM2(Z3). �


The Main Result

Dummit, David S. and Foote, Richard M., Abstract Algebra, Wiley& Sons, Hoboken, New Jersey, 2004

Robinson, Derek J. S., A Course in the Theory of Groups, SecondEdition, Springer, New York, 1996

The Term Equivalence Problem in M2(Z3)

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