MAT00050M Assignment
Creation date:2024-05-09 17:32:35
Assignment
MAT00050M Assignment
Page 1 (of 4)
MAT00050M
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Page 2 (of 4)
MAT00050M
1 (of 4). Consider the transformation semigroup TX with X = {1, 2, 3, 4, 5}.
(a) Show that the set
T =
{(
1 2 3 4 5
1 1 3 5 5
)
,
(
1 2 3 4 5
3 3 5 1 1
)
,
(
1 2 3 4 5
5 5 1 3 3
)}
forms a subgroup of TX . [8]
(b) Find the maximal subgroup M of TX containing T and list its elements. [9]
(c) To which well-known group is T isomorphic? To which well-known group is
M isomorphic? Write down explicit isomorphisms. [8]
2 (of 4). Consider the monogenic semigroup 〈a〉, where a has index n and period r.
(a) Describe all the principal ideals of 〈a〉. [7]
(b) Prove that all ideals of 〈a〉 are principal. [6]
(c) Let I be any ideal of 〈a〉. Prove that 〈a〉/I is monogenic, and describe all the
possible values the index and period of its generator can take. [6]
(d) Determine the values of n and r for which all congruences of 〈a〉 are Rees
congruences. [6]
Page 3 (of 4) Turn over
MAT00050M
3 (of 4). Consider the semigroup (R2×2, ·) of all 2-by-2 matrices with real entries under
multiplication, and let
S =
{(
1 0
0 0
)
,
(
0 1
0 0
)
,
(
0 0
1 0
)
,
(
0 0
0 1
)
,
(
0 0
0 0
)}
.
(a) Prove that S forms a subsemigroup of R2×2. [5]
(b) Determine the relations L,R,H and D on S and draw the egg-box diagrams
of all D-classes of S. [10]
(c) Deduce that S is completely 0-simple. [3]
(d) Give a Rees matrix semigroup isomorphic to S, making use of the proof of
Rees’s theorem. [7]
4 (of 4). Let S, T be inverse semigroups, and denote the inverse of s by s′.
(a) Prove that if ϕ : S → T is a semigroup morphism, then for any s ∈ S, we
have s′ϕ = (sϕ)′. [6]
(b) Deduce that if ρ is a congruence on S, then for any s, t ∈ S, we have s ρ t
implies s′ ρ t′. [6]
(c) Let ρ be any congruence on S. Prove that ρ ⊆ H if and only if for all
idempotents e, f of S, whenever e ρ f we have e = f . [8]
(d) Give an example of a nontrivial inverse semigroup which has no nontrivial
congruences ρ satisfying ρ ⊆ H. [5]
Page 4 (of 4) End of examination.
SOLUTIONS: MAT00050M
1. (a) Put α =
(
1 2 3 4 5
1 1 3 5 5
)
, β =
(
1 2 3 4 5
3 3 5 1 1
)
, γ =
(
1 2 3 4 5
5 5 1 3 3
)
.
One can check by computation that α is as an identity element in T , and
βγ = γβ = α, furthermore β2 = γ, γ2 = β. So T is closed under multiplica-
tion, and every element has an inverse with respect to α, (β−1 = γ, γ−1 = β). 8 Marks
(b) We know by the maximal subgroup theorem that M is the (common)H-class
of elements of T . The kernel partition of elements of T is {{1, 2}, {3}, {4, 5}},
and the image is {1, 3, 5}. The elements ofM are all the maps with this kernel
and image, so
M = T ∪
{(
1 2 3 4 5
1 1 5 3 3
)
,
(
1 2 3 4 5
5 5 3 1 1
)
,
(
1 2 3 4 5
3 3 1 5 5
)}
.
9 Marks
(c) We have T ∼= Z3 as that is the only group of order 3 up to isomorphism. An
isomorphism is ϕ : T → Z3, α 7→ 0, β 7→ 1, γ 7→ 2.
The group M is isomorphic to S3 via ψ : M → S3,(
1 2 3 4 5
1 1 3 5 5
)
ψ =
(
1 2 3
1 2 3
)
,
(
1 2 3 4 5
1 1 5 3 3
)
ψ =
(
1 2 3
1 3 2
)
,
(
1 2 3 4 5
3 3 1 5 5
)
ψ =
(
1 2 3
2 1 3
)
,
(
1 2 3 4 5
3 3 5 1 1
)
ψ =
(
1 2 3
2 3 1
)
,(
1 2 3 4 5
5 5 1 3 3
)
ψ =
(
1 2 3
3 1 2
)
,
(
1 2 3 4 5
5 5 3 1 1
)
ψ =
(
1 2 3
3 2 1
)
. 8 Marks Total: 25 Marks
2. (a) Put S = 〈a〉. Let k ∈ N, then we have b ∈ S1akS1 if and only if b =
axakay = ak+x+y for some x, y ∈ N0 (where a0 = 1), that is, if and only if
b = al for some l ≥ k.
We claim that
am ∈ S1akS1 ⇐⇒ m ≥ n or m ≥ k.
Recall that elements of S are uniquely of the form am with 1 ≤ m ≤ n+r−1.
If m < n, then for any l ∈ N, am = al implies m = l, therefore if m < n,
m < k, then we have am /∈ S1akS1 as am does not arise as al for any l ≥ k.
5
SOLUTIONS: MAT00050M
However, if m ≥ n, then am = am+kr, and as m + kr ≥ k, we have am =
am+kr ∈ S1akS1 by the first paragraph. If m ≥ k then of course we also have
am ∈ S1akS1 by the first paragraph. This proves our claim.
So S1akS1 = {am : min(k, n) ≤ m ≤ n + r − 1}. If k ≤ n, then this is
just the set {ak, ak+1, . . . , an+r−1}, if k > n then this is (independently of k)
{an, an+1, . . . , an+r−1}.
7 Marks
(b) Let I be any ideal, and let ak ∈ I (where 1 ≤ k ≤ n + r − 1) be such
that k is minimal. We will prove that I = S1akS1. We of course have
S1akS1 ⊆ I , so in particular k is the minimal exponent in S1akS1 as well,
thus k ≤ n by part (a). But then S1akS1 = {ak, ak+1, . . . , an+r−1} by part
(a) and I ⊆ {ak, ak+1, . . . , an+r−1} by the minimality of k, which proves the
claim.
6 Marks
(c) From parts (a) and (b), we have I = {ak, ak+1, . . . , an+r−1} for some 1 ≤
k ≤ n. Recall that S/I ∼= S \ I ∪ {0} = {a, . . . , ak−1, 0}, where 0 is a zero
element, and we have am = 0 for any m ≥ k. It is clear that a generates S/I ,
the index of a is k, and the period of a is 1.
6 Marks
(d) By exercise 3/6, we have that the image of any surjective morphism ϕ : 〈a〉 →
T is a monogenic semigroup generated by b = aϕ, and its index can be
any n′ ∈ N with n′ ≤ n and its period any r′ ∈ N with r′ | r. By the
homomorphism theorem, the congruences of 〈a〉 are exactly the kernels of
such maps ϕ, and thus we need to find for which n and r are these kernels all
Rees congruences. If kerϕ = ρI for some ideal I , then
〈b〉 = T ∼= 〈a〉/ kerϕ = 〈a〉/ρI
and so by part (c) we must have r′ = 1. This happens for all surjective mor-
phisms ϕ : 〈a〉 → T if and only if the only possibility for r′ ∈ N with r′ | r is
r′ = 1, that is if r = 1.
6 Marks Total: 25 Marks
3. (a) We need to prove that S is closed under taking products. The zero matrix
is the zero element of R2×2, so any product involving it will be contained in
S. We therefore only need to check the products of the other four matrices.
This can be done brute force but also follows from the following observation:
the product of two matrices each containing at most one nonzero entry will
also contain at most one nonzero entry, the product of the respective entries.