SOLUTIONS

a * a * b * a * b = a * (a * b) * a * b = a * (b * a) * a * b = (a * b) * a * a * b = b * a * a * a * b.

Suppose M is such a monoid. For all x,y in M we have

Hence

In particular, M is commutative.

We define multiplication on S {e} as follows:

For all x, y in S we define x * y to be the product of x and y as in S. Furthermore, as e should be the identity, we define x * e = e * x = x. Finally e * e is defined to be e.

To check associativity, distinguish various cases. For instance, if a, b, c are all in S, then (a * b) * c = a * (b * c), since associativity already holds in S. If a = e and b, c are in S, then we get (e * b) * c = b * c and e * (b * c) = b * c by the definition of multiplication with e. Remaining cases are verified in a similar way.

If S contains already a unit element e', then this element will not be a unit anymore in (S {e},*,e). Indeed e' * e = e and not e'.

The monoid (Z/2Z, ·, 1) is the extension of the semigroup (0+2Z, ·).

Let M = (m(i,j)) be an n × n multiplication table for a binary operation on n elements. Spot i,j contains the product of the i-th and j-th element (in that order).

Let assoc = comm = TRUE.

do for i from 1 to n
do for j from 1 to n
if m(i,j) <> m(j,i) then comm = FALSE endif
do for k from 1 to n
if   m(m(i,j),k) <> m(i,m(j,k)) then assoc = FALSE endif
enddo
enddo
enddo

Let (M_i, *_i,e_i) be a monoid (i = 1,2). Now consider the direct product M₁ × M₂. To prove that M₁ × M₂ is a monoid, we have to check that the multiplication * is associative and that (e₁, e₂) is an identity element.

Associativity:

Identity:

and

Let M = (Z/6Z,+) and consider the submonoids M₁ = {0, 3} and M₂ = {0, 2, 4}. The union of these submonoids is not a monoid since

is not contained in the union.

Since both S₁ and S₂ are closed under multiplication so is their product. If (a₁,a₂), (b₁,b₂) S₁ × S₂, then a₁ *₁ b₁ S₁ and a₂ *₂ b₂ S₂, so that

(a₁,a₂) * (b₁,b₂) = (a₁ *₁ b₁, a₂ *₂ b₂) S₁ × S₂.

Similarly, since S₁ contains the unit element of M₁ and S₂ contains the unit element of M₂, the product S₁ × S₂ contains the unit element of M₁ × M₂.

Any integral linear combination of the a_i/b_i is an integral multiple of 1/(b₁ ··· b_n). Hence it can never be equal to 1/p.

It follows that Q cannot be finitely generated.

Suppose f and g are elements of F. The map f *g: x -> f(x) * g(x) is also a map from X to M. So, F is closed under multiplication.

For all f, g, h in F and all x in X we have

and hence the operation is associative.

Let 1 be the constant map 1(x) = e. For all f in F and all x in X we have

and so 1 * f = f = f * 1.

Thus we have shown that this is indeed a monoid structure.

First note that c² c (otherwise you easily derive that c² = c⁶). Since c² e, we also find c e.

• The first candidate then is that c³ = e. This is consistent with the conditions and we find the monoid (in fact group) C_0,3.
• Suppose c³ e. Then c³ c, since otherwise we square and find c⁶ = c², contradicting the assumption. If c³ = c², we find
  c⁶ = (c³)² = (c²)² = c⁴ = c³ · c = c² · c = c³ = c²,

  again a contradiction. So c, c², c³ are all distinct. In case c³ = c⁴, we find the monoid C_3,1.

• Still assume c³ e. In case c³ c⁴ and c⁴ = c⁵, we find C_4,1 (the case C_1,3 is excluded, since it would lead to c⁴ c⁸). In case c³ c⁴ and c⁴ c⁵, but c³ = c⁵, we find C_3,2 (again, the case C_1,4 cannot occur, because it would lead to c² = c⁶).
• Continuing in this way we find the remaining possibilities C_3,4, C_4,2, C_4,4.

Rewrite the indices using c^j = c^{k+((j - k) mod n)}, i.e., if j k + n and j - k = a mod n (with 0 < n), then replace j by k + a. Here, the idea is that upon writing j as k + (j - k) we replace the term in brackets by a, since a and j - k differ by a multiple of n.

• f(M) is a submonoid of M'.

  f(e) = e' since f is a morphism so e' is in f(M). Furthermore, for x and y in M we have

  f(x) * f(y) = f(x * y).

  This implies that f(M) is closed under multiplication and hence is a submonoid.

• The kernel is a submonoid of M.

  Assume that x and y are in the kernel of f. Then

  f(x *y) = f(x) * f(y)e' = f(x) = e' * e' = e'.

  This implies that x *y is also is the kernel. Since f(e) = e', the identity e is also in the kernel. Thus, we conclude that the kernel is a submonoid of M.

If M = {1,a,b} is not cyclic then we have the following possibilities:

1. aa = 1, bb = 1. Then a and b are invertible. If ab = a then (multiply by a) b = 1, which is impossible. Arguing in this way we find ab = ba = 1.
2. aa = a, bb = 1. Arguing as in the previous item we find ab = ba = a. Think of {1,0,-1} with multiplication.
3. aa = 1, bb = b. Look at the previous case.
4. aa = a, bb= b. Then up to symmetry ab = a. Bu then ba = a or b. In the first case think of the matrices {(1,0;0,1),(0,0;0,0), (1,0;0,0)} with matrix multiplication. In the second case {(1,0;0,1),(1,1;0,0), (1,0;0,0)}.

Let m + nZ have inverse element t + nZ. Then

and hence there is an s such that

The existence of such a relation implies that gcd(m,n) = 1.

Conversely, if gcd(m,n) = 1, then there exist integers s and t such that

and t is an inverse of m mod n.

An element (a, b) of the product monoid is invertible if and only if there is an element (x, y) with

Such an element (x, y) exists if and only if x is an inverse of a in M₁ and y is an inverse of b in M₂.

If k > 0 then the only invertible element in C_k,n is the identity element. If k = 0 then C_0,n is isomorphic to (Z/lZ,+). So all elements are invertible.

Consider the group Z/2Z×Z/2Z×Z/2Z. Its order is 8 but it contains no element of order 4. In S₄, a group of order 24, there is no element of order 12. Just look at the cycle decomposition of an arbitrary element to check this.

Let x be an element occurring in a column more than once. So, there exist distinct g and g' and an element h in G such that x = gh = g'h.

Multiplying on the right by the inverse of h we see g = g', a contradiction.

The proof that x appears only once in each row is similar.

Since A^T is the inverse of A for every A in O_n(R), we find that O_n(R) is a subset of GL_n(R). The identity matrix is clearly an element of O_n(R). Furthermore, O_n(R) is closed under multiplication since for all A and B in O_n(R)

If A is in O_n(R), then A^-1 O_n(R). To see this, we observe that A^-1 = A^T and that

This shows that inverses of elements in O_n(R) are also in O_n(R).

Hence O_n(R) is a subgroup of GL_n(R).

The left cosets of S₃ in S₄ are:

• S₃,
• (1,4)S₃ = {(1,4),(1,2,4),(1,3,4),(1,4)(2,3),(1,2,3,4),(1,3,2,4)},
• (2,4)S₃ = {(2,4),(2,1,4),(2,3,4),(2,4)(1,3),(2,1,3,4),(2,3,1,4)},
• (3,4)S₃ = {(3,4),(3,2,4),(3,1,4),(3,4)(1,2),(3,2,1,4),(3,1,3,4)}.

• S₃,
• S₃(1,4) = {(1,4),(1,4,2),(1,4,3),(1,4)(2,3),(1,4,2,3),(1,4,3,2)},
• S₃(2,4) = {(2,4),(2,4,1),(2,4,3),(2,4)(1,3),(2,4,1,3),(2,4,3,1)},
• S₃(3,4) = {(3,4),(3,4,2),(3,4,1),(3,4)(1,2),(3,4,2,1),(3,4,1,2)}.

Let x = (1,2)(3,4,5). Then x² = (3,5,4), x³ = (1,2), x⁴ = (3,4,5), x⁵ = (1,2)(3,5,4), x⁶ = e.

Let c = c₁c₂ ··· c_m where the c_i are disjoint cylces of length l_i. Hence, if c^d = c₁^dc₂^d ··· c_m^d = e, then l_i divides d. SO, if d is the order of c, then d is a common multiple of the cycle structure of a disjoint cycle decomposition of c.

On the other hand, if for any common multiple k of the cycle structure we have c^k = c₁^kc₂^k ··· c_m^k = e. Hence, the order d of c equals the lowest common multiple of the cycle structure.

1. Let h be an element in H. The set {hⁱ | i = 1, ...} is a subset of the finite subset H, and therefore also finite.
2. From the previous item we infer that there exist r and s with r > s such that h^r = h^s. This implies that h^{(r - s)} is the identity. So the identity is also in H.
3. h^{(r - s -1)} is the inverse of h. Thus, the inverse of h is contained in H.
4. H is a subgroup of G: It contains the identity, is closed under multiplication and taking inverses.

Consider the cyclic subgroup of G generated by g. Let r be the order of this subgroup. By Lagrange's Theorem we know that r divides m. If r < m, then there is a prime divisor p of m such that r divides m/p. But that implies that g^(m/p) is a power of g^r = e, which contradicts the hypothesis. Hence r = m and G is cyclic.

The map f : Z/6Z -> Z/2Z × Z/3Z defined by f(a) = (a, a) is a morphism. Moreover, one easily checks that it is surjective and injective. Hence, it is an isomorphism. The groups Z/6Z and Z/2Z × Z/3Z are commutative, the group S₃ not. So Z/6Z and Z/2Z × Z/3Z are not isomorphic to the group S₃.

f(0) = g⁰ = e and f(a + b) = g^{(a + b)} = g^a * g^b = f(a) * f(b) and hence f is a morphism.

If G has order n, then gⁿ = e. Hence f(a + tn) = g^{(a + nt)} = g^a. Hence the map f: Z/nZ -> G defined by f(a + nZ) = g^a is well defined. Moreover, by the same reasoning as above, this map is a morphism. It is easily seen to be surjective and, since Z/nZ and G have the same order, it is even an isomorphism.

For all x, y and z in R we have y * x = y + x - yx = x + y - xy = x * y. Furthermore (x * y) * z = (x + y - xy) * z = x + y - xy + z - (x + y - xy)z = x + y + z - xy - xz - yz - xyz and x * (y) * z) = x + (y + z - yz) - x(y + z - yz) = x + y + z - xy - xz - yz - xyz. So * is also associative.

The element e = 0 is the unit element for this multiplication.

1. If Z × Z is cyclic then there exists an element (a,b) such that every (c,d) is an integral multiple of (a,b). It is clear that a and b must be nonzero. If we choose (c,d) = (a,b + 1) we get a contradiction.

   3 · (3,4) - 4 · (2,3) = (1,0) and 3 · (2,3) - 2 · (3,4) = (0,1) can be obtained from (2,3) and (3,4). Then every element can be obtained from (2,3) and (3,4).

2. (a,b) and (c,d) generate Z × Z if for all (u,v) there exist integers s and t) such that (u,v) = s(a,b) + t(c,d). So u = as + ct and v = bs + dt. This implies (ad - bc)s = du - cv (multiply the first relation by d and subtract the second equality multiplied by c) and (ad - bc)t = av - bu. Hence, we can find the desired s and t for all u and v if and only if ad - bc = 1 or -1.

1. The set consisting of only the zero matrix does not contain the identity matrix of M_n(R). Hence this is not a submonoid.
2. The set consisting of only the identity matrix is a submonoid.
3. The set of all matrices of determinant 1 is a submonoid. Indeed, the identity matrix has determinant 1 and the product of two matrices with determinant 1 has also determinant 1.
4. The set of matrices with trace 0 does not contain the identity matrix. So it is not a submonoid.
5. The set of upper triangular matrices is a submonoid. Indeed, the identity matrix is upper triangular, just like the product of two upper triangular matrices.

Notice that 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32.

• Mod 3 we find: 2⁰ = 1, 2¹ = 2, 2² = 1. Thus the submonoid generated by 2 is isomorphic to C_0,2
• Mod 4: 2⁰ = 1, 2¹ = 2, 2² = 0, 2³ = 0, 2⁴ = 0, 2⁵ = 0, etc. Thus the monoid generated by 2 is C_2,1.
• Mod 5: 2⁰=1, 2¹=2, 2²=4, 2³=3, 2⁴=1. The monoid generated by 2 is isomorphic with C_0,4.

The multiplication monoid Z/8Z contains the following submonoids: {0,1,2,3,4,6}, {0,1,3,5,7}, {0,1,2,4,5,6} and {0,1,2,4,6,7}. Any pair of elements is in one of these submonoids. Hence Z/8Z cannot be generated by 2 elements. The submonoid generated by 2, 3 and 5 contains 0 = 2³, 1 (as unit), 2, 3, 4 = 2², 5, 6 = 2·3 and 7= 3·5, i.e., this submonoid is the whole monoid.

1. In a product M × M' the element (a,b) is invertible if and only if a and b are invertible. So the invertible elements (with respect to multiplication) of Z/2Z × Z/3Z are (1,1) and (1,2).
2. In Q[X]/(X²) an element f + (X²) is invertible if and only if gcd(f,X²) = 1. So, for f we can take the elements of the form aX + b, with nonzero b.
3. In Z/16Z an element k+16Z is invertible if and only k is relatively prime with 16. So, precisely all the elements of the form k+16Z where k is odd are invertible in the multiplication monoid of Z/16Z.

The operation *' is associative since

(M,*',e) has identity e (this is trivial).

1. U₂₆ consists of those elements that are relatively prime to 26. So it has 12 elements, namely 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23 and 25.
2. It has the following generators: 7 + 26Z, 11 + 26Z, 15 + 26Z, 19 + 26Z.

1. f(0,0) = 0 and f(x₁ + x₂, y₁ + y₂) = x₁ + x₂ - 2(y₁ + y₂) = x₁ - 2 y₁ + x₂ - 2y₂ = f(x₁, y₁) + f(x₂, y₂). Hence f is a morphism. The image is Z, as every z in Z equals f(z,0).
2. The map f(k) = g^2k is a morphism since f(a + b) = g^{2(a + b)} = g^2ag^2b = f(a)f(b) and f(0) = g⁰ = e.

   If the order of g is 6 the image is {e, g², g⁴}, a subgroup of G isomorphic to Z/3Z.

   If the order of g is 7 then the image is <g>, which is isomorphic to Z/7Z.

3. The group Z/4Z contains one element, namely 2, of order 2 and two elements, 1 and 3, of order 4. So f is completely determined by the image of 1. Hence we have the following possibilities for a morphism f:
   1. f(1) = 0, then f maps all elements to 0,
   2. f(1) = 1, then f is the identity,
   3. f(1) = 2, then f(2) = 2 + 2 = 0 and f(3) = 2 + 2 + 2 = 2,
   4. f(1) = 3, then f(2) = 3 + 3 = 2 and f(3) = 3 + 3 + 3 = 1.
   In cases 2 and 4 f is an isomorphism.
4. Suppose x and y are elements if G. Then f(g(x * y)) = f(g(x) * g(y)) = f(g(x) * f(g(y)). Furthermore f(g(e)) = f(e) = e. Hence f _° g is a morphism. If both f and g are bijective, then so is f _° g. Thus if both f and g are isomorphisms, then so is f _° g.

1. If G contains an element g of order 4 then f(gⁿ) = n + 4Z is an isomorphism from G to Z/4Z.
2. Suppose that G does not contain elements of order 4. Then, by Lagrange's Theorem, all non-identity elements have order 2. In particular, G is abelian, see Exercise 2. Let g and h be two distinct elements of order 2 in G. Then G = {e, g, h, gh}. Define f : G -> Z/2Z × Z/2Z by f(e) = (0,0), f(g) = (1,0), f(h) = (0,1) and f(gh) = (1,1). Then f is an isomorphism (check this by comparing the two multiplication tables).

For m = 2 to p - 1 compute the powers of m mod p. If the powers m^k are not equal to 1 mod p for all k less than p - 1, then m is a generator. If there is a k smaller than p - 1 with m^k = 1 mod p, then increase m.

Can you think of (easy) ways to improve this? Can you bring the divisors of p - 1 into play more prominently?