GROUP THEORY BY THE SPMPS 2013 GROUP THEORY CLASS Abstract. This paper presents theorems proven by the Group Theory class of the 2013 Summer Program in Mathematical Problem Solving. It also includes the course s central definitions. The ideas in the proofs were created by Samuel Acosta, Joel Infante, Tia Jones, Mona Naser, Peter Ngaba, Alan Ramos, Jayden Reaves, and Crisleidy Tejada. The definitions are standard. The paper was edited by Ben Blum-Smith. 1. Definition of a group Definition 1 (Associativity). An operation on a set S is associative if (and only if) for any x, y, z in S, (x y) z is always the same as x (y z) For example, addition of numbers is associative, because for any numbers x, y, z, it is true that (x + y) + z = x + (y + z) Also, multiplication of numbers is associative, because for any x, y, z, it is true that (xy)z = x(yz) Definition 2 (Group). A set G with an operation is called a group if (and only if): (1) The operation is well-defined on the set, i.e. the set is closed under the operation. (2) The operation is associative: for any x, y, z in S, (x y) z = x (y z) (3) There is an element of G, call it e, that is an identity for. In other words, for any x in G, x e = x and e x = x 1
2 BY THE SPMPS 2013 GROUP THEORY CLASS (4) For each element x in G, there is another (or possibly the same) element y in G such that and x y = e y x = e For each x, the y that satisfies this condition is called the inverse of x. 2. Key examples Theorem 1. The operation of composition is associative on the LAX group. Proof. Given three actions A, B and C from the LAX group, both (AB)C and A(BC) come out to the net result of doing A then B then C. The only difference is the order in which you simplify. Theorem 2. The set of integers Z forms a group with respect to addition. Proof. Z is a set. The operation + is well-defined on the integers because an integer plus an integer is an integer. Addition is always associative. The identity for + is 0 because for any integer x, x + 0 = 0 and 0 + x = 0. Every integer has an inverse, namely its negation: for any integer x, Therefore (Z, +) is a group. x + ( x) = 0 Theorem 3. The set of nonzero real numbers R \ {0} forms a group with respect to multiplication. Proof. R \ {0} is a set. The operation of is well-defined on R \ {0} because if a, b are real numbers and ab = 0, then either a or b has to be zero. It follows that if a, b are real numbers that are not zero, then ab 0 as well. Thus any two nonzero real numbers have a product that is also nonzero. Multiplication is always associative. 1 is an identity for multiplication because for any real number x, 1x = x1 = x. Finally, every nonzero real number x has a reciprocal, 1/x, which is its inverse because x 1 x = 1 x x = 1 Therefore (R \ {0}, ) is a group.
GROUP THEORY 3 3. Fundamental properties of a group Theorem 4. In any group, for any elements a, b in the group, there is always an element x in the group such that ax = b Proof. First, if there is such an x, it must be equal to a 1 b. This is because if ax = b, then by composing on the left of each side with a s inverse, we get a 1 (ax) = a 1 b Because the group operation is associative, the left side of the equation is equal to (a 1 a)x. But a 1 a is the identity, because a 1 is a s inverse. So the equation becomes 1x = a 1 b where 1 stands for the identity in the group, whatever it is. But then 1x = x since 1 is the identity, so the equation becomes x = a 1 b This proves that if there is an x in the group such that ax = b, it must be that x = a 1 b. Second, the element a 1 b does exist in the group. We know this because a 1 exists in the group, since every element of a group has an inverse, and then a 1 b exists in the group, because the group operation is well-defined, so you will definitely get some group element by combining a 1 with b. Thirdly, the element a 1 b is in fact the element that is needed to work for x in the equation ax = b, because a(a 1 b) = (aa 1 )b = 1b = b where the first equality is from the associative property, the second is from the definition of inverses, and the third is from the definition of the identity. Lemma 1. If ab = ad then b = d. Proof. If ab = ad, we can compose on the left with a s inverse to find that By the associative property, this means that a 1 (ab) = a 1 (ad) (a 1 a)b = (a 1 a)d By the definition of inverses, a 1 a = I where I is the identity, so this becomes Ib = Id But Ib = b and Id = d by the definition of the identity, so this becomes b = d
4 BY THE SPMPS 2013 GROUP THEORY CLASS Theorem 5. No row of a Cayley table contains any element more than once. Proof. Lemma 1 says that ad cannot equal ab unless d = b. Therefore, b is the only element that will give you ab when composed with a. Therefore, a s row in the Cayley table does not contain ab more than once. Since a can be any element, and ab can be any entry in its row, this menas no row contains any element more than once. We can prove the same result for columns of a Cayley table using the lemma that if ba = da then b = d. The proof is the same except with composition by a 1 on the right instead of the left, and columns taking the place of rows. 4. Some theorems about groups An object with three-fold rotational symmetry has a symmetry group with Cayley table that looks like this: e r r 2 e e r r 2 r r r 2 e r 2 r 2 e r Theorem 6. All groups of order 3 are isomorphic to the above group. Proof. Suppose we have any group with three elements. Call the elements a, b and c and suppose a is the identity. Then we can use this to fill in 5 of the entries in the Cayley table: b b c c Now from theorem 5 and the comments following it, we know that each element occurs only once in each row and column. bc cannot equal c because there is already a c in its column and it cannot equal b because there is already a b in its row. It must be a. Similar comments apply to cb. Thus we can fill in the Cayley table further: b b a c c a Now bb = b 2 must be c because there is already a b and an a in its row, so we can fill this in: b b c a c c a
GROUP THEORY 5 Similarly, c 2 must be b because there is already an a and a c in its row: b b c a c c a b Thus we have a complete Cayley table for the group. This applies to any group of order 3. Now substitute a = e, b = r, c = r 2 to see that this group is isomorphic to the 3-fold rotation group. Theorem 7. It is possible to get an n-element set into any order using only transpositions. Proof. You transpose the elements of the set based on the order that you want your set to be in. You have to find the element that s first in the order you want, and transpose it with the element that was in its spot. Then you ignore that element, and move the element that s second to its spot by transposing it with the element that is in its place. Then you ignore the first two and you keep going. Here is an example to illustrate the method. If you have elements A, B, C, D in the order DABC and you want them in alphabetical order, use the transposition (1, 2), transposing the first and second, to put A in the right spot: ADBC Then use (2, 3), transposing the second and third, to put B in the right spot: ABDC Then use (3, 4), transposing the third and fourth, to put C in the right spot: ABCD Since D is the only element left, it is automatically in the right spot. Another way to state this result is that S n, the group of all possible permutations of an n-element set, is generated by transpositions. The example given in the proof shows that (1, 4, 3, 2), the transposition needed to restore DABC to alphabetical order, can be composed from (1, 2), (2, 3), and (3, 4): (1, 4, 3, 2) = (1, 2)(2, 3)(3, 4)