FINAL REPORT: INTRODUCTION TO GROUPS

ERNESTO D. CALLEROS PROFESSOR DAVID LARSON MATH 482 25 APRIL 2013 TEXAS A&M UNIVERSITY

A group in an algebra setting is a type of set that, together with some operation, obeys a predefined set of rules called axioms. In order to understand these rules and to be able to define them precisely, it helps to understand binary operations first.

Definition 1. A binary operation ∗ on a set S is a function mapping S × S into S. For each (a, b) ∈ S × S, we will denote the element ∗(a, b) of S by a ∗ b.

To emphasize certain conditions for a binary operation on a set, we give the following remarks: Remark 2. By definition, a binary operation ∗ on a set S is well-defined; that is, ∗ assigns exactly one element to each possible ordered pair of elements in S.

Remark 3. For a binary operation ∗ on a set S, S is closed under ∗; that is, for all a, b ∈ S, we also have a ∗ b ∈ S.

We proceed with some examples.

Example 4. Determine if usual addition is a binary operation on the sets Z, Q, {1,2,3}.

Solution. Given a, b ∈ Z, a + b is a well-defined element of Z. Also, Z is closed under addition, so addition is a binary operation on Z. Similarly, addition is a binary operation on Q. However, 2 + 2 = 4 ∈/ {1, 2, 3}, so addition is not closed on {1, 2, 3}, and thus is not a binary operation on this set.

Example 5. For a finite set, such as S={a, b, c}, a binary operation can be defined by using a table in which the elements of the set are listed across the top as heads of columns and at the left as heads of rows. For example, TABLE 1 below defines the binary operation * on the set S by the following rule: (ith entry on the left)*(jth entry on the top) = (entry in the ith row and jth column)

∗ a b c a b c b b a c b c c b a

Table 1.

1 2 ERNESTO D. CALLEROS PROFESSOR DAVID LARSON MATH 482 25 APRIL 2013 TEXAS A&M UNIVERSITY

We can easily see that, since exactly one element is inside each inner box of the table above, Remark 2 is fulfilled. Also, since all the elements inside the inner boxes are elements of S, Remark 3 is fulfilled, as well. Thus, ∗ is a binary operation on S.

Some operations have certain properties with respect to certain sets. One of these properties is associativity.

Definition 6. A binary operation ∗ on a set S is associative if, for all a, b, c ∈ S, (a ∗ b) ∗ c = a ∗ (b ∗ c).

In the next example we will see that the composition of functions is associative on the set of func- tions that take the same codomain as the domain. To be precise, we first define what it means to compose two functions.

Definition 7. Let R,S, and T be sets and g : R → S and g : S → T be functions. Their composi- tion is the function f ◦ g : R → T defined by (f ◦ g)(a) = f(g(a)) for all a ∈ R.

Example 8. Verify that composition of functions is a binary operation on the set F (S)= {f : S → S | f is a function}. Then check that this operation is associative.

Solution. If f : S → S and g : S → S, then by definition their composition is a function f ◦ g : S → S, so (f ◦ g) ∈ F (S). Thus, it is clear that composition is a binary operation on F (S). Now let’s show that composition is associative. We need to show that (f ◦ g) ◦ h = f ◦ (g ◦ h). Let a ∈ S. By definition of composition, we obtain: ((f ◦ g) ◦ h)(a) = (f ◦ g)(h(a)) = f(g(h(a))). On the other hand, (f ◦ (g ◦ h))(a) = f((g ◦ h)(a)) = f(g(h(a))).

Therefore, ((f ◦ g) ◦ h)(a)= (f ◦ (g ◦ h))(a), and since a was an arbitrary element of S, we conclude that (f ◦ g) ◦ h = f ◦ (g ◦ h) and composition is associative.

Another property that some binary operations have with respect to certain sets is the existence of an identity element.

Definition 9. Let ∗ be an operation on a set S. We say that e ∈ S is an identity element for ∗ if, for all a ∈ S, e ∗ a = a ∗ e = a.

The following example illustrates this definition.

Example 10. For + on Z, 0 is an identity element because for any n ∈ Z, 0 + n = n + 0 = n. For · on Z, 1 is an identity element because for any n ∈ Z, 1 · n = n · 1 = n. For ◦ on F (S), the function id : S → S defined by id(x) = x for all x ∈ S is an identity element FINAL REPORT: INTRODUCTION TO GROUPS 3

because f ◦ id = id ◦ f = f for all f ∈ F (S).

It turns out that if an identity element for a binary operation exists, then it must be unique. We write this fact in the following proposition.

Proposition 11. Suppose e1 and e2 are identity elements for a binary operation ∗ on a set S. Then e1 = e2.

Proof. Regarding e1 as an identity element, we must have e1 ∗ e2 = e2. However, regarding e2 as an identity element, we must have e1 ∗ e2 = e1. We thus obtain e1 = e2, showing that an identity element must be unique.

Now that we have a better understanding of binary operations and their properties, we will take a look at groups. What are they, and why are we interested in studying them? Let us answer the second question first. Problems in real life sometimes lead to linear equations of the form a ∗ x =b involving some binary operation ∗ with an unknown element x in some set S. For instance, in the case of addition on the set of integers, we may have an equation of form a + x = b. This additive equation always has an integer solution, and the solution is unique. However, finding a unique solution - or any solution at all - is not always possible for every binary operation on a given set. For instance, consider the binary operation ∗ defined as in Example 5 on the set S={a, b, c}. As we can see, the equation a ∗ x = a has no solution in S. This leads us to try to figure out what kinds of sets, and under what binary operations, will guar- antee us a unique solution to such linear equations. It turns out that a group, a certain kind of set together with a binary operation, guarantees the desired existence and uniqueness of solutions.

Just before we give a precise definition of a group, let us introduce the following definition.

Definition 12. Let G be a set, closed under a binary operation ∗, with identity element e. We say that a ∈ G has an inverse if there exists an a0 ∈ G such that a0 ∗ a = a ∗ a0 = e.

Definition 13. A group (G,∗) is a set G, closed under a binary operation ∗, such that the following axioms are satisfied: G1) ∗ is associative in G. G2) G has an identity element for ∗. G3) Every element in G has an inverse in G. We proceed with some examples.

Example 14. Verify that (Z,+) is a group.

Solution. + is associative in Z, since for any a, b, c ∈ Z, (a + b) + c = a + (b + c). We see that 0 is an identity element. Lastly, given a ∈ Z, −a is its inverse, since (−a) + a = a + (−a) = 0. 4 ERNESTO D. CALLEROS PROFESSOR DAVID LARSON MATH 482 25 APRIL 2013 TEXAS A&M UNIVERSITY

It is now time to look at an example in which checking the group axioms might not be readily obvious.

Example 15. We start this example by defining an operation called multiplication modulo n. + Let n ∈ Z and Un = {x ∈ Zn, | gcd(x, n) = 1}. Define ∗ : Un × Un → Un by

a ∗ b = ab (mod n).

With the following set of Lemmas, we will verify that (Un, ∗) is a group.

We start with a useful fact.

Fact: “If a, b ∈ Z, then there exist s, t ∈ Z such that gcd(a, b) = as + bt. Furthermore, gcd(a, b) is the smallest positive integer of the form as0 + bt0 for some s0, t0 ∈ Z.”

+ Lemma 16. Let x, n ∈ Z . Then gcd(x (mod n),n) = gcd(x,n).

Proof. To prove this lemma, it suffices to show gcd(x, n)|gcd(x (mod n), n) and gcd(x (mod n), n)|gcd(x, n). Using the first part of this fact, we obtain: gcd(x (mod n), n) = gcd(x + kn, n) for some k ∈ Z = (x + kn)s + nt for some s, t ∈ Z = xs + (ks + t)n. (1). Recall that given w, z ∈ Z, then gcd(w, z)|(wa + bz) for all a, b ∈ Z. Thus, gcd(x, n)|(xs + (ks + t)n), so it follows by (1 ) that gcd(x, n)|gcd(x (mod n), n), as desired. Similarly, it can be shown that gcd(x (mod n), n)|gcd(x, n). Therefore, we can conclude that gcd(x (mod n), n) = gcd(x, n).

Lemma 17. Un is closed under ∗.

Proof. Let x1, x2 ∈ Un. Then gcd(x1, n) = 1 = gcd(x2, n). We need to show: x1 ∗ x2 ∈ Un, or equivalently, gcd(x1 ∗ x2, n) = 1. We would have to prove gcd(x1x2 (mod n), n) = 1, but, by Lemma 16, it suffices to show gcd(x1x2, n) = 1. Since gcd(x1, n) = 1 = gcd(x2, n), then using the first part of the above fact, we take s1, s1, s2, s2 ∈ Z with x1s1 + nt1 = 1 and x2s2 + nt2 = 1. We have:

1 = 1 · 1

= (x1s1 + nt1)(x2s2 + nt2)

= x1s1x2s2 + x1s1nt2 + nt1x2s2 + nt1nt2

= (x1x2)(s1s2) + n(x1s1t2 + t1x2s2 + t1nt2)

Thus, we have found s0 = s1s2 ∈ Z and t0 = (x1s1t2 + t1x2s2 + t1nt2) ∈ Z such that (x1x2)s0 + nt0 = 1. It follows, by the second part of the above fact, that gcd(x1x2, n) = 1, and Un is closed under ∗.

+ Lemma 18. ∗ is well-defined. That is, for n ∈ Z and a, b, c, d ∈ Un, if a ≡ c (mod n) and b ≡ d (mod n), then ab (mod n) = cd (mod n). FINAL REPORT: INTRODUCTION TO GROUPS 5

Proof. We are given that a ≡ c (mod n). This means that a − c = kn for some k ∈ Z, so a = c + kn. Similarly, since b ≡ d (mod n), there exists h ∈ Z such that b − d = hn, so d = b − hn. We obtain: ab − cd = (c + kn)b − c(b − hn) = cb + knb − cb + chn = knb − chn. This implies that ab − kbn = cd + chn. Taking modulo n of both sides of this equation, we obtain: ab (mod n) = cd (mod n), as desired. This shows that ∗ is well-defined.

Lemma 19. ∗ is associative.

Proof. Let a, b, c ∈ Un. We will show that (a ∗ b) ∗ c = a ∗ (b ∗ c). We have: (a ∗ b) ∗ c = (ab (mod n)) ∗ c = (ab + kn) ∗ c for some k ∈ Z = (ab + kn)c (mod n) = (abc + knc)(mod n) = abc (mod n). On the other hand, we also have: a ∗ (b ∗ c) = a ∗ (bc (mod n)) = a ∗ (bc + qn) for some q ∈ Z = a(bc + qn)(mod n) = (abc + knc)(mod n) = abc (mod n). Therefore, (a ∗ b) ∗ c = abc (mod n)= a ∗ (b ∗ c), and it follows that ∗ is associative.

Lemma 20. Un has an identity element for ∗.

Proof. Since 1 ∗ a = 1 · a (mod n) = a (mod n) = a · 1 (mod n) = a ∗ 1 and 1 ∈ Un, then 1 is an identity element of (Un, ∗).

Lemma 21. Inverses exist in Un.

0 0 0 Proof. Let x ∈ Un. We want to show there exists x ∈ Un such that x ∗ x = x ∗ x = 1. Since gcd(x, n) = 1, then, by the first part of the fact mentioned earlier, choose s, t ∈ Z such that

xs + nt = 1 (2 ). Taking modulo n of both sides of (2 ), we obtain: 1 = xs (mod n) = (x (mod n)) ∗ (s (mod n)) = x ∗ (s (mod n)). Since xs = sx, then we also have (s (mod n)) ∗ x = 1. Thus, letting x0 =(s (mod n)), we have: 0 0 0 x ∗ x = x ∗ x = 1. To complete the proof, all that remains to be shown is that x is in Un. Since 6 ERNESTO D. CALLEROS PROFESSOR DAVID LARSON MATH 482 25 APRIL 2013 TEXAS A&M UNIVERSITY

0 s = x + wn for some w ∈ Z, then using (2 ), we have: 1 = xs + nt = x(x0 + wn) + tn = xx0 + xwn + tn = xx0 + (xw + t)n.

0 0 Therefore, gcd(x , n) = 1, since x ∈ Z and (wx+t) ∈ Z. Thus, x ∈ Un and group axiom G3 is satisfied.

Overall, we have shown with the previous lemmas that (Un, ∗) is a group. It is called the group of units in the integers modulo n, and it is a very important group in mathematics. FINAL REPORT: INTRODUCTION TO GROUPS 7

References

Fraleigh, John B. A First Course In Abstract Algebra. Reading, MA: Addison-Wesley Pub., 1967. 20-38. Print.