Introduction to abstract algebra: definitions, examples, and exercises

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1 Introduction to abstract algebra: definitions, examples, and exercises Travis Schedler January 21, Definitions and some exercises Definition 1. A binary operation on a set X is a map X X X, (x, y) x y, also denoted xy. The operation is called the multiplication or composition. Definition 2. A magma (X, ) is a set together with a binary operation (a b is the operation for a, b X). Definition 3. A semigroup is a magma whose operation is associative. In other words, we could say that a semigroup is an associative magma. Definition 4. An identity, or unit, element of a magma X is an element e such that ea = a = ae for all a X. Note that, if an identity e exists, it is unique, since if e and e are both identity elements, then e = ee = e. Definition 5. A monoid is a semigroup for which there exists an identity. Definition 6. A group G is a monoid for which, for every element a G, there exists b G such that ab = e = ba. Such an element b is unique, since if ab = e = b a as well, then b = be = bab = eb = b. Definition 7. An abelian (or commutative) group is a group for which the operation is commutative. Definition 8. A ring (R, +, ), is a set R equipped with two operations, called addition + and multiplication, such that (R, +) is an abelian group and (R, ) is a monoid. The unit for addition is denoted 0 and the unit for multiplication is denoted 1. The additive inverse of an element a is denoted a, so ( a) + a = 0 = a + ( a). 1

2 Definition 9. A commutative ring R is a ring for which the multiplication is commutative: ab = ba for all a, b R. Note that, since addition is always commutative, when we say the ring is commutative, it is clear commutativity is applying to the multiplication rather than the addition. Definition 10. A division ring (R, +, ) is a ring such that every nonzero element a has a multiplicative inverse. In other words, the monoid R, defined as the set of nonzero elements of R under multiplication, is a group. Definition 11. A field is a commutative division ring. Definition 12. A homomorphism (or morphism) of magmas f : (X, cdot) (Y, ) is a map f : X Y such that f(ab) = f(a)f(b) for all a, b X. The set of homomorphisms from X to Y is denoted Hom(X, Y ) (the same holds when magmas are replaced by any other of the structures below for which we define homomorphisms). Definition 13. A homomorphism (or morphism) of semigroups is a map f : X Y which is a homomorphism of magmas. In other words, the condition to be a homomorphism is the same as for magmas; we only require that the source and target are associative. Definition 14. A homomorphism (or morphism) of monoids is a map f : X Y which is a homomorphism of magmas (equivalently, of semigroups) such that f(e X ) = e Y, where e X X is the unit in X and e Y Y is the unit in Y. Definition 15. A homomorphism (or morphism) of groups is a map f : G H which is a homomorphism of monoids. Exercise 16. (i) Show that a homomorphism of groups also has the property that f(a 1 ) = f(a) 1 for all a G. (ii) Show that the condition that f(e G ) = e H is redundant for a group. More precisely, show that, if G and H are groups and f : G H is a morphism of magmas, then actually f(e G ) = e H and hence f is a morphism of groups (here e G G and e H H are the identity elements). Thus, a homomorphism of groups is the same as a map f such that f(ab) = f(a)f(b) for all a, b G. Exercise 17. Show that this is not true for a monoid: give an example of monoids (M, ) and (N, ) with units e M and e N, and a morphism of semigroups M N which does not satisfy f(e M ) = e N, i.e., is not a morphism of monoids. Hint: You can try N=a monoid with two elements e, x such that x 2 = x and e is the identity.

3 Definition 18. An endomorphism of X is a homomorphism from X to itself. They are denoted End(X). These are always equipped with composition: f g End(X) is given by (f g)(x) = f(g(x)). Exercise 19. If A is an abelian group, show that End(A) forms a ring where, for ϕ, ψ End(A), the sum is given by (ϕ + ψ)(a) = ϕ(a) + ψ(a), and the product is as usual given by composition. Why does the formula for sum not work when A is not abelian? 2 Some examples Example 20. Some examples of fields: Q, R, C. Example 21. Z is a commutative ring but not a division ring (hence not a field.) Z 0 is a commutative monoid under addition (with identity 0) and a commutative monoid under multiplication (with identity 1). Z >0 is a commutative semigroup under addition (but not a monoid) and a commutative monoid under multiplication (with identity 1). Example 22. The nonzero real numbers, R, form an abelian group under multiplication. The same is true for Q and C, or more generally F for any field F (again denoting the nonzero elements of F under multiplication). Example 23. The n n matrices over R, Mat n (R) (or, Mat n (R) where R is any ring) form a ring under matrix addition and multiplication. The additive identity is the zero matrix, and the multiplicative identity is the identity matrix. Example 24. The invertible n n matrices over R, GL n (R) (or GL n (R) for any ring R) form a group which is not abelian. Example 25. The m n matrices over any ring form an abelian group under matrix sum, but the matrix multiplication does not make sense when m n. Example 26. A student asked for an example of a division ring which is not a field (i.e., not commutative). Here is an example: the quaternions H := {a + bi + cj + dk a, b, c, d R}, with the multiplication which is R-linear satisfying I 2 = J 2 = K 2 = 1 and IJ = K, JK = I, KI = J. 3 Isomorphisms and permutations Let X and Y have any type of structure. Definition 27. An isomorphism is an invertible morphism, i.e., a morphism f : X Y such that there is an inverse morphism g : Y X, with f g the identity on Y and g f the identity on X.

4 In the case of sets, an invertible map of sets f : X Y is the same as a bijective map, i.e., one which is one-to-one and onto. Definition 28. A permutation of a set X is an invertible (equivalently bijective) map p : X X. Let Perm(X) denote the set of permutations of X. Exercise 29. Show that Perm(X) is a group under composition. One of the main constructions of the course is that of a group action: this is an action G X X, (g, x) g x of a group G on a set X satisfying (gh) x = g (h x) and e x = x for all x X. Exercise 30. Show that a group action is the same as a homomorphism a : G Perm(X), with a(g) the map (a(g))(x) = g x. Hint: The main step is to see that the condition e x = x for all x X is equivalent to the condition that the map a(g) is bijective (hence a permutation) for all g G. To show one implication, if e x = x for all x X, then a(g) a(g 1 ) = a(g g 1 ) = a(e) is the identity. To show the other implication, if a(e) is bijective, then since also a(e) a(e) = a(e 2 ) = a(e), we conclude that a(e)(a(e)(x)) = a(e)(x) for all x X, but a(e) : X X is surjective, so that a(e)(x) = X, and hence the identity says that a(e) is the identity. Exercise 31. Show that, similarly to maps of sets, a homomorphism f : X Y of any of the types of structures above (magmas, semigroups, monoids, groups, abelian groups, rings, division rings, commutative rings, fields) is invertible if and only if it is bijective. Hint: The main step is to show that, if f is bijective, and g : Y X is the inverse map of sets, then g is also a homomorphism of the structure in question. Example 32. Here is an example of a type of morphism for which being bijective is weaker than being invertible: differentiable functions from R to R. For example, if f : R R is the function f(x) = x 3, this is bijective and differentiable. But the inverse function f 1 is f 1 (x) = 3 x, which is not differentiable at zero. So, if we restrict our attention to differentiable functions, then not all bijective differentiable functions are invertible (by differentiable functions). 4 Modules and vector spaces The analogue of a group action on a set for rings is an action of a ring on an abelian group. This is called a module: Definition 33. A module over a ring R is an abelian group M equipped with an action R M M, (r, m) r m, satisfying the axioms (rs)(m) = r(sm), 1(m) = m, (r+s)(m) = rm+sm, r(m+n) = rm+rn, r, s R, m, n M.

5 Exercise 34. Analogously to Exercise 30, show that a module is the same as a ring homomorphism R End(M), where End(M) is the ring of abelian group endomorphisms of M. See Exercise 19 for why End(M) is a ring. Definition 35. A vector space V over a field F is the same thing as a module over F (it is just called a vector space in the case the ring is actually a field).