Lecture 3: Homomorphisms and Isomorphisms

Lecture 3: Homomorphisms and Isomorphisms 3 Homomorphisms and Isomorphisms 3.1 Review Last time, we discussed subgroups and cyclic groups. A subgroup of a group is essentially a subset of that group that is compatible with the group or multiplicative structure on it. A cyclic subgroup of an element g in a group is essentially the subgroup consisting of all the powers of g. 3.2 Homomorphisms Now that we understand a little bit more about groups and their structures, the natural next step is to look at maps between groups. Guiding Question How can we understand groups by considering maps between diferent groups? What kinds of maps can provide useful insight into various groups? First, we defne a type of map that is compatible with the group structure on both groups. Defnition 3.1 Given groups G and G′ , a homomorphism between them is a map f : G −→ G′ such that: • For all a, b ∈ G, f (ab) = f (a)f (b). • The identity element is mapped to the identity: f (eG ) = eG′ . • Inverses are preserved under the mapping: f (a)−1 = f (a−1 ) for all a ∈ G. Essentially, each of these conditions requires that the map preserve the group structure (multiplication, identity, inverse) from the domain G to the codomain G′ . Either f can be applied to a product, or the product can be taken after f is applied, and it should yield the same element f (ab) = f (a)f (b).11 In fact, only the frst condition is really necessary, and implies the second two.12 Proposition 3.2 If f (ab) = f (a)f (b), then f (eG ) = eG′ and f (a)−1 = f (a−1 ). Proof. For the frst part, take f (eG ·eG ) = f (eG ) = f (eG )·eG′ by the defnition of eG′ . Since f is a homomorphism, this will also be equal to f (eG )f (eG ). Multiplying on both sides by f (eG )−113 gives f (eG ) = e′G . The second part is similar. Take a ∈ G. Then f (a) · f (a−1 ) = f (a · a−1 ) = f (eG ) = e′G , and multiplying on the left by f (a)−1 gives f (a−1 ) = f (a)−1 . 3.3 Examples Let’s see some examples. Example 3.3 The determinant det : GLn (R) −→ (R × , ×) is a homomorphism from invertible matrices to the real numbers under multiplication, since det(AB) = det(A) det(B). 11 In other words, a homomorphism will commute with multiplication in that they can be applied in either order. This results in a commutative diagram. 12 In some way, this shows that the multiplication is the essential part of the group structure, and the identity and inverse properties are simply there to make sure nothing is able to go wrong with the multiplication. 13 This must exist by the group property of invertibility. Lecture 3: Homomorphisms and Isomorphisms Example 3.4 The exponential exp : (C, +) −→ (C × , ×) taking z −→ ez is a homomorphism, since ea+b = ea eb . Let the standard basis vectors of Rn be ⃗e1 = (1, 0, · · · , 0)t , ⃗e2 = (0, 1, · · · , 0)t , and so on, where ⃗ei is the vector consisting of a 1 in the ith entry and 0s elsewhere. For a permutation p ∈ Sn , let Ap be the permutation matrix taking ⃗ei 7−→ ⃗ep(i) . In particular, the ith column of Ap will be ⃗ep(i) .   0 0 1 For example, for p(123), Ap = 1 0 0. 0 1 0 Example 3.5 The mapping φ : Sn −→ GLn (R) p ∈ Sn 7−→ Ap , where Ap is the permutation matrix, is a homomorphism. This is because Ap (Aq (⃗ei )) = Ap (⃗eq(i) ) = ⃗ep(q(i)) , and Apq (⃗ei ) = ⃗epq(i) = ⃗ep(q(i)) , which matches, so Ap Aq = Apq . There is also another important homomorphism from Sn to another group. Example 3.6 Let sign = det ◦φ take Sn −→ R × by taking the determinant of the permutation matrix. This mapping sign is also a homomorphism, since φ and det are both homomorphisms. 14 In fact, sign(p) = ±1. It is always possible to write any permutation as a composition of transpositions :   0 1 p = τ1 τ2 · · · τr for transpositions τi . The determinant of a transposition matrix is −1, since det = −1, 1 0 so sign(p) = (−1)r where r is the number of transpositions making up p. In fact, if p = τ1 · · · τr = τ1′ · · · τs′ , r = s modulo 2, since the sign homomorphism can be applied on either side. For example, for S3 , e, (123), and (132) all have a sign of +1, while (12), (13), and (23), the transpositions, all have a sign of −1. Notice that R × = GL1 (R), since 1×1 invertible matrices are simply nonzero real numbers. These two examples provide a hint as to why homomorphisms are so useful: matrices/linear mappings and GLn are generally wellunderstood, so if there is a homomorphism from a group to GLn , the knowledge from GLn can then be used to learn more about that particular group. This idea is the core theme of a branch of math called representation theory.15 Example 3.7 For any G and any x ∈ G, let fx : Z −→ G n 7−→ xn . This is a homomorphism because xa+b = xa xb , and is related to the cyclic subgroups of G. Last time, in class, we studied cyclic subgroups ⟨g⟩ using Z and essentially used this homomorphism. In general, homomorphisms allow us to study complicated groups with simpler groups. 14 Permutations 15 These that swap two elements and leave all other elements fxed. examples actually provides the so-called permutation representation and sign representation of Sn . Lecture 3: Homomorphisms and Isomorphisms Theorem 3.8 Let f be a homomorphism from G −→ G′ . Then im(f )a is a subgroup of G′ . a The image of f consists of all the elements in G′ that are mapped to by f. This theorem is not surprising; the whole point of a homomorphism is that it plays nicely with the group structure, and the whole point of a subgroup is that it also plays nicely with the group structure. Example 3.9 For example, im(fx ) from Example 3.7 is ⟨x⟩. Proof. Consider y, y ′ ∈ im(f ). Then there exist x, x′ ∈ G such that y = f (x) and y ′ = f (x′ ). Then yy ′ = f (x)f (x′ ) = f (xx′ ) ∈ im(f ). The inverse and identity conditions are verifed similarly.16 Defnition 3.10 The kernel of f is ker(f ) := {x ∈ G : f (x) = eG′ }. Theorem 3.11 The kernel of a homomorphism f is also a subgroup. Proof. If x, x′ ∈ ker(f ), then f (xx′ ) = f (x)f (x′ ) = eG′ eG′ = eG′ , so xx′ ∈ ker(f ). Also, f (eG ) = eG′ so −1 eG ∈ ker(f ). Lastly, if x ∈ ker(f ), then f (x−1 ) = f (x)−1 = e−1 ∈ ker(f ) as well. G′ = eG′ , so x The image and kernel of each of the previous examples can be seen to be subgroups. The fact that f is a homomorphism is imperative to the proofs of either fact, and these two theorems demonstrate that a homomorphism does in fact respect the group structure. Example 3.12 Consider det : GLn (R) −→ (R × , ×). Since the determinant for invertible matrices can take on any nonzero value, the image of the determinant is all of R × . The kernel of the determinant is SLn (R), the special linear group consisting of the n×n matrices with determinant 1. Example 3.13 For exp : (C, +) −→ (C × , ×), the image is all of C × , and the kernel is 2πiZ ⊆ C, since e2πik = 1. Example 3.14 For φ : Sn −→ GLn (R) p ∈ Sn 7−→ Ap , the image is the set of permutation matrices in GLn (R), whereas ker(φ) = {e}, the identity permutation. Example 3.15 The image of the sign homoomorphism sign = det ◦φ is {±1} ∈ R × . The kernel defnes a new group, called the alternating group An := ker(sign). 16 For inverse, consider y ∈ im(f ). Then there exists x such that y = f (x). From the defnition of a homomorphism, y −1 = f (x)−1 = f (x−1 ) ∈ im(f ). For identity, f (eG ) = eG′ , so eG′ ∈ im(f ). Lecture 3: Homomorphisms and Isomorphisms For example, A3 = {e, (123), (132)} ⊆ S3 . Example 3.16 The kernel of fx is {n : xn = eG }, which was used in the previous class, and is dZ where d is the order of x if it is fnite, and {0} if the order of x is infnite. Defnition 3.17 A mapping f : G −→ G′ is an isomorphism if it is a bijective homomorphism. In some sense, if two groups are isomorphic (that is, if there exists an isomorphism between them), they are essentially the same group, because there are the exact same number of elements and the multiplication relationships between the elements will be exactly the same. Usually, in group theory, groups are considered with respect to the isomorphism classes. Example 3.18 The exponential map from the real numbers under addition onto the positive real numbers under multiplication exp : (R, +) −→ (R>0 , ×) t 7−→ et is an isomorphism. Given an isomorphism f : G −→ G′ , f −1 : G′ −→ G is also an isomorphism, since f −1 (yy ′ ) = f −1 (y)f −1 (y ′ ). If there exists an isomorphism between G and G′ , this is denoted as G ∼ = G′ .