The First Isomorphism Theorem

The First Isomorphism Theorem 3-22-2018 The First Isomorphism Theorem helps identify quotient groups as known or familiar groups. I ll begin by proving a useful lemma. Proposition. Let φ : be a group map. φ is injective if and only if kerφ = {1}. Proof. ( ) Suppose φ is injective. Since φ(1) = 1, {1} kerφ. Conversely, let g kerφ, so φ(g) = 1. Then φ(g) = 1 = φ(1), so by injectivity g = 1. Therefore, kerφ {1}, so kerφ = {1}. ( ) Suppose kerφ = {1}. I want to show that φ is injective. Suppose φ(a) = φ(b). I want to show that a = b. φ(a) = φ(b) φ(a)φ(b) 1 = φ(b)φ(b) 1 φ(a)φ(b 1 ) = 1 φ(ab 1 ) = 1 ence, ab 1 kerφ = {1}, so ab 1 = 1, and a = b. Therefore, φ is injective. Eample. (Proving that a group map is injective) Define f : R 2 R 2 by Prove that f is injective. f(,y) = (3+2y,+y). As usual, R 2 is a group under vector addition. I can write f in the form f ([ ]) = y [ 3 2 1 1 ][ ]. y Since f has been represented as multiplication by a constant matri, it is a linear transformation, so it s a group map. To show f is injective, I ll show that the kernel of f consists of only the identity: kerf = {(0,0)}. Suppose (,y) kerf. Then [ 3 2 ][ ] = y [ ] 0. 0 1 1 [ ] 3 2 Since det = 1 0, I know by linear algebra that the matri equation has only the trivial 1 1 solution: (,y) = (0,0). This proves that if (,y) kerf, then (,y) = (0,0), so kerf {(0,0)}. Since (0,0) kerf, it follows that kerf = {(0,0)}. ence, f is injective. Theorem. (The First Isomorphism Theorem) Let φ : be a group map, and let π : /kerφ bethequotientmap. Thereisanisomorphism φ : /kerφ imφsuchthatthefollowingdiagramcommutes: π ց φ /kerφ φ im φ 1

Proof. Since φ maps onto imφ and kerφ kerφ, the universal property of the quotient yields a map φ : /kerφ imφ such that the diagram above commutes. Since φ is surjective, so is φ; in fact, if φ(g) imφ, by commutativity φ(π(g)) = φ(g). It remains to show that φ is injective. By the previous lemma, it suffices to show that ker φ = {1}. Since φ maps out of /kerφ, the 1 here is the identity element of the group /kerφ, which is the subgroup kerφ. So I need to show that ker φ = {kerφ}. owever, this follows immediately from commutativity of the diagram. For gkerφ ker φ if and only if φ(gkerφ) = 1. This is equivalent to φ(π(g)) = 1, or φ(g) = 1, or g kerφ i.e. ker φ = {kerφ}. Eample. (Using the First Isomorphism Theorem to show two groups are isomorphic) Use the First Isomorphism Theorem to prove that R {1, 1} R+. R is the group of nonzero real numbers under multiplication. R + is the group of positive real numbers under multiplication. {1, 1} is the group consisting of 1 and 1 under multiplication (it s isomorphic to Z 2 ). I ll define a group map from R onto R + whose kernel is {1, 1}. Define φ : R R + by φ() =. φ is a group map: φ(y) = y = y = φ()φ(y). If z R + is a positive real number, then φ(z) = z = z. Therefore, φ is surjective: imφ = R +. Finally, φ clearly sends 1 and 1 to the identity 1 R +, and those are the only two elements of R which map to 1. Therefore, kerφ = {1, 1}. By the First Isomorphism Theorem, Note that I didn t construct a map the isomorphism for me. R {1, 1} = R kerφ imφ = R+. R {1, 1} R+ eplicitly; the First Isomorphism Theorem constructs Eample. R 2 is a group under componentwise addition and R is a group under addition. Let = { ( } 5, π) R. Prove that R2 R. 2

Define f : R 2 R by Note that f f(,y) = π+ 5y. ([ ]) = [π [ ] 5]. y y Since f can be epressed as multiplication by a constant matri, it s a linear transformation, and hence a group map. Let ( 5, π). Then f[ ( 5, π)] = f( 5, π) = π( 5)+ 5( π) = 0. Therefore, ( 5, π) kerf, and hence kerf. Let (,y) kerf. Then f(,y) = 0 π+ 5y = 0 5y = π y = π 5 ence, (,y) = (, π ) = 1 ( 5, π). 5 5 Therefore, kerf. ence, kerf =. Let z R. Note that ( ) 1 f π z,0 = π 1 π z + 5 0 = z. ence, imf = R. Thus, R 2 = R2 kerf imf = R. Eample. Z Z is a group under componentwise addition and Z is a group under addition. Prove that Define f : Z Z Z by f can be represented by matri multiplication: ence, it s a group map. Let n(12,17) = (12n,17n) (12,17). Then Thus, (12,17) kerf. Z Z (12,17) Z. f(,y) = 17 12y. ([ ]) [ ] = [17 12]. y y f((12n,17n) = 17(12n) 12(17n) = 0. 3

Let (,y) kerf. Then f(,y) = 0 17 12y = 0 17 = 12y Now 17 12y but (12,17) = 1. By Euclid s lemma, 17 y. Say y = 17n. Then 17 = 12(17n), so = 12n. Therefore, Thus, kerf (12,17). ence, (12,17) = kerf. Let z Z. Note that Multiplying by z, I get Then This proves that imf = Z. ence, (,y) = (12n,17n) = n(12,17) (12,17). 1 = (17, 12) = 5 17+7 ( 12). z = 17(5z) 12(7z). f(5z,7z) = 17(5z) 12(7z) = z. Z Z (12,17) = Z Z imf = Z. kerf Eample. R R R is a group under componentwise addition. Consider the subgroup Prove that R R R R R. = (R R is a group under componentwise addition.) Define f : R R R R R by { } (1,2,3) R. f(,y,z) = (y 2,z 3). Note that f y = z [ 2 1 0 3 0 1 ] y z. Since f is defined by matri multiplication, it is a linear transformation. ence, it s a group map. Let (1,2,3) = (,2,3). Then ence, (,2,3) kerf, and kerf. Let (,y,z) kerf. Then f(,2,3) = (2 2,3 3) = (0,0). f(,y,z) = (0,0) (y 2,z 3) = (0,0) 4

Equating the first components, I have y 2 = 0, so y = 2. Equating the second components, I have z 3 = 0, so z = 3. Thus, (,y,z) = (,2,3). Therefore, kerf, and so = kerf. Let (a,b) R R. Then f(0,a,b) = (a 2 0,b 3 0) = (a,b). ence, imf = R R. Thus, R R R = R R R kerf imf = R R. The first equality follows from = kerf. The isomorphism follows from the First Isomorphism Theorem. The second equality follows from imf = R R. Proposition. If φ : is a surjective group map and, then φ(). Proof. 1, so 1 = φ(1) φ(), and φ(). Let a,b, so φ(a),φ(b) φ(). Then φ(a)φ(b) 1 = φ(a)φ(b 1 ) = φ(ab 1 ) φ(), since ab 1. Therefore, φ() is a subgroup. (Notice that this does not use the fact that is normal. ence, I ve actually proved that the image of a subgroup is a subgroup.) Now let h, a, so φ(a) φ(). I want to show that hφ(a)h 1 φ(). Since φ is surjective, h = φ(g) for some g. Then hφ(a)h 1 = φ(g)φ(a)φ(g) 1 = φ ( gag 1). But gag 1 because is normal. ence, φ ( gag 1) φ(). It follows that φ() is a normal subgroup of. Theorem. (The Second Isomorphism Theorem) Let,, <. Then. Proof. I ll use the First Isomorphism Theorem. To do this, I need to define a group map. To define this group map, I ll use the Universal Property of the Quotient. The quotient map π : is a group map. By the lemma preceding the Universal Property of the Quotient, = kerπ. Since, it follows that kerπ. Since π : is a group map and kerπ, the Universal Property of the Quotient implies that there is a group map π : given by π(g) = g. If g, then π(g) = g. Therefore, π is surjective. 5

I claim that ker π =. First, if h (so h ), then π(h) = h =. Since is the identity in, it follows that h ker π. Conversely, suppose g ker π, so π(g) =, or g =. The last equation implies that g, so g. Thus, ker π =. By the First Isomorphism Theorem, = ker π im π =. There is also a Third Isomorphism Theorem (sometimes called the Modular Isomorphism, or the Noether Isomorphism). It asserts that if < and, then. You can prove it using the First Isomorphism Theorem, in a manner similar to that used in the proof of the Second Isomorphism Theorem. c 2018 by Bruce Ikenaga 6