CHAPTER 4 Group Homomorphisms and Isomorphisms c WWLChen,1991,1993,2013. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Consider the groups (Z4, +) and (Z⇤ , ). They have the group tables below. 10 · + 0 1 2 3 1 7 9 3 · 0 0 1 2 3 1 1 7 9 3 1 1 2 3 0 7 7 9 3 1 2 2 3 0 1 9 9 3 1 7 3 3 0 1 2 3 3 1 7 9 Note that for the group table on the right, we have deliberately permuted the elements of Z10⇤ in order to highlight the similarity between the two groups. Indeed, if we match the elements 0, 1, 2, 3 of Z4 respectively with the elements 1, 7, 9, 3 of Z10⇤ , then although the two groups look di↵erent, they are “essentially the same”. 4.1. Formal Definition Definition. Suppose that (G, ) and (H, ) are groups. A function φ : G H is said to be a group homomorphism if the following⇤ condition◦ is satisfied: ! (HOM) For every x, y G,(x y)φ = xφ yφ. 2 ⇤ ◦ Definition. Suppose that (G, ) and (H, ) are groups. A function φ : G H is said to be a group isomorphism if the following⇤ conditions◦ are satisfied: ! (IS1) φ : G H is a group homomorphism. (IS2) φ : G ! H is one-to-one. (IS3) φ : G ! H is onto. ! Definition. We say that two groups G and H are isomorphic if there exists a group isomorphism φ : G H. ! Examples. (1) (Z4, +) and (Z10⇤ , ) are isomorphic. To see this, define φ : Z4 Z10⇤ by taking 0φ = 1, 1φ = 7, 2φ = 9 and 3φ = 3. · ! (2) (Z2, +) and ( 1 , ) are isomorphic. To see this, define φ : Z2 1 by taking 0φ = 1 and 1φ = 1. {± } · !{± } − (3) It is not difficult to show that (R, +) and (R+, ) are groups. Here R+ denotes the set of positive · real numbers. These two groups are isomorphic. To see this, define φ : R R+ by writing xφ =ex ! for every x R.Thenφ is a group homomorphism, as ex+y =exey for every x, y R. On the other 2 2 hand, it is well known that the exponential function is a one-to-one and onto function from R to R+. (4) The set 10 01 10 0 1 G = , , , , 01 10 −0 1 10− ⇢✓ ◆ ✓ − ◆ ✓ − ◆ ✓ ◆ together with matrix multiplication, forms a group. This group is isomorphic to (Z4, +). To see this, define φ : G Z4 to send the four matrices above to 0, 1, 2, 3respectively. ! 17 18 4. GROUP HOMOMORPHISMS AND ISOMORPHISMS (5) The set 10 10 10 10 G = , , , , 01 0 1 −01 −0 1 ⇢✓ ◆ ✓ − ◆ ✓ ◆ ✓ − ◆ together with matrix multiplication, forms a group. This group is isomorphic to (Z⇤, ). To see this, 8 · define φ : G Z⇤ to send the four matrices above to 1, 3, 5, 7respectively. ! 8 (6) Suppose that G is an abelian group, and let n Z.Defineφ : G G by writing xφ = xn for every x G.Thenφ is clearly a group homomorphism2 from G to itself.! For every x, y G,we clearly have2 (xy)φ =(xy)n = xnyn =(xφ)(yφ). 2 (7) Consider the multiplicative group ⇤ (R) of invertible 2 2 matrices with entries in R. M2,2 ⇥ Consider also the set R 0 ; this forms a group under multiplication. Now define a function \{ } φ : ⇤ (R) R 0 by writing Aφ =det(A), the determinant of A, for every matrix A ⇤ (R). M2,2 ! \{ } 2M2,2 This is clearly a group homomorphism, since det(AB)=det(A)det(B) for every A, B ⇤ (R). 2M2,2 (8) This is an example we almost take for granted. Consider the groups (Z, +) and (Z4, +). We define a function φ : Z Z4 as follows. For every n Z, we can clearly write n =4q + r,where ! 2 q, r Z and 0 6 r<4. It is well known that for given n, the integers q and r are uniquely determined. Now2 let nφ = r. So, for example, 4φ = 0, 707φ = 3 and ( 22)φ = 2. It can easily be shown that − φ : Z Z4 is a group homomorphism. This is called reduction modulo 4. ! (9) Similarly, for every n N 1 , reduction modulo n is a group homomorphism from Z to Zn. 2 \{ } 4.2. Some Properties of Homomorphisms Proposition 4.1. Suppose that (G, ) and (H, ) are groups, with identity elements eG and eH respectively. Suppose further that φ : G ⇤ H is a group◦ homomorphism. Then e φ = e . ! G H Proof. Simply note that e φ =(e e )φ = e φ e φ. G G ⇤ G G ◦ G Theorem 4.2. Suppose that (G, ) and (H, ) are groups, and that φ : G H is a group homo- morphism. Then the set ⇤ ◦ ! Gφ = xφ : x G { 2 } is a subgroup of H. Definition. Suppose that (G, ) and (H, ) are groups, and that φ : G H is group homomor- phism. Then the set Gφ is called the⇤ image of◦ the group homomorphism. ! Proof of Theorem 4.2. Clearly Gφ is non-empty. Let xφ,yφ Gφ. In view of Theorem 1.5, it 2 suffices to show that xφ (yφ)0 Gφ. Clearly y0φ Gφ.Sinceφ is a homomorphism, we must have ◦ 2 2 yφ y0φ =(y y0)φ = e φ = e by Proposition 4.1. Similarly y0φ yφ =(y0 y)φ = e φ = e .It ◦ ⇤ G H ◦ ⇤ G H follows that (yφ)0 = y0φ, so that xφ (yφ)0 = xφ y0φ =(x y0)φ Gφ, as required. ◦ ◦ ⇤ 2 Example. Suppose that we do not know that 10 01 10 0 1 (4.1) , , , 01 10 −0 1 10− ⇢✓ ◆ ✓ − ◆ ✓ − ◆ ✓ ◆ forms a group under matrix multiplication. We can remedy this in the following way if we know that (Z4, +) is a group. We define a function φ : Z4 ⇤ (R)bywriting !M2,2 10 01 10 0 1 0φ = , 1φ = , 2φ = , 3φ = ; 01 10 −0 1 10− ✓ ◆ ✓ − ◆ ✓ − ◆ ✓ ◆ note Example (4) above. We then show that φ is a homomorphism from the group (Z4, +) to the group ⇤ (R), and deduce our desired result by using Theorem 4.2 to conclude that (4.1) is a M2,2 subgroup of ⇤ (R). M2,2 This simple example suggests the following improvement of our idea. To be precise, we do not need the knowledge that H, given by ⇤ (R) in this special example, is a group. M2,2 Proposition 4.3. Suppose that (G, ) is a group, and that is a binary operation on a set S. Suppose further that the function φ : G ⇤ S satisfies the condition◦ that (x y)φ = xφ yφ for every x, y G. Then the set Gφ = xφ : x G!, together with the binary operation⇤ , forms◦ a group. 2 { 2 } ◦ 4.3. NORMAL SUBGROUPS 19 Sketch of Proof. All we need to do is to show that (Gφ, ) forms a group. However, we already know that (G, ) is a group. We can therefore use the property,◦ that (x y)φ = xφ yφ for every x, y G, to “bring⇤ the group structure over” from G to Gφ. ⇤ ◦ 2 Definition. Suppose that (G, ) and (H, ) are groups, and that φ : G H is group homomor- ⇤ ◦ ! phism. Suppose further that eH is the identity element in H.Thentheset ker φ = x G : xφ = e { 2 H } is called the kernel of the group homomorphism. Examples. (1) As in Example (8) above, consider the groups (Z, +) and (Z4, +), and the group homomorphism reduction modulo 4. The kernel of this group homomorphism is clearly the set 4x : x Z . { 2 } 2 (2) Consider the group (Z10⇤ , ). Define φ :(Z10⇤ , ) (Z10⇤ , )bywritingxφ = x for every x Z10⇤ . It can easily be shown that this· is a group homomorphism.· ! Furthermore,· ker φ = 1, 9 . 2 (3) Suppose that G is a finite group of order n. Show that the function φ : G{ }G,definedby xφ = xn for every x G, is a group homomorphism and that ker φ = G. ! 2 Theorem 4.4. Suppose that (G, ) and (H, ) are groups, and that φ : G H is a group homo- morphism. Then ker φ is a subgroup⇤ of G. ◦ ! Proof. Clearly e ker φ. Let x, y ker φ. In view of Theorem 1.5, it clearly suffices to show that 2 2 x y0 ker φ.Sinceφ is a homomorphism, we must have ⇤ 2 y0φ = e y0φ = yφ y0φ =(y y0)φ = e φ = e H ◦ ◦ ⇤ G H by Proposition 4.1. It follows that (x y0)φ = xφ y0φ = e e = e , as required. ⇤ ◦ H ◦ H H Using an argument similar to the proof of Lagrange’s theorem, we can prove the following result. Proposition 4.5. Suppose that (G, ) and (H, ) are groups, and that φ : G H is a group homomorphism. Suppose further that G⇤is finite. Then◦ Gφ divides G . ! | | | | Proof. Define a relation on G in the following way. For every x, y G, x y if xφ = yφ.It is not difficult to check that R is an equivalence relation, and that ker φ 2is oneR of the equivalence classes. Furthermore, the numberR of equivalence classes is Gφ . Suppose now that K is one of the | | equivalence classes. Let a K. For every x ker φ,wehave(a x)φ = aφ xφ = aφ eH = aφ,so that (a x) a,whencea 2x K.Wenowdefine2 f :kerφ K⇤ by writing◦ f(x)=a◦ x for every x ker φ⇤.