Why Read This Book? - Index of

280 Chapter 8 Groups
the way elements of H behave in the presence of elements of G, and not simply
how they behave among themselves. So, if H1 is also a group, then the fact that
H is normal in G will imply H is normal in H1 as well. For if g −1 hg ∈ H for all
g ∈ G, then certainly the same is true for all g ∈ H1. Now if G1 is a group, we
know that H is a subgroup of G1. However, it might be that H is not normal in G1.
Even though g −1 hg ∈ H for all g ∈ G, there might be some g ∈ G1 − G for which
g −1 hg /∈ H.
Example 8.5.18 Beginning with S4, create the following subgroups. First, let
H ={(1), (12)}. Since (12) is its own inverse, H is a subgroup of S4. Let G =
{(1), (12), (34), (12)(34)}. By Exercise 8.4.6, G is a subgroup of S4. Furthermore,
G is abelian. By Exercise 8.5.15, H is a normal subgroup of G because G is abelian.
However, by the next exercise, H is not normal in S4. �
EXERCISE 8.5.19 Show that H ={(1), (12)} is not normal in S4 by finding a
conjugate of (12) that is not in H.
8.6 Group Morphisms
Now that we have a basic understanding of groups and some of their internal
structure, let’s turn our attention outward to a special type of function from one
group to another. The special feature we want these functions to have is that they
preserve the binary operation.
Definition 8.6.1 Suppose (G, ∗,eG, −1 ) and (H, ·,eH, −1 ) are groups, and suppose
φ : G → H is a function with the property that φ(x ∗ y) = φ(x)·φ(y) for all
x, y ∈ G. Then φ is called a homomorphism, or simply a morphism from G to H.
If φ is one-to-one, it is called a monomorphism.Ifφ is onto, it is called an epimorphism.Ifφ
is both one-to-one and onto, it is called an isomorphism, and we write
G ∼ = H, which is read “G is isomorphic to H.” If φ : G → G is an isomorphism,
φ is called an automorphism.
EXERCISE 8.6.2 Let Z be the group of integers under addition and E the group
of all even integers under addition. Assuming that each of the following is a
function, prove the following.
(a) Show that φ1 : Z → E defined by φ1(n) = 2n is an isomorphism.
(b) Show that φ2 : Z → E defined by φ2(n) = 4n is a monomorphism that is not
an isomorphism.
(c) Show that φ3 : Z → Z defined by φ3(n) =−nis an automorphism.
(d) Show that φ4 : Z → Z defined by φ4(n) = 2n + 2 is not a morphism.