Why Read This Book? - Index of

276 Chapter 8 Groups
where we can replace h1a −1
2
a1h1a −1
2
−1
= a1a2 h2, and the fact that a1a −1
2
with a−1
2 h2 for some h2 ∈ H. If we have this, then
∈ H allows us to conclude that
(a1b1)(a2b2) −1 = a1b1b −1
2 a−1
−1 −1
2 = a1h1a2 = a1a2 h2 ∈ H (8.68)
So here is the demand on H that will get us over the hump of showing that the
operation on G/H is well defined when G is not abelian. Since a −1
2 in our analysis
above could be any element of the group, and h1 could be any element of H,we
arrive at the following definition of a special sort of subgroup where we can be
sure that the operation on G/H is well defined.
Definition 8.5.1 Suppose H is a subgroup of a group G. If for all g ∈ G and
h ∈ H, there exists h1 ∈ H such that hg = gh1, then H is called a normal subgroup
of G, and we write H ⊳ G.
Definition 8.5.1 only allows you to swap hg for gh1. It does not explicitly allow
you to swap gh with some h1g. But you can show that it does.
EXERCISE 8.5.2 Suppose H is a normal subgroup of G. Then for all g ∈ G and
h ∈ H, there exists h1 ∈ H such that gh = h1g. 15
All our work here shows that the operation defined in Eq. (8.65) is well defined
if H is a normal subgroup of G. Furthermore, your other work from Exercise 8.3.4
where you did not exploit the abelian nature of G completes the proof that G/H
is a group, and we arrive at the following.
Theorem 8.5.3 Suppose G is a group and H is a normal subgroup of G. Then
G/H with binary operation ∗ defined by (Ha) ∗ (Hb) = H(ab) is a group with
identity He and inverses (Ha) −1 = Ha −1 .
There are ways other than Definition 8.5.1 to define normal subgroup, and
different authors approach the idea in different ways. As we will see in Section 8.6,
normality can be naturally defined in terms of mappings between groups. But we
have chosen our definition, so any equivalent forms will have to be demonstrated
as theorems. If we reread the analysis that led us to Definition 8.5.1, another way
to state what we needed might jump out at us. Rather than state that there exists
h2 ∈ H such that a −1
2 h2 = h1a −1
2 , we simply could have insisted that there exists
h2 ∈ H such that h2 = a2h1a −1
−1
2 , or simply that a2h1a2 ∈ H. This has a natural
appeal because testing a subgroup for normality then reduces simply to making
sure that the criterion in the following theorem is satisfied.
15 Apply Definition 8.5.1 to h −1 g −1 .