Why Read This Book? - Index of

8.5 Normal Subgroups 277
Theorem 8.5.4 Suppose H is a subgroup of G. Then H is normal if and only if
for all h ∈ H and g ∈ G, g −1 hg ∈ H.
It is one thing to say that a subgroup is closed under the operations of the
group. But the fact that g −1 hg ∈ H for all h ∈ H and g ∈ G adds a new dimension
to the closure of H by saying, in a sense, that elements of H cannot be kicked
outside of H by boxing them inside things of the form g −1 ✷g. For a given h ∈ H,
an expression of the form g −1 hg is called a conjugate of h. Theorem 8.5.4 says that
H is normal if and only if the conjugates of all its elements lie in H. Thus choosing
any h ∈ H and g ∈ G, it might be that g −1 hg is different from h, but at least it’s
still in H. This is the way you will want to argue the following.
EXERCISE 8.5.5 The alternating group An is a normal subgroup of Sn.
If G is not abelian and a subgroup H is not normal, then a conjugate of an
element of H might or might not be in H. For example, using ρ = (1234) ∈ D8 and
(12) ∈ S4,
(12) −1 (1234)(12) = (12)(1234)(12) = (1342) /∈ D8
(8.69)
which by Theorem 8.5.4 proves that D8 is not normal as a subgroup of S4. On the
other hand, using (123) = (13)(12) ∈ A4 and (14) ∈ S4,
(14) −1 (123)(14) = (14)(123)(14) = (234) = (24)(23) ∈ A4
which is a consequence of the fact that A4 is a normal subgroup of S4.
(8.70)
EXERCISE 8.5.6 Use Theorem 8.5.4 to show that H ={(1), (123), (132)} is a
normal subgroup of S3.
EXERCISE 8.5.7 Is H ={(1), (12)} a normal subgroup of S3? Prove or disprove.
EXERCISE 8.5.8 From Exercise 8.4.6, H ={(1), (12), (34), (12)(34)} is a subgroup
of S4. Prove or disprove that H is normal in S4.
If G is abelian, conjugation is not particularly interesting.
Theorem 8.5.9 A group G is abelian if and only if every h ∈ G has no conjugates
other than itself.
The proof of Theorem 8.5.9 should be immediately clear, for the conditions
that gh = hg for all g, h ∈ G and g −1 hg = h for all g, h ∈ G are identical.
If we think of g as fixed and allow h to take on all values in H, we create what
we call a conjugate of the subgroup H. Notationally, we write this as
g −1 Hg ={g −1 hg : h ∈ H} (8.71)