Download Semidirect Products

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Semidirect Products
A semidirect product is a groups having two complementary subgroups one
of which is normal.
Definition 10.1. Two subgroups H, K ≤ G are called complementary (to
each other) if
1. H ∩ K = 1
2. HK = G [HK = {hk | h ∈ H, k ∈ K}]1
For example, if G is a group of order pa q b then the p-Sylow subgroup P
and q-Sylow subgroup Q are complementary. The complement of H is not
unique, even if both H, K are normal. [E.g., take G = R2 as an additive
group and let H, K be any two distinct lines through the origin.]
Proposition 10.2. If H, K ≤ G are complementary then each element of G
can be written uniquely in the form
g = hk
where h ∈ H, k ∈ K.
Proof. We are assuming that G = HK so each g = hk. So we only need
−1
the uniqueness. But g = h1 k1 = h2 k2 implies h−1
2 h1 = k2 k1 . This is in
H ∩ K = 1 so h1 = h2 and k1 = k2 .
Proposition 10.3. If K E G then any complementary subgroup H is iso≈
morphic to Q = G/K. Furthermore the isomorphism q : H −
→ Q is the
restriction to H of the quotient map G ³ Q.
Proof. The kernel of the map φ : H → Q induced by the quotient map is
K ∩ H = 1. Thus φ is injective. To show that φ is onto take any element
gK of Q = G/K. Then g = hk so gK = hkK = hK = φ(h) ∈ im(φ).
Definition 10.4. G is a semidirect product of K by Q if K E G and there
exists H ≤ G is a complement of K isomorphic to Q.
By Proposition 10.3, Q ∼
=H∼
= G/K. Thus a semidirect product of K by
Q is an extension of K by Q.
Example 10.5. Sn is a semidirect product of An by Z/2 where the complement of An is given, e.g., by H = h(12)i.
Example 10.6. The dihedral group D2n is a semidirect product of Z/n = hsi
by Z/2 where H = hti is the complement of hsi.
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Note that, in general, HK is not a subgroup of G unless either H or K is normal. For
example H = h(12)i, K = h(234)i generate all of S4 but HK has only 6 elements.
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Example 10.7. If n is odd, the cyclic group Z/2n is also a semidirect product
of Z/n by Z/2 where the complement of K = 2Z/2n is the 2-Sylow subgroup
of Z/2n.
HW4.ex05: Prove that Q (the quaternion group of order 8) is not a semidirect
product of Z/4 by Z/2.
Theorem 10.8. G is a semidirect product of K by Q = G/K iff the quotient
map q : G ³ Q splits, i.e., if there exists a homomorphism s : Q → G so
that qs = idQ . [s is called a section of q.]
Proof. The image of the section s, if it exists, is a complement for K so G is
a semidirect product. Conversely, if G is a semidirect product, then K has
a complement H and the inverse of the isomorphism φ : H → Q is a section
s.
Definition 10.9. Suppose that θ : Q → Aut(K) (x 7→ θx ) is a homomorphism. Then we say that a semidirect product G of K by Q realizes θ if the
conjugation action of H = s(Q) on K is given by θ, i.e.,
γs(x) = θx
for all x ∈ Q.
The basic theorem about semidirect products is that they exist and are
uniquely determined (up to isomorphism) by K, Q, θ. I will outline the usual
proof and then explain a fancy categorical proof (using holomorphs and pullbacks) which although more complicated has a certain appeal.
Definition 10.10. Given K, Q and θ : Q → Aut(K) let K oθ Q be the
cartesian product K × Q with the following multiplication rule:
(a, x)(b, y) = (aθx (b), xy).
We need to check that this is a group operation.
1. (associativity) Use the fact that θ is a homomorphism (θxy = θx θy ):
(a) ((a, x)(b, y))(c, z) = (aθx (b), xy)(c, z) = (aθx (b)θxy (c), xyz)
(b) (a, x)((b, y)(c, z)) = (a, x)(bθy (c), yz) = (aθx (b)θx θy (c), xyz)
2. (left identity) (1, 1)(a, x) = (1θ1 (a), x) = (a, x)
3. (left inverse) (θz (a)−1 , z)(a, x) = (θz (a)−1 θz (a), zx) = (1, 1) if z = x−1 .
Theorem 10.11. K oθ Q is a semidirect product of K by Q realizing θ.
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Proof. We identify K with the set of all (a, 1) where a ∈ K. This is a normal
subgroup of K oθ Q:
(∗, x)(c, 1)(∗, x−1 ) = (∗θx (c)∗, xx−1 ) ∈ K
H = {(1, x) | x ∈ Q} is a subgroup complementary to K and isomorphic to
Q. Thus K oθ Q is a semidirect product of K by Q. It is also clear that this
realizes the action θ. [Replace ∗ by 1 in the calculation above.]
Theorem 10.12. Any semidirect product G of K by Q realizing θ is isomorphic to K oθ Q.
Proof. By definition 10.9 there is a section s : Q → G of the quotient map
q : G → Q so that θx = γs(x) . Let ψ : K oθ Q → G be given by ψ(a, x) =
as(x). This is a homomorphism since
ψ(a, x)ψ(b, y) = as(x)bs(y) = aγs(x) (b)s(x)s(y) = aθx (b)s(xy) = ψ(aθx (b), xy)
and ψ is a bijection by Proposition 10.2.
There is a “universal” semidirect product of K called the “holomorph”
of K.
Definition 10.13. For any group K its holomorph Hol(K) is defined to be
the group of permutations of K generated by
1. automorphisms ψ : K → K
2. left translations Lg : K → K (x 7→ gx)
In other words,
­
®
Hol(K) = Aut(K), K `
where K ` = {Lg | g ∈ K}.
Note that Aut(K) normalizes K ` since
ψLg ψ −1 = Lψ(g)
Consequently, K ` E Hol(K) and Hol(K) = K ` Aut(K).
Theorem 10.14. Hol(K) ∼
= K oid Aut(K) where id : Aut(K) → Aut(K) is
the identity homomorphism.
Proof. The elements of Hol(K) can be written uniquely in the form Lg ψ and
multiplication is given by:
Lg ψLh φ = Lg Lψ(h) ψφ = Lgψ(h) ψφ
which corresponds to the multiplication rule in K oid Aut(K)
The semidirect product K oθ Q is now given by the “pull-back” construction:
K oθ Q = {(x, f ) | x ∈ Q, f = Lg ψ ∈ Hol(K), θ(x) = ψ}
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