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Home , Semidirect product

10 Semidirect Products

A semidirect is a groups having two complementary 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 of paqb then the p-Sylow 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 −1 the uniqueness. But g = h1k1 = h2k2 implies h2 h1 = k2k1 . 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 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 comple- ment of An is given, e.g., by H = h(12)i.

Example 10.6. The D2n is a semidirect product of Z/n = hsi by Z/2 where H = hti is the complement of hsi.

1Note 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.

1 Example 10.7. If n is odd, the 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 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 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 homomor- phism. 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 ( isomorphism) by K, Q, θ. I will outline the usual proof and then explain a fancy categorical proof (using holomorphs and pull- backs) 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 = θxθy):

(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)θxθy(c), xyz)

2. (left identity) (1, 1)(a, x) = (1θ1(a), x) = (a, x)

−1 −1 −1 3. (left inverse) (θz(a) , z)(a, x) = (θz(a) θz(a), zx) = (1, 1) if z = x .

Theorem 10.11. K oθ Q is a semidirect product of K by Q realizing θ.

2 Proof. We identify K with the set of all (a, 1) where a ∈ K. This is a of K oθ Q:

−1 −1 (∗, x)(c, 1)(∗, x ) = (∗θx(c)∗, xx ) ∈ 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 isomor- phic 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 “” 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. 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

−1 ψLgψ = 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 Lgψ and multiplication is given by:

LgψLhφ = LgLψ(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” construc- tion: K oθ Q = {(x, f) | x ∈ Q, f = Lgψ ∈ Hol(K), θ(x) = ψ}

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