Skip to main content

Section 7.2 Semidirect Products

Subsection External Semidirect Products

β€œDon’t give up and always keep on believing in your product.”
―Niels van Deuren
We now discuss an important generalization for the direct product and a new method of constructing a new groups from the action of one group on another.
Suppose G is a group with subgroups H⊴G and K≀G such that H∩K={e}. Then we still have HK≀G; let’s see what we would need the multiplication on the cartesian product HΓ—K to be in order for ΞΈ:HΓ—Kβ†’HK defined by ΞΈ(h,k)=hk to still be a group homomorphism:
ΞΈ(h1,k1)ΞΈ(h2,k2)=h1k1h2k2=h1h2β€²k1k2=ΞΈ(h1h2β€²,k1k2),
where h2β€²βˆˆH is such that k1h2k1βˆ’1=h2β€².
This means that we would need to have (h1,k1)(h2,k2)=(h1h2β€²,k1k2) for ΞΈ to be a homomorphism. This motivates the following definition.

Definition 7.11. External Semidirect Products.

Let H and K be groups and let ρ:Kβ†’Aut(H) be a homomorphism. The (external) semidirect product induced by ρ is the set HΓ—K with the binary operation defined by
(h,k)(hβ€²,kβ€²)=(hρ(k)(hβ€²),kkβ€²).
This group is denoted by Hβ‹ŠΟK.

Discussion 7.1.

Decide amongst yourselves whether it should be spelled "semidirect", "semi-direct", "semi direct".
Before we prove that the construction above actually gives a group, let’s compute a few examples.

Example 7.12. Semidirect, Sans the Semi.

Given H and K we could always take ρ to be the trivial homomorphism, so that ρ(y)(x)=x for all y∈K and x∈H. Then Kβ‹ŠΟH is just the usual direct product:
(y1,x1)(y2,x2)=(y1y2,x1x2).

Example 7.13. .

Fix a group G, a normal subgroup H⊴G and a subgroup K≀G. Then the function
ρ:Kβ†’Aut(H)
given by ρ(x)(y)=xyxβˆ’1 for x∈K,y∈H is a homomorphism. Thus K acts on H via automorphisms.

Example 7.14. .

Let K=⟨x⟩ be cyclic of order 2 and H=⟨y⟩ be cyclic of order n for any nβ‰₯1. There is an automorphism of K that sends y to yβˆ’1. This automorphism is its own inverse; i.e., it has order 2. Therefore, by Proposition 2.28, there is a group homomorphism
ρ:Kβ†’Aut(H)
with ρ(x)(y)=yβˆ’1. We may thus form the group
G:=Hβ‹ŠΟK.
The elements of G are (yi,xj) for 0≀i≀nβˆ’1 and 0≀j≀1, in particular |G|=2n. Set
y~=(y,e)∈G and x~=(e,x)∈G.
y~n=(y,eK)n=(yn,eK)=(eH,eK)=eG,
x~2=(eH,x)2=(eH,x2)=(eH,eK)=eG,
x~y~x~y~=(eH,x)(y,eK)(eH,x)(y,eK)=(ρ(x)(y),x)(ρ(x)(y),x)=(yβˆ’1,x)(yβˆ’1,x)=(yβˆ’1y,e)=eG.
Looks familar!
Indeed, by the universal mapping property for D2n we have a homomorphism
θ:D2n→G
such that ΞΈ(r)=(y,eK) and ΞΈ(s)=(x,eH). Moreover, ΞΈ is onto since
ΞΈ(risj)=(yi,xj) for all 0≀i≀nβˆ’1,0≀j≀1
and since |D2n|=|G|=2n it follows that ΞΈ is a bijection. So the dihedral group is a semidirect product, in which the two component groups are cyclic of orders n and 2 respectively:
D2nβ‰…βŸ¨yβŸ©β‹ŠΟβŸ¨x⟩
and ρ is the inversion homomorphism as described above.

Proof.

  1. The proof is straightforward but a bit messy. For associativity, note that
    (y1,x1)((y2,x2)(y3,x3))=(y1,x1)(y2ρ(x2)(y3),x2x3)=(y1ρ(x1)(y2ρ(x2)(y3)),x1x2x3)=(y1ρ(x1)(y2)(ρ(x1)∘ρ(x2))(y3),x1x2x3)=(y1ρ(x1)(y2)ρ(x1x2)(y3),x1x2x3)
    On the other hand
    ((y1,x1)(y2,x2))(y3,x3)=(y1ρ(x1)(y2),x1x2)(y3,x3)=(y1ρ(x1)(y2)ρ(x1x2)(y3),x1x2x3).
    This gives associativity.
    The fact that (e,e) is a two-sided identity follows from the fact that ρ(e)(y)=idH(y)=y.
    Finally
    (y,x)(ρ(xβˆ’1)(yβˆ’1),xβˆ’1)=(yρ(x)(ρ(xβˆ’1)(yβˆ’1)),e)=(y(ρ(x)∘ρ(xβˆ’1))(yβˆ’1),e)=(yρ(e)(yβˆ’1),e)=(yyβˆ’1,e)=(e,e),
    and similarly
    (ρ(xβˆ’1)(yβˆ’1),xβˆ’1)(y,x)=(e,e).
  2. Define a funtion
    i:Hβ†’Hβ‹ŠΟK
    as i(y)=(y,e). Then i is a homomorphism, since
    i(y1)i(y2)=(y1,e)(y2,e)=(y1ρ(e)(y2),ee)=(y1y2,e)=i(y1y2).
    The map is clearly injective and hence its image is isomorphic to H. In fact, the image is normal since the second component of
    (y,x)(y2,e)(ρ(xβˆ’1)(yβˆ’1),xβˆ’1)
    is clearly e. Let us write this image as
    Hβ€²:={(y,e)∣y∈H}⊴Hβ‹ŠΟK.
    The function
    j:Kβ†’Hβ‹ŠΟK
    defined by j(x)=(e,x) is also an injective homomorphism and thus its image
    Kβ€²:={(e,x)∣x∈H}≀Hβ‹ŠΟK
    is isomorphic to K. Kβ€² is typically not normal, however. Finally, it is easy to see that Hβ€²Kβ€²=Hβ‹ŠΟK and Hβ€²βˆ©Kβ€²={e}. Putting this all together we have
    • Hβ€²βŠ΄Hβ‹ŠΟK,
    • K′≀Hβ‹ŠΟK,
    • Hβ€²Kβ€²=Hβ‹ŠΟK, and
    • Hβ€²βˆ©Kβ€²={e}.
  3. Consider the projection onto the second factor Ο€2:Hβ‹ŠΟKβ†’K given by Ο€2(y,x)=x. This is a goup homomorphism since the second component of (y1,x1)(y2,x2) is x1x2 and is surjective by definition. Now
    Ker(Ο€2)={(y,eK)∣y∈H}=Hβ€²β‰…H.
    By the First Isomorphism Theorem for Groups we conclude that (Hβ‹ŠΟK)/Hβ€²β‰…K.

Subsection Internal Semidirect Products

β€œI’ve often said there’s nothing better for the inside of a man than the outside of a horse.”
―Ronald Reagan
We can turn this around.
 1 
the semidirect product, not the Reagan quote.

Proof.

We have
ΞΈ((y1,x1)(y2,x2))=ΞΈ(y1cx1(y2),x1x2)=y1x1y2x1βˆ’1x1x2=y1x1y2x2=ΞΈ(y1,x1)ΞΈ(y2,x2)
and thus ΞΈ is a homomorphism. It’s kernel is {(y,x)∣y=xβˆ’1,y∈H,x∈K}, which is just {e} since H∩K={e}. The image of ΞΈ is clearly KH=G. This proves ΞΈ is an isomorphism. It is obvious that ΞΈ(Kβ€²)=K and ΞΈ(Hβ€²)=H.

Definition 7.17. Internal Semidirect Products.

In this situation of the Theorem 7.16, we will say that G is the internal semi-direct product of H and K.

Example 7.18. .

Returning to D2n, let H=⟨s⟩ and K=⟨r⟩. Then H≀G, K⊴G, HK=G and H∩K={e}. So, G is isomorphic to a semi-direct product, as we already showed.

Example 7.19. .

Let G=Sn, K=An and H=⟨(12)⟩. Then K⊴G, H≀G, KH=G and K∩H={e}. It follows that
Snβ‰…Anβ‹ŠΟC2
where C2=⟨x⟩ is cyclic of order 2 and the action ρ:C2β†’Aut(An) sends x to conjugation by (12).
It is important to be aware that for a fixed pair of groups H and K, different actions of H on K via automorphisms can result in isomorphic semi-direct products. Indeed, determining when Kβ‹ŠΟHβ‰…Kβ‹ŠΟβ€²H is in general a tricky business. The previous example shows this.

Example 7.20. .

Let G=Sn and K=An again, but this time take Hβ€²=⟨(13)⟩=(123)⟨(12)⟩(123)βˆ’1 (assuming nβ‰₯3). Then we get
Snβ‰…Anβ‹ŠΟβ€²C2
where C2=⟨x⟩ is cyclic of order 2 and the action ρ′:C2β†’Aut(An) sends x to conjugation by (13).
The actions ρ and ρ′ are not identical. For example, assuming nβ‰₯4 we have
ρ(x)(12)(12)(34)=(12)(34) maps 1↦2
ρ′(x)(12)(34)=(13)(12)(34)(13) maps 1↦4.
Anβ‹ŠΟHβ‰…Anβ‹ŠΟβ€²Hβ€²
since each is isomorphic to Sn.
On HW 8 you will give a more conceptual reason for why these two semidirect products turned out to be isomorphic: it is because H and Hβ€² are conjugate in Sn. More generally, below is a criterion for a two semidirect products to be isomorphic.

Subsection Groups of Order pq

β€œI did not know I was in my prime until afterwards.”
―Mason Cooley

Proof.

Let G be a group of order 6. Cayley’s Theorem gives that there exist elements x,y∈G with |x|=2 and |y|=3. Let K=⟨x⟩ and H=⟨y⟩. Since [G:H]=2, H is a normal subgroup of G and since H∩K is a common subgroup of H and K Lagrange’s Theorem gives that |H∩K|∣gcd(|H|,|K|)=1. Thus H∩K={e} and since |HK|=|H||K||H∩K|=6=|G| we deduce that HK=G. Theorem 7.16 now gives that G is the internal semidirect product of H and K. More to the point, Gβ‰…Hβ‹ŠΟK, where ρ:Kβ†’Aut(H) gives the action of K on H by Conjugation.
We now analyze the possibilities for ρ. By Proposition 2.23, Aut(H)β‰…Aut(Z/3)β‰…(Z/3Γ—,β‹…)=({Β±1},β‹…). There are two possibilities for the image of ρ: either Im(ρ)={idH} or Im(ρ)=Aut(H).
If Im(ρ)={idH}, then ρ(x)=cx=idH (which implies xy=yx) and Hβ‹ŠΟK=HΓ—K. Therefore, in this case Gβ‰…HΓ—Kβ‰…C3Γ—C2β‰…C6, where the last isomorphism uses the Sunzi’s Remainder Theorem.
If Im(ρ)=Aut(H), then ρ(x) is the map yi↦yβˆ’1 and by an earlier example for this ρ we have Hβ‹ŠΟKβ‰…D6, so Gβ‰…D6.
Finally, C6≇D6 because the former is abelian and the latter is not.
Let’s repeat the previous example for classifying groups of order pβ‹…q with p,q distinct primes into isomorphism classes.

Proof.

Let G be a group of order pq and let H,K be Sylow subgroups of order q and p respectively. We see that H is a normal subgroup using Theorem 5.16, since [G:H]=p is the smallest prime that divides |G|.
Furthermore, since H∩K is a subgroup of both H and K we have by Lagrange’s Theorem that |H∩K|∣gcd(q,p)=1, so that H∩K={eG}. From here it follows that
|HK|=|H||K||H∩K|=qβ‹…p1=pq=|G|
and so HK=G. Theorem 7.16 now yields that
Gβ‰…Hβ‹ŠΟKβ‰…Cqβ‹ŠΟCp
for some homomorphism ρ:Kβ†’Aut(H), equivalently ρ:Cpβ†’Aut(Cq). By Proposition 2.28 to give such a homomorphism ρ is equivalent to giving an element z∈Aut(Cq) so that zp=id, which will give ρ(yj)=zj for Cp=⟨y⟩. Thus |z|∣p yielding that either |z|=1 or |z|=p.
Case 1: if |z|=1 then ρ is the trivial homomorphism and thus Gβ‰…CqΓ—Cpβ‰…Cpq.
Case 2: if |z|=p then it must be the case by Lagrange that p∣|Aut(Cq). By Corollary 2.24 we know that Aut(Cq)β‰…Cpβˆ’1 is a cyclic group. Therefore we have that p∣(qβˆ’1) if and only if there exists an element z∈Aut(Cq) of order p by Theorem 2.26 (2). Moreover any such element z generates a subgroup of Aut(Cq) of order p. Since there is a unique subgroup of a cyclic group of a given order by Theorem 2.26 (2) we see that the image of ρ is independent of the choice of z. Thus by Theorem 7.21 we conclude that all subgroups resulting from any choice of z of order p are isomorphic.
Moreover, from the explicit presentation of semidirect products of cyclic groups given in a homework problem we see that the resulting group is non-abelian; in particular it is not isomorphic to Cpq.