semidirect product of groups

The goal of this exposition is to carefully explain the correspondence between the notions of external and internal semi–direct products of groups, as well as the connection between semi–direct productsMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and short exact sequencesMathworldPlanetmathPlanetmath.

Naturally, we start with the construction of semi–direct products.

Definition 1.

Let H and Q be groups and let θ:QAut(H) be a group homomorphismMathworldPlanetmath. The semi–direct product HθQ is defined to be the group with underlying set {(h,q)hH,qQ} and group operationMathworldPlanetmath (h,q)(h,q):=(hθ(q)h,qq).

We leave it to the reader to check that HθQ is really a group. It helps to know that the inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of (h,q) is (θ(q-1)(h-1),q-1).

For the remainder of this article, we omit θ from the notation whenever this map is clear from the context.

Set G:=HQ. There exist canonical monomorphismsMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath HG and QG, given by

h(h,1Q), hH
q(1H,q), qQ

where 1H (resp. 1Q) is the identity elementMathworldPlanetmath of H (resp. Q). These monomorphisms are so natural that we will treat H and Q as subgroupsMathworldPlanetmathPlanetmath of G under these inclusions.

Theorem 2.

Let G:=HQ as above. Then:


Let p:GQ be the projection map defined by p(h,q)=q. Then p is a homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath with kernel H. Therefore H is a normal subgroup of G.

Every (h,q)G can be written as (h,1Q)(1H,q). Therefore HQ=G.

Finally, it is evident that (1H,1Q) is the only element of G that is of the form (h,1Q) for hH and (1H,q) for qQ. ∎

This result motivates the definition of internal semi–direct products.

Definition 3.

Let G be a group with subgroups H and Q. We say G is the internal semi–direct product of H and Q if:

  • H is a normal subgroup of G.

  • HQ=G.

  • HQ={1G}.

We know an external semi–direct product is an internal semi–direct product (Theorem 2). Now we prove a converseMathworldPlanetmath (Theorem 5), namely, that an internal semi–direct product is an external semi–direct product.

Lemma 4.

Let G be a group with subgroups H and Q. Suppose G=HQ and HQ={1G}. Then every element g of G can be written uniquely in the form hq, for hH and qQ.


Since G=HQ, we know that g can be written as hq. Suppose it can also be written as hq. Then hq=hq so h-1h=qq-1HQ={1G}. Therefore h=h and q=q. ∎

Theorem 5.

Suppose G is a group with subgroups H and Q, and G is the internal semi–direct product of H and Q. Then GHθQ where θ:QAut(H) is given by


By Lemma 4, every element g of G can be written uniquely in the form hq, with hH and qQ. Therefore, the map ϕ:HQG given by ϕ(h,q)=hq is a bijection from G to HQ. It only remains to show that this bijection is a homomorphism.

Given elements (h,q) and (h,q) in HQ, we have


Therefore ϕ is an isomorphismMathworldPlanetmathPlanetmath. ∎

Consider the external semi–direct product G:=HθQ with subgroups H and Q. We know from Theorem 5 that G is isomorphic to the external semi–direct product HθQ, where we are temporarily writing θ for the conjugationMathworldPlanetmath map θ(q)(h):=qhq-1 of Theorem 5. But in fact the two maps θ and θ are the same:


In summary, one may use Theorems 2 and 5 to pass freely between the notions of internal semi–direct product and external semi–direct product.

Finally, we discuss the correspondence between semi–direct products and split exact sequences of groups.

Definition 6.

An exact sequenceMathworldPlanetmathPlanetmathPlanetmathPlanetmath of groups


is split if there exists a homomorphism k:QG such that jk is the identity map on Q.

Theorem 7.

Let G, H, and Q be groups. Then G is isomorphic to a semi–direct product HQ if and only if there exists a split exact sequence


First suppose GHQ. Let i:HG be the inclusion map i(h)=(h,1Q) and let j:GQ be the projection map j(h,q)=q. Let the splitting map k:QG be the inclusion map k(q)=(1H,q). Then the sequence above is clearly split exact.

Now suppose we have the split exact sequence above. Let k:QG be the splitting map. Then:

  • i(H)=kerj, so i(H) is normal in G.

  • For any gG, set q:=k(j(g)). Then j(gq-1)=j(g)j(k(j(g)))-1=1Q, so gq-1Imi. Set h:=gq-1. Then g=hq. Therefore G=i(H)k(Q).

  • Suppose gG is in both i(H) and k(Q). Write g=k(q). Then k(q)Imi=kerj, so q=j(k(q))=1Q. Therefore g=k(q)=k(1Q)=1G, so i(H)k(Q)={1G}.

This proves that G is the internal semi–direct product of i(H) and k(Q). These are isomorphic to H and Q, respectively. Therefore G is isomorphic to a semi–direct product HQ. ∎

Thus, not all normal subgroups HG give rise to an (internal) semi–direct product G=HG/H. More specifically, if H is a normal subgroup of G, we have the canonical exact sequence


We see that G can be decomposed into HG/H as an internal semi–direct product if and only if the canonical exact sequence splits.

Title semidirect product of groups
Canonical name SemidirectProductOfGroups
Date of creation 2013-03-22 12:34:49
Last modified on 2013-03-22 12:34:49
Owner djao (24)
Last modified by djao (24)
Numerical id 10
Author djao (24)
Entry type Definition
Classification msc 20E22
Synonym semidirect product
Synonym semi-direct product