Groups, Direct and Semidirect Products

Direct Product, Direct Sum

Let G and H be arbitrary groups, and let J be a larger group whose elements, as a set, are defined by the cross product of G and H. Let g1 and g2 be members of G, while h1 and h2 are members of H. The composite group J is the direct product of G and H, if the product of any two elements in J is equal to g1g2,h1h2. In other words, the groups G and H run in parallel. We've already seen examples of this. The group Z7Z7 contains 49 elements, two parallel copies of the integers mod 7. For instance: 2,5 + 3,3 = 5,1. Project J onto G or H by throwing away the "other" component. This is a group homomorphism, hence either G or H can act as the kernel of J, with H or G playing the role of factor group.

Apply the above several times to take the direct product of a finite number of groups.

The direct product of an infinite set of groups is also well defined. We don't have to worry about the axiom of choice, because every set has a special element, the group identity. The direct product includes the element 1,1,1,1,1,1… which acts as the identity element for the composite group. Group operations are performed per component.

The direct sum is essentially defined in the same way as the direct product: group operations are still performed per component, but almost all components are set to the identity element. See more on the direct sum here.

As an example, consider the direct product or sum of Z_p for all primes p. The prime cycles run in parallel, and independent of each other. The element 0,0,0,0… is the group identity. In the direct product, the element 1,2,3,4,5,6,7,8,9… is well defined. In the direct sum, almost all components must be 0, as in 1,2,3,0,5,0,0,0,0… zeros thereafter.

Semidirect Product

The semidirect product is a bit trickier. Let G and H be arbitrary groups as above, and let a group homomorphism map H into the automorphisms of G. In other words, every element x in H is associated with an automorphism on G. Call this automorphism f_x.

Use the "group action" convention, where the automorphism defined by xy is the automorphism of y followed by the automorphism of x. This seems backwards, but it's actually more convenient. Naturally, f₁ = 1, the trivial automorphism, and the automorphism f_1/x is the inverse automorphism of f_x.

If automorphisms have been assigned to the generators of H, then we have the entire map from H into the automorphisms of G. If H is cyclic, for instance, generated by x, we only need know f_x. Then f_x*x is f_x applied twice, and so on.

Now let's define the semidirect product. The product of two elements g₁,h₁ * g₂,h₂ has the following formula.

g₁*f_h1(g₂), h₁h₂

Is J a group? Well 1,1 is certainly the two-sided identity. Let's bring in g₃,h₃ and verify associativity. The second component, from the group H, is clearly associative; we only need look at the first component.

First component of (J₁ * J₂) * J₃ =

g₁ f_h1(g₂) f_h1h2(g₃) =

g₁ f_h1(g₂) f_h1(f_h2(g₃)) =

g₁ f_h1( g₂f_h2(g₃) ) =

First component of J₁ * (J₂ * J₃)

Does g₁,h₁ have an inverse? Let f be the automorphism associated with h₁. Apply f inverse to the inverse of g₁ and call this g₂. Sure enough, g₁f(g₂) produces the identity in G, hence g₂,h₂ is the inverse of g₁,h₁, and J is a group.

A simple homomorphism maps J onto H, with G as kernel. Thus J is the "semidirect product of G by H".

Necessary and Sufficient Conditions

Let J be a group with kernel G and factor group H. Thus J is an "extension of G by H". But is J a semidirect product?

If it is, then J contains a copy of H. Furthermore, the group H that lives inside J maps, by our group homomorphism, onto the factor group H. Seen another way, we can select a cosrep for every coset of G in J, and these cosreps form a closed subgroup of J. This subgroup is a mirror of H inside J.

Conversely, let H be a set of cosreps of G in J, such that H is a closed subgroup inside J. Of course H also represents and defines the factor group.

Let x be any element of H. since G is the kernel, xG/x equals G, as a set. That is, conjugation by x permutes the elements of G. Apply this to g₁g₂, and the result is the same as xg₁/x * xg₂/x. Therefore conjugation by x implements a group automorphism on G. Call this automorphism f_x. This holds for every x in J, so it certainly holds for every x in H.

Since left and right cosets are the same, let the members of H represent right cosets. The elements of J can now be labeled unambiguously. Each element is g_i,h_i for some g_i in G and some h_i in H.

Now consider the product g₁,h₁ * g₂,h₂. Realize that x,y is the same as x*y, and you're almost there.

g₁,h₁ * g₂,h₂ =

g₁ * h₁ * g₂ * h₂ =

g₁ * h₁ * g₂ / h₁ * h₁ * h₂ =

g₁ * f_h1(g₂) * h₁ * h₂ =

g₁ f_h1(g₂), h₁h₂

The group J, with kernel G and factor group H, is a semidirect product iff a copy of H exists inside J, and maps onto the factor group H. Such a group is called "split exact".

Consider the cyclic group of order p² for some prime p. Use the integers mod p². The multiples of p form the kernel, and the integers mod p form the factor group. We have G and H in hand, but is Z_p² a semidirect product? Let x be an integer that is 1 mod p, such that x generates a copy of H inside J. The elements x, 2x, 3x all map to 1 2 3 etc in the factor group, but px is not 0. The subgroup is not closed. This is a group that has a kernel and quotient, but is not a semidirect product.

Relatively Prime Subgroups

Let J be a finite group, with normal subgroup G, and subgroup H, such that |G| and |H| are relatively prime. Let c and d be elements of H that happen to lie in the same coset of G. Thus c/d also lies in G. This means the order of c/d divides into |G| and |H|, which are relatively prime. Thus c/d is the group identity, and c = d. The members of H represent distinct cosets of G in J. In other words, the members of H represent the factor group J/G. Multiplication in H corresponds to multiplication in J/G, and H is in fact a copy of the factor group, living in J. The group is split exact - a semidirect product.

Next suppose G is normal in J, and |G| and |J/G| are relatively prime, and J/G is known to be cyclic. Let c be an element of G that generates J/G. If |J/G| = u, c^u is back in G, and has some order v in G. Replace c with c^v. Since v and u are coprime, we are merely selecting a new generator for J/G. Also, c^u is the identity element in G, and in J. Thus c generates a copy of the factor group inside J, and J is split exact - a semidirect product.