Direct Products

The Cartesian product, with operations defined coordinatewise, allowed us to construct new rings from known ones, as direct sum. The same is true for groups. Direct products are an easy way to construct larger groups from smaller ones. This construction yields all finite abelian groups.

Definition 7.17.1 (External Direct product of two groups) Let G (with operation ⋅) and H (with operation *) be groups. Define an operation ∘ on G X H by

Then G X H is a group. If G and H are abelian, then so is G X H. If G and H are finite, then so is G x H and |G x H| = |G| |H|. For G and H abelian, with groups operations written as addition, often the direct product is written instead as a (direct) sum.

The notion of direct product can be extended to more than two groups.

Definition 7.17.2 (Generalized External direct product). If G₁,G₂,...,G_n are groups. The (direct) product is defined as the set

along with the coordinatewise operation, ∘, on the Cartesian product G₁ x G₂ x ... x G_n defined as

It is easy to verify that G₁ x G₂ x ... x G_n, is a group under this operation: If e_i is the identity element of G_i, then (e₁, e₂, e_n) is the identity element of G₁ x G₂ x ... x G_n, and (a⁻¹, a₂⁻¹, ... ,a_n⁻¹) is the inverse of (a₁, a₂, ..., a_n). This group is called the external direct product of G₁,G₂,...,G_n. When each G_i is an additive abelian group, the external direct product of G_i, is sometimes called the direct sum and denoted G₁ ⊕ G₂ ⊕ ... ⊕ G_n

More in depth, if for the sake of simplicity we consider only G₁ x G₂, we can verify the properties of

1. Closure: We have (g₁,g₂)(g₁',g₂') = (g₁g₁', g₂g₂') ∈ G₁ x G₂, because g₁g₁' ∈ G₁, with g₁, g₁' ∈ G₁ because G₁ is a group. Similarly g₂g₂' ∈ G₂.

2. Associativity: If (g₁,g₂), (g₁', g₂'), (g₁'', g₂'') ∈ G₁ x G₂ then

   [(g₁,g₂) (g₁', g₂')] (g₁'', g₂'') =
   (g₁g₁', g₂g₂') (g₁'', g₂'') = ([g₁ g₁']g₁'', [g₂g₂'] g₂'')=
   (g₁ [g₁'g₁''], g₂[g₂' g₂''])=
   (g₁,g₂) (g₁'g₂'', g₂'g₂'') =
   (g₁,g₂) [(g₁', g₂') (g₁'', g₂'')]

3. If (g₁,g₂) ∈ G₁ x G₂, then (e_G1,e_G2) (g₁,g₂) = (e_G1g₁,e_G2g₂) = (g₁,g₂). Therefore (e_G1,e_G2) is the left identity of G₁ x G₂;

4. Existence of left inverse (g₁, g₂)⁻¹ is (g₁⁻¹, g₂⁻¹).

Notice that neither G₁ nor G₂ is a subgroup of G₁ x G₂, but G₁ x G₂ does contain isomorphic replicas of each, namely,

Since

every element of G is the product of an element of G₁* and one of G₂*. Moreover two such elements commute.

Examples 7.17.3

1. Let G₁ = ℤ₂ and G₂ = ℤ₃. Then

   ℤ₂ x ℤ₃ = {(0̅, 0̅), (0̅, 1̅), (0̅, 2̅), (1̅, 0̅), (1̅, 1̅), (1̅, 2̅)}

   the elements order of ℤ₂ x ℤ₃ are respectively 1, 3, 3, 2, 6, 6. Which is a cyclic group of order 6, thus

   ℤ₂ x ℤ₃ ≃ ℤ₆

2. Let G₁ = ℤ₂ and G₂ = ℤ₄. The external product ℤ₂ x ℤ₄ is made by 8 elements, but this time there is no element with order 8, hence we are not dealing with the cyclic i.e. ℤ₂ x ℤ₄ ≠ ℤ₈. ■

Definition 7.17.4 (Internal direct product of subgroups). Let G be a group and H, K be normal subgroups of G such that G = HK and H ∩ K = {1}. Then G is called the internal direct product of H and K. If G is an additive abelian group, then it is called the internal direct sum of H and K and denoted by G = H ⊕ K.

Theorem 7.17.5 Let H and K two groups, and let G = H x K, their external direct product. Then G has two subgroups H̄, K̄, such that

1. H̄ ≃ H, K̄ ≃ K;

2. H̄ ⊲̲ G, K̄ ⊲̲ G;

3. G = H̄K̄;

4. H̄ ∩ K̄ = {(e_H, e_K)} = trivial subgroup of H x K

Proof. Let

Consider the mapping φ: H x K ⟶ K by (h, k) ⟼ k. It is not hard to show that this map is a homomorphism from G onto K. Obviously the map is onto K. Then

φ is a homomorphim of H x K onto K. The kernel of φ is {(h,k): (h,k) ⟼ e_K} = {(h, e_K): h ∈ H} = H₁. This establishes that H̄ is a normal subgroup of G and that G/H̄ ≃ K by the group isomorphism theorem. In an identical fashion using the map (h,k) ⟼ h, it can be shown that K̄ is a normal subgroup of G and G/K̄ ≃ H.

Since H̄ and K̄ are normal subgroups of G = H x K, then it follows from Lemma 7.12.4 that H̄K̄ ⊂ G is a subgroup. Let (h, k) be any element of G = H x K. Then (h, k) = (h, e_K) (e_H, k) ∈ H̄K̄. This shows G ⊂ H̄K̄. Hence G = H̄K̄.

Finally, consider H̄ ∩ K̄. If (x, y) ∈ H̄ ∩ K̄ then (x ,y) ∈ H̄ and (x, y) ∈ K̄. But if (x, y) ∈ H̄ then y = e_K and if (x, y) ∈ K̄ then x = e_H. This implies that (x,y) = (e_H, e_K). Then all conditions listed above are satisfied.  □

If a group has two subgroups H̄ and K̄ satisfying the conditions b), c), d) of the previous theorem, then G is said internal direct product of its subgroups H̄ and K̄. We just proved that a a group which is an external direct product of two groups is isomorphic to the internal product.

The previous theorem says that inside any direct product of two groups there are two normal subgroups whose product is the whole group and whose intersection is trivial. We now see that the converse of this is true as well providing a method for recognizing that a group G is isomorphic to the direct product of two groups.

Theorem 7.17.6 Let G a group and let H and K two subgroups of G, such that

1. H and K are normal subgroups of G;

2. G = HK;

3. H ∩ K = {e};

Then G ≃ H x K.

Before proving the theorem we introduce two lemmas.

Lemma 7.17.7 Let G a group and let H and K two subgroups of G, such that

Then, for every h ∈ H and every k ∈ K it results hk = kh.

Proof. Let x = khk⁻¹. From x = (hkh⁻¹)k⁻¹, it is clear that since hkh⁻¹ ∈ K for the normality of K in G then x ∈ K. From x = h(kh⁻¹k⁻¹), for the normality of H in G, x belongs to H. From H ∩ K{e}, it follows that x = hkh⁻¹k⁻¹ = e, thus hk = kh. □

Lemma 7.17.8 Let G a group and let H and K two subgroups of G, such that

1. G = HK;

2. H ∩ K = {e};

Then every g ∈ G is written uniquely as product of an element h ∈ H by an element k ∈ K.

Proof. By condition a) every g ∈ G can be written as hk for some h ∈ H by an element k ∈ K. To prove the uniqueness of the writing, suppose we have

Then h⁻¹hkk'⁻¹ = h⁻¹h'k'k'⁻¹, from which kk'⁻¹ = h⁻¹h' ∈ H ∩ K = {e}. Thus h = h' and k = k' is a unique writing.  □

We now prove the theorem.

Proof of Theorem 7.17.6. We define the following map

The application is surjective for condition b) of the statement. It is injective by lemma 7.17.8 and it is also a homomorphism by lemma 7.17.7. We thus proved that G is isomorphic to the direct product H by K.  □

The theorems just proved show us that the internal direct product HK is isomorphic to the direct product H x K; We shall use the symbols HK and H x K as well as the terms "internal direct product" and "direct product" interchangeably.

The last theorem says that to check whether a group is isomorphic to a direct product, it is enough to check the conditions of the theorem. We give some examples.

Example 7.17.9 The group (ℤ, +) is not a direct product of two non-trivial groups since the intersection of its subgroups is always non-trivial. ■

Example 7.17.10 The symmetric group S₃ is not a direct product, since it possesses a single normal subgroup. ■

Example 7.17.11 ℤ₁₂ is isomorphic to the direct product ℤ₃ x ℤ₄, since the two subgroups

sono normal with intersection reduced to the class 0̅, and are such that ℤ₁₂ = H + K. Moreover H ≃ ℤ₃ and K ≃ ℤ₄.  ■

Example 7.17.12 The dihedral group D₄ is not a direct product, since two whatever non-trivial normal subgroups of D₄ intersect to give a non-trivial subgroup, containing the element r². ■

Example 7.17.13 The orthogonal group O₃(R) is isomorphic to the direct product of SO₃ (orthogonal matrices with determinant 1) by ℤ₂. More generally

Example 7.17.14 (ℂ, +) is isomorphic to (ℝ, +) x (ℝ, +).  ■

The following properties hold:

Proposition 7.17.15. Let G₁ x G₂ x ... x G_k. Then

1. if every factor has finite order,

   |G₁ x G₂ x ... x G_k| = |G₁| |G₂| ⋅⋅⋅ |G_k|;

2. G₁ x G₂ x ... x G_k is abelian if and only if for every G_i is abelian;

3. |(g₁, g₂, ..., g_k)| = lcm (|g₁|,|g₂|, ..., |g_k|), the order is infinite if one of the g_i has inifnite order;

4. if every G_i has finite order, G₁ x G₂ x ... x G_k is cyclic if and only if every G_i is cyclic and (|G_i, G_j|) = 1, for every i ≠ j.

Proof. Points (a), and (b) are obvious. (c) let m = lcm (|g₁|,|g₂|, ..., |g_k|) and let t the order of (g₁, g₂, ..., g_k). From

it results that t is a multiple of every |g_i|, hence m| t. On the other hand from

we have t | m. Thus t = m.

(d) (⇒) That G_i are cyclic is obvious, because isomorphic to a subgroup of a cyclic group. Let g_i a generator of G_i, the element g = (g₁, g₂, ..., g_k) has greatest order among all the elements of G₁ x G₂ x ... x G_k and by point (c) its order is lcm(g₁, g₂, ..., g_k). If it were (|G_i|, |G_j|) = d ≠ 1 for some i,j, i ≠ j then the order of (g₁, g₂, ..., g_k) would be

and it couldn't be a generator of of G₁ x G₂ x ... x G_k.

To understnad the last passage: Suppose gcd(m,n) = d, then g^mn/d = (g^m)^n/d = e.

(⇐) If every G_i is cyclic and (|G_i|, |G_j|) for every i, j, i ≠ j, then letting g_i a generator of G_i for every i = 1, ..., k, the element (g₁, g₂, ..., g_k) has order |g₁| |g₂| ⋅⋅⋅ |g_k|, hence G₁ x G₂ x ... x G_k is cyclic. □

Corollary 7.17.16. ℤ_r x ℤ_s is isomrphic to ℤ_rs (that is cyclic) if and only if (r,s) = 1.

Theorem 7.17.6 can be generalized to the product of more than two groups as follows

Theorem 7.17.17. Let G a group and let N₁, N₂, ...., N_k subgroups of G such that

1. N_i ⊲̲ G;

2. G = N₁N₂⋅⋅⋅N_k;

3. N_i ∩ (N₁ N₂ ⋅⋅⋅ N_{i −1}N̂_iN_i+1 ⋅⋅⋅ N_k) = {e} ∀i = 1,...,k.

Then G is isomorphic to the direct product N₁ x N₂ x ⋅⋅⋅ N_k

Proof. We define the map

The map f is surjective because by hypothesis (b), each element of G can be written in the form a₁ a₂···ak for some a_i ∈ N_i, i = 1,2, ... ,k. The map f is injective because when f(a₁, a₂, ..., a_k) = f(b₁, b₂, ..., b_k), we obtain that a₁a₂···a_k = b₁b₂ ··· b_k, so that by the uniqueness of writing each element of G as a product, we have a₁ = b₁, a₂ = b₂, ... , a_k = b_k.

By Lemma 7.17.7, we then have that a_ib_j = b_ja_i for all a_i ∈ N_i and b_j ∈ N_j when i ≠j. We use this fact to show that f is a homomorphism

Therefore f: N₁ × N₂ ×··· × N_k → G is an isomorpism.  □

We've seen that if a group is a direct product then there exist in G two subgroups H K such that G = HK and H ∩ K = {e}. And to prove this we used the fact kh' = h'k of Lemma 7.17.6

We suppose now that the group possesses a normal subgroup N and another subgroup (not necessarily normal) H such that

in this case (7.17.1) does not hold anymore i.e. in general nh ⋅ n'h' ≠ nn' ⋅ hh'. But it is always true that

where hn'h⁻¹, being conjugates of an element of N, which is normal. We have now that the product of two elements nh and n'h' of NH does not equal (as in the case of both subgroups normal) the product of nn' by hh', but the product of n(hn'h⁻¹) by hh'; The difference is that in place of n' there must be the conjugate of n' through the element h. Thus for every h ∈ H the following map is defined

an automorphism of N. Note that in the case of H normal as well, this automorphism results the identical automorpshim: indeed we have hn' = n'h, from which hn'h⁻¹ = n'.

We can define the following map

which is a homomorpshim. Equation 7.17.2, can be written in terms of the last homomorphism as

Note that, in case of H normal in G, Φ(h)(n') = n' we find 7.17.1.

Let now H and K two arbitrary groups and let Φ a homomorpshim from K to Aut(H). In the cartesian product H x K we can define the following operation

H x K with this operation is a group, and the subsets

are two normal subgroups in H x K of which H₁ is isomorphic to H and K₁ to K.

Definition 7.17.18. Let H and K two groups and Φ a homomorphism from K to Aut(H). The cartesian product H x K along with the operation 7.17.3 is named semidirect product of H x K through the homomorphism Φ, indicated by

The analogous theorem for direct product can be given for semidirect products:

Theorem 7.17.19 Let N and H two subgroups of a group G. If N is normal in G, and G = NH and N ∩ H = {e}, then G is isomorphic to the semidirect product N ∐_Φ H, where Φ: H ⟶ Aut(N) is the automorphism defined by

Proof. The map

is a surjective homomoerphism, whose kernel is the neutral element, hence a isomorphism.   □

Examples 7.17.20 Let G = S₃, N = ⟨(1,2,3)⟩, and H = ⟨(1,2)⟩. Then it results

where Φ((1,2)(1,2,3)) = (1,2)(1,2,3)(1,2)⁻¹ = (1,3,2). The map between N ×_Φ H and S₃ is the following

so with a direct product, (a,x)(b,y) = (ab, xy) with a semidirect product, (a,x)(b,y) = (a φ_x(b), xy).