Direct sum of groups

Properties

If G = H + K, then it can be proven that:

• for all h in H, k in K, we have that h ∗ k = k ∗ h
• for all g in G, there exists unique h in H, k in K such that g = h ∗ k
• There is a cancellation of the sum in a quotient; so that (H + K)/K is isomorphic to H

The above assertions can be generalized to the case of G = ΣH_i, where {H_i} is a finite set of subgroups:

• if i ≠ j, then for all h_i in H_i, h_j in H_j, we have that h_i ∗ h_j = h_j ∗ h_i
• for each g in G, there exists a unique set of elements h_i in H_i such that

g = h₁ ∗ h₂ ∗ ... ∗ h_i ∗ ... ∗ h_n

• There is a cancellation of the sum in a quotient; so that ((ΣH_i) + K)/K is isomorphic to ΣH_i.

Note the similarity with the direct product, where each g can be expressed uniquely as

g = (h₁,h₂, ..., h_i, ..., h_n).

Since h_i ∗ h_j = h_j ∗ h_i for all i ≠ j, it follows that multiplication of elements in a direct sum is isomorphic to multiplication of the corresponding elements in the direct product; thus for finite sets of subgroups, ΣH_i is isomorphic to the direct product ×{H_i}.

Direct summand

Given a group ${\displaystyle G}$, we say that a subgroup ${\displaystyle H}$ is a direct summand of ${\displaystyle G}$ if there exists another subgroup ${\displaystyle K}$ of ${\displaystyle G}$ such that ${\displaystyle G=H+K}$.

In abelian groups, if ${\displaystyle H}$ is a divisible subgroup of ${\displaystyle G}$, then ${\displaystyle H}$ is a direct summand of ${\displaystyle G}$.

Examples

• If we take $G=\prod _{i\in I}{H}_i$ it is clear that ${\displaystyle G}$ is the direct product of the subgroups $H_{i_0}\times \prod _{i\ne i_0}{H}_i$.

• If ${\displaystyle H}$ is a divisible subgroup of an abelian group ${\displaystyle G}$ then there exists another subgroup ${\displaystyle K}$ of ${\displaystyle G}$ such that ${\displaystyle G=K+H}$.
• If ${\displaystyle G}$ also has a vector space structure then ${\displaystyle G}$ can be written as a direct sum of ${\displaystyle \mathbb{R}}$ and another subspace ${\displaystyle K}$ that will be isomorphic to the quotient ${\displaystyle G/K}$.

Equivalence of decompositions into direct sums

In the decomposition of a finite group into a direct sum of indecomposable subgroups the embedding of the subgroups is not unique. For example, in the Klein group ${\displaystyle V_4\cong C_2\times C_2}$ we have that

${\displaystyle V_4=⟨(0,1)⟩+⟨(1,0)⟩,}$ and

${\displaystyle V_4=⟨(1,1)⟩+⟨(1,0)⟩.}$

However, the Remak-Krull-Schmidt theorem states that given a finite group G = ΣA_i = ΣB_j, where each A_i and each B_j is non-trivial and indecomposable, the two sums have equal terms up to reordering and isomorphism.

The Remak-Krull-Schmidt theorem fails for infinite groups; so in the case of infinite G = H + K = L + M, even when all subgroups are non-trivial and indecomposable, we cannot conclude that H is isomorphic to either L or M.

Generalization to sums over infinite sets

To describe the above properties in the case where G is the direct sum of an infinite (perhaps uncountable) set of subgroups, more care is needed.

If g is an element of the cartesian product Π{H_i} of a set of groups, let g_i be the ith element of g in the product. The external direct sum of a set of groups {H_i} (written as Σ_E{H_i}) is the subset of Π{H_i}, where, for each element g of Σ_E{H_i}, g_i is the identity ${\displaystyle e_{H_i}}$ for all but a finite number of g_i (equivalently, only a finite number of g_i are not the identity). The group operation in the external direct sum is pointwise multiplication, as in the usual direct product.

This subset does indeed form a group, and for a finite set of groups {H_i} the external direct sum is equal to the direct product.

If G = ΣH_i, then G is isomorphic to Σ_E{H_i}. Thus, in a sense, the direct sum is an "internal" external direct sum. For each element g in G, there is a unique finite set S and a unique set {h_i ∈ H_i : i ∈ S} such that g = Π {h_i : i in S}.