Direct sum of groups

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 ${\displaystyle G}$.

• If we take ${\displaystyle G=\prod _{i\in I}H_{i}}$ it is clear that ${\displaystyle G}$ is the direct product of the subgroups ${\displaystyle H_{i_{0}}\times \prod _{i\not =i_{0}}H_{i}}$.

• 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 ${\displaystyle G/K}$.

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

${\displaystyle V_{4}=\langle (0,1)\rangle +\langle (1,0)\rangle ,}$ and

${\displaystyle V_{4}=\langle (1,1)\rangle +\langle (1,0)\rangle .}$

However, the Remak-Krull-Schmidt theorem states that given a finite group G = ∑A_i = ∑B_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.

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 ∏{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 ∑_E{H_i}) is the subset of ∏{H_i}, where, for each element g of ∑_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 = ∑H_i, then G is isomorphic to ∑_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 = ∏ {h_i : i in S}.

• Direct sum
• Coproduct
• Free product
• Direct sum of topological groups