If G is the direct sum of subgroups H and K, then we write G = H + K; if G is the direct sum of a set of subgroups {H_{i}}, we often write G = ∑H_{i}. Loosely speaking, a direct sum is isomorphic to a direct product of subgroups.
In abstract algebra, this method of construction can be generalized to direct sums of vector spaces, modules, and other structures; see the article direct sum for more information.
This notation is commutative; so that in the case of the direct sum of two subgroups, G = H + K = K + H. It is also associative in the sense that if G = H + K, and K = L + M, then G = H + (L + M) = H + L + M.
A group which can be expressed as a direct sum of nontrivial subgroups is called decomposable; otherwise it is called indecomposable.
If G = H + K, then it can be proven that:
The above assertions can be generalised to the case of G = ∑H_{i}, where {H_{i}} is a finite set of subgroups.
Note the similarity with the direct product, where each g can be expressed uniquely as
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, ∑H_{i} is isomorphic to the direct product ×{H_{i}}.
The direct sum is not unique for a group; for example, in the Klein group, V_{4} = C_{2} × C_{2}, we have that
However, it is the content of the RemakKrullSchmidt theorem that given a finite group G = ∑A_{i} = ∑B_{j}, where each A_{i} and each B_{j} is nontrivial and indecomposable, then the two sums are equivalent up to reordering and isomorphism of the subgroups involved.
The RemakKrullSchmidt theorem fails for infinite groups; so in the case of infinite G = H + K = L + M, even when all subgroups are nontrivial and indecomposable, we cannot then assume that H is isomorphic to either L or M.
If we wish to describe the above properties in the case where G is the direct sum of an infinite (perhaps uncountable) set of subgroups, we need to be a bit more careful.
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 ∑_{E}{H_{i}}) is the subset of π{H_{i}}, where, for each element g of ∑_{E}{H_{i}}), g_{i} is the identity e_{Hi} 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.
It should be readily apparent that this subset does indeed form a group; and for a finite set of groups H_{i}, the external direct sum is identical to the direct product.
Then 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. We have that, for each element g in G, there is a unique finite set S and unique {h_{i} in H_{i} : i in S} such that
g = π {h_{i} : i in S}
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