Direct sum of groups

In mathematics, a group G is called the direct sum ^[1][2] of two subgroups H₁ and H₂ if

• each H₁ and H₂ are normal subgroups of G
• the subgroups H₁ and H₂ have trivial intersection (i.e., having only the identity element $e$ in common), and
• G = <H₁, H₂>; in other words, G is generated by the subgroups H₁ and H₂.

More generally, G is called the direct sum of a finite set of subgroups {H_i} if

• each H_i is a normal subgroup of G
• each H_i has trivial intersection with the subgroup <{H_j : j not equal to i}>, and
• G = <{H_i}>; in other words, G is generated by the subgroups {H_i}.

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 weak 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 of modules 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 non-trivial subgroups is called decomposable; otherwise it is called indecomposable.

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 = ∑H_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 unique set of {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 ((∑H_i) + K)/K is isomorphic to ∑H_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, ∑H_i is isomorphic to the direct product ×{H_i}.

Direct summand

Given a group $G$, we say that a subgroup $H$ is a direct summand of $G$ (or that splits form $G$) if and only if there exist another subgroup $K\leq G$ such that $G$ is the direct sum of the subgroups $H$ and $K$

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

Examples

• If we take

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

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

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, V₄ = C₂ × C₂, we have that

V₄ = <(0,1)> + <(1,0)> and

V₄ = <(1,1)> + <(1,0)>.

However, it is the content of the Remak-Krull-Schmidt theorem 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 then assume 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 ∏{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 $e_{{H_{i}}}$ 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 identical 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 unique {h_i in H_i : i in S} such that g = ∏ {h_i : i in S}.

See also

• direct sum
• coproduct
• free product
• Direct sum of topological groups

References

1. Homology. Saunders MacLane. Springer, Berlin; Academic Press, New York, 1963.
2. László Fuchs. Infinite Abelian Groups