# Direct sum of groups

In mathematics, a group G is called the direct sum 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 $e$ of G in common),
• 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 {Hi} if

• each Hi is a normal subgroup of G,
• each Hi has trivial intersection with the subgroup <{Hj : ji}>,
• G = <{Hi}>; in other words, G is generated by the subgroups {Hi}.

If G is the direct sum of subgroups H and K then we write G = H + K, and if G is the direct sum of a set of subgroups {Hi} then we often write G = ∑Hi. 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 direct sum is commutative up to isomorphism. That is, if G = H + K then also G = K + H and thus 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, and if a group cannot be expressed as such a direct sum then 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 = ∑Hi, where {Hi} is a finite set of subgroups:

• if ij, then for all hi in Hi, hj in Hj, we have that hi*hj = hj*hi
• for each g in G, there exists a unique set of elements hi in Hi such that
g = h1*h2* ... * hi * ... * hn
• There is a cancellation of the sum in a quotient; so that ((∑Hi) + K)/K is isomorphic to ∑Hi

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

g = (h1,h2, ..., hi, ..., hn).

Since hi*hj = hj*hi for all ij, 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, ∑Hi is isomorphic to the direct product ×{Hi}.

## Direct summand

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

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 exists another subgroup $K$ of $G$ such that $G=K+H$ .
• If $G$ also has a vector space structure then $G$ can be written as a direct sum of $\mathbb {R}$ and another subspace $K$ 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 $V_{4}\cong C_{2}\times C_{2}$ we have that

$V_{4}=\langle (0,1)\rangle +\langle (1,0)\rangle ,$ and
$V_{4}=\langle (1,1)\rangle +\langle (1,0)\rangle .$ However, the Remak-Krull-Schmidt theorem states that given a finite group G = ∑Ai = ∑Bj, where each Ai and each Bj 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 ∏{Hi} of a set of groups, let gi be the ith element of g in the product. The external direct sum of a set of groups {Hi} (written as ∑E{Hi}) is the subset of ∏{Hi}, where, for each element g of ∑E{Hi}, gi is the identity $e_{H_{i}}$ for all but a finite number of gi (equivalently, only a finite number of gi 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 {Hi} the external direct sum is equal to the direct product.

If G = ∑Hi, then G is isomorphic to ∑E{Hi}. 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 {hiHi : iS} such that g = ∏ {hi : i in S}.