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 $e$ 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 {H_i} if

each H_i is a normal subgroup of G,
each H_i has trivial intersection with the subgroup <{H_j : j ≠ i}>,
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, and if G is the direct sum of a set of subgroups {H_i} then 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 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 = ∑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 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 ((∑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[edit]

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[edit]

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[edit]

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 = ∑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.

Generalization to sums over infinite sets[edit]

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 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}.

References[edit]

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

[1] Homology. Saunders MacLane. Springer, Berlin; Academic Press, New York, 1963.

[2] László Fuchs. Infinite Abelian Groups

[1]

[2]

ANHDUC - WIKI 2

Monday, August 31, 2020