By Sandor Szabo, Arthur D. Sands

Decomposing an abelian workforce right into a direct sum of its subsets ends up in effects that may be utilized to numerous components, comparable to quantity conception, geometry of tilings, coding idea, cryptography, graph conception, and Fourier research. Focusing as a rule on cyclic teams, **Factoring teams into Subsets explores the factorization concept of abelian teams. **

The ebook first indicates easy methods to build new factorizations from previous ones. The authors then talk about nonperiodic and periodic factorizations, quasiperiodicity, and the factoring of periodic subsets. additionally they learn how tiling performs an immense position in quantity thought. the following a number of chapters hide factorizations of limitless abelian teams; combinatorics, comparable to Ramsey numbers, Latin squares, and intricate Hadamard matrices; and connections with codes, together with variable size codes, blunders correcting codes, and integer codes. the ultimate bankruptcy bargains with numerous classical difficulties of Fuchs.

Encompassing the various major components of the factorization idea, this publication explores difficulties within which the underlying factored staff is cyclic.

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**Additional info for Factoring groups into subsets**

**Example text**

Then G = H + A + B. Now A ⊆ K and the sum K + B is direct. 1 (on page 38), it follows that A + B is non-periodic, as required. We are able to extend a bad factorization to a larger group from a subgroup. 3 If a proper subgroup H of a group G is k-bad, then G is both k-bad and (k + 1)-bad. PROOF There exists a factorization H = A1 +· · ·+Ak , where each subset Ai is non-periodic. 1 (on page 38), there exists a non-periodic subset C such that G = H + C. Then G = A1 + · · · + Ak + C and so G is © 2009 by Taylor & Francis Group, LLC Non-periodic factorizations 41 (k + 1)-bad.

Then the sum (i) A1 + B is a (|B| − 1)-fold, (ii) A + B1 is a (|A| − 1)-fold, (iii) A1 + B1 is a (|A| − 1)(|B| − 1)-fold © 2009 by Taylor & Francis Group, LLC New factorizations from old ones 25 factorization of G. PROOF (i) Note that A1 + B = (G\ A)+ B = (G+ B)\ (A+ B). The sum G + B is a |B|-fold factorization of G. The sum A + B is a 1-fold factorization of G. Therefore the sum A1 + B is a (|B| − 1)-fold factorization of G. The (ii) and (iii) cases can be settled in a similar way. We extend the concept of factorization to the case of more than two factors.

Let G be a finite abelian group and let A, B be subsets of G. If each element g ∈ G can be represented in exactly k ways in the form g = a + b, a ∈ A, b ∈ B, then we say that the sum A + B is a k-fold factorization of G. In other words, for each g ∈ G, there are k distinct pairs (a1 , b1 ), . . , (ak , bk ), ai ∈ A, bi ∈ B such that g = a1 + b1 = · · · = ak + bk . Further, g = a + b, a ∈ A, implies that (a, b) ∈ {(a1 , b1 ), . . , (ak , bk )}. 14 Let f : G → H be a homomorphism from G onto H such that k = |Kerf | is finite.