Computation with finitely presented groups by Charles C. Sims

By Charles C. Sims

The booklet describes tools for operating with components, subgroups, and quotient teams of a finitely provided staff. the writer emphasizes the relationship with primary algorithms from theoretical computing device technology, quite the idea of automata and formal languages, from computational quantity thought, and from computational commutative algebra. The LLL lattice aid set of rules and numerous algorithms for Hermite and Smith basic types are used to review the Abelian quotients of a finitely provided crew. The paintings of Baumslag, Cannonito, and Miller on computing non-Abelian polycyclic quotients is defined as a generalization of Buchberger's Gröbner foundation tips on how to correct beliefs within the necessary team ring of a polycyclic team.

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A word U in X'* is freely reduced if U contains no subword of the form x"x-a. If X is finite, then JXJ is called the rank of F. 4 Presentations 23 generated by a set Y with JYJ # IXI. However, this cannot happen. 3. Let U be a word in X'*. Thus U = xi' ... xm , where the xi are in X and the exponents ai are ±1. * xi °' . Then the inverse of [U] in F is [U-1]. Of course, U-1 is not an inverse for U in X}* unless m = 0. Clearly (U-1)-1 = U. Suppose S is a subset of X}* x X±*. By Grp(X I S) we shall mean G = Mon(X} I R US), where R = FGRe1(X).

Proof. We shall prove the statement about well-ordering, leaving the rest as an exercise. Suppose -< is a well-ordering on X. Let U1 >- U2 >- U3 >be a strictly decreasing sequence of words in X*. Since i UZ I > J UZ+11 for all i, from some point on all of the UZ have the same length m. 2, the lexicographic ordering of Xm is a well-ordering. Therefore the sequence must terminate. There is also a right-to-left version of the length-plus-lexicographic ordering. An ordering -< of X * is translation invariant if U -< V implies that AUB AVB for all A and B in X*.

Assume that there is an integer m such that the minimal generating set of each of the Ij has at most m elements. 7 =I,, Ion >n. Proof. Let I=UI;. It is easy to see that I is an ideal. Let U be the minimal generating set for I, and suppose that it is possible to find distinct elements U1,... , U,,,+i in U. There is an index j such that Ii contains all of the U. Let V be the minimal generating set for Ij. By assumption, not all of the Uz are in V, and so some Ui contains an element V of V as a proper subword.

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