Varieties of Representations of Finitely Generated Groups by Alexander Lubotzky

By Alexander Lubotzky

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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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