By Gilman R.H., Hermiller S., Holt D.F.

We end up finitely generated crew G is almost unfastened if and provided that there exists a producing set for G and к > zero such that each one k-locally geodesic phrases with appreciate to that producing set are geodesic.

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**Example text**

3 One-dimensional representations Let G be a finite group. The commutator subgroup, denoted by [G, G], is the group generated by elements of the form xyx- 1 y- 1 for any x, y E G. It is easy to see that [G, G] is a normal subgroup. The quotient G /[G, G] is called the abelization of G. 1. The number of one-dimensional irreps of G is o( G /[G, G]) = o(G)jo[G,G] and, in particular, it divides the order of G. Let A= Gj[G, G]. The idea of the proof will be to show that {U E G I dim(U) = 1} is in one-one correspondence with A.

Let P2 be the projection P 2(rp1, 11'2) = (0, rp 2). Then P 2i£ n Y is an invariant subspace for U, so for V, so either {0} or 7-l. If it was all of 7-l, then P 2i£ C Y and thus CP2i£ = (1- P 2)i£ c Y, soY is il violating nontriviality. Thus, P 2i£ n Y = {0}. Similarly, define P 1 and note that P 1Y is invariant for V and so it must be all of 7-l. Thus by this pair of remarks, for any rp E 7-l, there is a unique W rp E 1{ so (rp, W rp) E Y; that is, W is linear and y = {(rp, Wrp) I rp E 7-l}. 52 III.

D Since x = y'2 obeys x 2 - 2 = 0, it is an algebraic integer but it is clearly not an integer. So it cannot be rational. Pythagoras' argument was just the proof above in the special case. More to the point, we will show that o( G)/ da E A. 2 will imply that it is an integer. To show o(G)/da is an algebraic integer, we'll need a rapid minicourse in elementary algebraic number theory, which we make as a sequence of propositions. Definition. Hom(Zn) will denote the set of matrices with coefficients in Z.