Algebra: Monomials and Polynomials by John Perry

By John Perry

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In fact, (N, +) = there does exist an isomorphism f between the two monoids. What would have to be true about f? We know that f preserves the identity; that is, f (0) = 1. After all, 0 is the identity of (N, +), while 1 is the identity of (N, ×). We also know that f preserves the operation, so for any x, y ∈ N, we would have to have f ( x + y ) = f ( x ) f ( y ). Let’s see if that’s actually possible. 17 The definition uses the variables x and y, but those are just letters that stand for arbitrary elements of M .

However, while N is a monoid under addition, it is not a group. Why not? The problem is with inverses. We know that every natural number has an additive inverse; after all, 2 + (−2) = 0. 1. Groups 25 Nevertheless, the inverse property is not satisfied because −2 ∈ N! It’s not enough to have an inverse in some set; the inverse be in the same set! For this reason, N is not a group. 4. Let n ∈ N+ . The set of invertible n × n matrices is a multiplicative group. We leave much of the proof to the exercises, but this fact is a consequence of properties you learn in linear algebra.

G −2 , g −1 , e, g , g 2 , . . We know that every cyclic group has the form 〈g 〉 for some g ∈ G. Is the converse also true that 〈g 〉 is a group for any g ∈ G? As a matter of fact, yes! 53. For every group G and for every g ∈ G, 〈g 〉 is an abelian group. 53, we need to make sure we can perform the usual arithmetic on exponents. 3. 54. Let G be a group, g ∈ G, and m, n ∈ Z. Each of the following holds: (A) g m g −m = e; that is, g −m = ( g m )−1 . (B) ( g m )n = g mn . (C) g m g n = g m +n . The proof will justify this argument by applying the notation described at the beginning of this chapter.

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