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Extra resources for Abstract Algebra [Lecture notes]

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R} such that w = vk . Let k be the smallest such integer. By repeated use of the end of the previous paragraph, v1 , . . , vr gives the same configuration as w = vk , v1 , . . , vk−1 , vk+1 , . . , vr . This last sequence starts with the same toppling as w1 , . . , ws , and then we only have to compare the shorter sequences of topplings imposed on the configuration obtained after applying the toppling from the vertex w. This proves that toppling the grains of sand on the graph in the prescribed manner is a well-defined operation.

Now verify that XR is a prime ideal. 7 Let R be a commutative ring with unity and let I be an ideal of R. Then I is a prime ideal if and only if R/I is a domain. Proof. Do it. We already know that Z/5Z is a domain, even a field, so by the proposition 5Z is a prime ideal. Furthermore, if we can prove that Z[X]/(X + 2, X 2 + 1) ∼ = Z/5Z, then we’ll have established that (X + 2, X 2 + 1) is a prime ideal in Z[X]. Note: every element f in Z[X] can be written as f = q(X + 2) + r for some q ∈ Z[X] and some r ∈ Z (by the division algorithm for polynomials), so every element in Z[X]/(X + 2, X 2 + 1) can be represented by an integer r.

K, gjl = e. Thus |(g1 , g2 , . . , gk )| is a multiple of |gi | for all i, whence it is a multiple of lcm{|g1 |, . . , |gk |}. But it couldn’t be anything smaller, for otherwise some gil wouldn’t be e. 5 Let G and H be finite groups such that G ⊕ H is cyclic. Then G and H are cyclic and |G| and |H| are relatively prime. Proof. Suppose that G ⊕ H is generated by (a, b). Then G is generated by a, so G is cyclic. Similarly H = b is cyclic. By the previous theorem, |a| · |b| = |G| · |H| = |G ⊕ H| = |(a, b)| = lcm{|a|, |b|}, so that the orders of a and b are relatively prime.

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