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By Irena Swanson

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Extra info for Abstract Algebra [Lecture notes]

Example text

This is the class equation of G. 6 Let p be a prime integer and n a positive integer. All groups of order pn have a non-trivial center. Proof. Let G be a group of order pn . 4 (Lagrange’s Theorem), for all x ∈ G, |CG (x)| and |G|/|CG (x)| are powers of p (possibly a zeroth power of p). If for some is not a multiple of p, it has to be 1, whence CG (x) = G, which proves that x, |C|G| G (x)| x ∈ Z(G). Then by the class equation, |Z(G)| = |G| − 20 |G| i |CG (xi )| is a multiple of p. 1 Let G1 , . .

If we can prove that g ∈ (Y − X 2 , Z − X 3 ), then we will have proved that f ∈ (Y − X 2 , Z − X 3 ). Thus by induction it suffices to prove that if f ∈ (ker ψ) ∩ Q[X], then f = 0. But this is easy! Say why! 7 Let X, Y, Z, T be variables over Q. 2, there exists a unique ring homomorphism ψ : Q[X, Y, Z] → Q[T ] such that ψ|Q is identity, ψ(X) = T 3 , ψ(Y ) = T 4 , ψ(Z) = T 5 . Find the kernel of ψ. This example is harder than the previous example! We will prove that ker ψ = (X 3 − Y Z, Y 2 − XZ, Z 2 − X 2 Y ).

For a vertex v, the degree |v| denotes the number of edges adjacent to v. Each vertex v can hold exactly |v| − 1 grains on it. If there are more grains on it, then the grains start spilling off to its neighbors and to the sink, meaning that an equal number of grains spills along each of the incident edges to its neighbors until fewer than |v| grains of sand lie on v. Those grains that move to the sink stay there, but those that topple to the visible vertices may cause overloads on its neighbors, causing spills from there, etc.

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