** Question (65c):** Why does N have such a description?

** Answer **:
Recall that elements of S_p are bijections of any set S consisting of p elements.
Consider the set S of congruence classes modulo p. Then the map x -> ax+b is a bijection from S to itself, as long as "a" is non-zero modulo p. The composition of any two linear maps is linear, and so the set N of such maps forms a subgroup of S_p of order (p-1)p (there are (p-1) choices for a, and p choices for b). The subgroup P generated by the map h: x -> x+1 has order p; the element h corresponds to the element (1,2,3,...,p). To see that P is normal in N, let g be the map given by x ->ax+b. Then g^-1 is the map given by x->(x-b)/a.
We compute that:

ghg^-1(x) = g(h(g^(-1)(x))) = g(h((x-b)/a)) = g((x-b)/a + 1) = a((x-b)/a + 1) + b = x + a = h^a.

** Question (66):**

** Answer **: Suppose that P is a p-Sylow subgroup of N. Then P is also a subgroup of G. Since the largest power of p dividing N is the largest power of p dividing G, it follows that P is a Sylow subgroup of G. In particular, any p-Sylow of N is a p-Sylow of G. It suffices to prove the converse. By the Sylow theorems, any p-Sylow P' of G is conjugate to P, and hence is of the form gPg^-1. Since P is a subgroup of N, this is contained in gNg^-1. Since N is normal, this is just N. Hence P' is contained in N.

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