Some Solutions for the Final practice problems Math 222 G. McNinch 1. (a). Since p - 1 is divisible by 2 (p is ODD), the order of the element sigma^2 is (p-1)/2, and that is thus the order of the subgroup H. So there are 2 cosets of H in G by Lagrange's theorem. (b). If p = 1 (mod 3), then sigma^3 has order (p-1)/3 and that is the order of H, so there are 3 cosets of H in G in this case. If p = 2 (mod 3), sigma^3 has order (p-1); in other words, sigma generates the whole group G. Thus H=G and there is only one coset of H in G in this case. If p = 3, sigma^3 is equal to sigma by the little Fermat theorem. Thus H = G again in this case. (c). There are 2 cosets of H in G in this case. (d). 10 (e). 10!/20 = 9!/2 2. (a). The order of the unit group of Z_61 is 60. 60 is not divisible by 7, so the unit group contains no elements of order 7. Since the unit group is cyclic and 2, 3, and 5 all divide 60, there are elements of each of these orders. (Actually in truth one doesn't need to know that the group is cyclic to know there are elements of these orders. But we proved haven't that.) (b). Let w be a unit of order 3 in Z_61. This exists by a. Now, 4, w4, and (w^2)4 are distinct roots of the polynomial f(T), which thus factors as f(T) = (T - 4)(T - w4)(t - (w^2)4) 3. [Note: I messed up and asked the same question in a and b. Sorry 'bout that.] Note that for a and b integers, then (1 + ap) (1 + bp) = 1 + (a+b)p + abp^2 = 1 + (a+b)p (mod p^2) Thus 1 -------- = (1 - ap) (1 + ap) This proves that each element of H is a unit in Z_{p^2}, and that the product of any 2 elements of H is again in H. Since (taking a = 0) we have 1 in H, H is evidently a subgroup of the unit group of Z_{p^2}. This takes care of (a) and (b). As to (c), notice that a = b (mod p) means that there is some integer k so that a - b = kp. Thus, (1 + ap) - (1 + bp) = (a-b)p = kp^2, so that (1 + ap) = (1 + bp) (mod p^2) as desired. (d). Note that the calculations for (b) show that f(a + b) = f(a)f(b). [Technically speaking, this means that f is a _group_ _homomorphism_.] (e). Since any element of H has the form (1 + ap) = f(a), any element of H is in the image of f; i.e. f is onto. Since (1 + ap) = (1 + bp) in Z_{p^2} if and only if (1 + ap) - (1 + bp) = (a-b)p = 0 (mod p^2) if and only if a = b (mod p) there are p distinct elements of H. There are also p distinct elements of Z_p. Hence, f is an onto function between two finite sets of the same size, so by the "pigeon-hole" argument, f is also a 1-1 function. [Again, technically this means that f is actually an _isomorphism_ of groups. Notice that it is an isomorphism between the _additive_ group Z_p and a subgroup of the multiplicative group of units in Z_{p^2}]. (f). Since the group of units of Z_{p^2} has order the value of the phi function of p^2, which is phi(p^2) = p^2 - p = p(p-1), and H has order p, Lagrange's theorem indicates that there are (p-1) distinct cosets of H in the group of units of Z_{p^2}. 4. (a). (This part doesn't have anything to do with the field F). Note that p^5 - 1 p - 1 ------- = p^3 + p + ------- p^2 - 1 p^2 - 1 hence p^5 - 1 = p - 1 (mod p^2 - 1) Similarly, p^5 - 1 p^2 - 1 ------- = p^2 + ------- p^3 - 1 p^3 - 1 hence P^5 - 1 = p^2 - 1 (mod p^3 - 1) Finally, it a similar calculation shows that p^5 - 1 = p - 1 (mod p^4 - 1). (b). Suppose that E is any subfield of F. If E is not all of F, then the order of E must be less than p^5. If E is not Z_p, the order of E must be at least p^2. So the only possibilities for the order of E are p^2, p^3, and p^4. If E has these orders, then its group of units must have order (respectively) m = p^2 - 1, p^3 - 1, or p^4 - 1. The group of units of E is a subgroup of the group of units of F, and part (a) shows that the group of units of F, which has order p^5 -1, contains no subgroup of order m. Thus, there is no such subfield E. [Remark: One can prove that (p^m - 1) is divisible by (p^n -1) if and only if m is divisible by n. This gives a condition for what subfields a field of order p^n can have.] 5. (a). If x is in the kernel of f, then x^5 = 1 by the definition of f (and of the notion of kernel). Hence if x is in the kernel, then x has order 5 (or 1 -- sorry, I goofed a bit in the statement. The identity element 1 is of course in the kernel and doesn't have order 5). (b). If 1 is the only element in the kernel, suppose that f(g) = f(h). Then f(g h^-1) = 1, so that g h^-1 is in the kernel, and is therefore 1. This means that g = h, hence f is a 1-1 function. (This proves one direction of the "if and only if" assertion.) On the other hand, if f is a 1-1 function, note that 1 is always in the kernel of f. If x is any element in the kernel if f, then f(x) = 1 and f(1) = 1. The definition of 1-1 function then implies that x = 1. (This proves the other direction). (c). If p -1 is not divisible by 5, then there the unit group of Z_p contains no elements of order 5. Hence the kernel of f must contain only the identity element. Hence by (b) the function f is 1-1. Since Z_p is finite, this means that the function f is onto. So if x is a unit in Z_p, there is a unit y in Z_p so that f(y) = y^5 = x; in other words, y is a 5th root of x as desired. If z is any other 5th root of x, then consider the element y/z. Notice that (y/z)^5 = y^5 / z^5 = x/x = 1. Hence, (y/z) is in the kernel of f. It follows that (y/z) = 1 or y = z. This proves that each element has a _unique_ 5th root.