(b) Since I R, we have IJ I. Similarly IJ J and we deduce that IJ I J.
(c) Since I + J = R, we may write i + j = 1 where i I and j J. If x I J, then x = xi + xj JI + IJ = IJ (because R is commutative). Therefore x IJ and we have proven that I J IJ. The result now follows from (b).
(d) Let a R\ 0. We need to prove that a has a multiplicative inverse. Using IJ = I J with I = J = aR, we see that aRaR = aR aR = aR, hence aras = a for some r, s R. Since a 0 and R is an integral domain, we may cancel a to obtain ars = 1. We have now shown that all nonzero elements of R have a multiplicative inverse, hence R is a field.
(b) The subfields E of F containing K which are Galois extensions of K correspond to the normal subgroups of Gal(F/K). Specifically if H is a normal subgroup of Gal(F/K), then the corresponding subfield is Fix(H), the elements of F which are fixed by all automorphisms of H. Furthermore we have Gal(Fix(H)/K) Gal(F/K)/H and [Fix(H) : K] = [Gal(F/K) : H]. Since S5 has a unique nontrivial normal subgroup, namely the alternating group A5, it follows that the subfield E required is Fix(A5). Then [E : K] = 2 and Gal(E/K) S5/A5 /2.
(b) Since G/A is a group of order 35, we can apply Sylow's theorems to see that G/A has exactly one subgroup of order 7, which by the subgroup correspondence theorem we may call H/A. Then H/A G/A, so H G. Now H is a group of order | A|| H/A| = 13·7, and we may apply Sylow's theorems for the prime 7 to deduce that H has exactly one subgroup of order 7; we shall call this subgroup B. Then B H; in fact we can assert more, namely that B G. To see this, let g G. Then gBg-1 is a subgroup of gHg-1 = H because H G, and since | gBg-1| = | B| = 7, we see that gBg-1 = B which establishes that B G. Similarly G has a normal subgroup of order 5, which we shall call C.
We now have that G has normal subgroups A, B, C of orders 13, 7 and 5 respectively. Since 13, 7 and 5 are coprime and their product is 455, we deduce that G A X B X C. Let a A be an element of order 13, let b B be an element of order 7, and let c C be an element of order 5. We want to show that abc is an element of order 455. Since the order of an element divides the order of the group, we certainly have the order of abc divides 455. Suppose the order of abc was less than 455. Then the order of abc would have to divide 455/13, or 455/7, or 455/5. Suppose the order of abc divided 455/13 = 35. Then (abc)35 = 1 and since a, b, c commute, we see that a35b35c35 = 1. But b35 = c35 = 1, hence a35 = 1. This is not possible because a has order 13. Similarly the order of abc cannot divide 455/7 and 455/5. We deduce that abc has order 455, hence < abc > = G and the result is proven.
|A||= Ra (R/qiR)ai|
|B||= Rb (R/qiR)bi|
|Rna (R/qiR)nai Rnb (R/qiR)nbi.|
Let H be a group of order 12. The number of Sylow 3-subgroups is 1 or 4. Suppose there is exactly one Sylow 3-subgroup B. Then B H and | H/B| = 4. Since groups of order 3 and 4 are abelian, we see that B and H/B are abelian and hence H is solvable. Suppose on the other hand that H has 4 Sylow 3-subgroups. If B1 and B2 are two distinct Sylow 3-subgroups, then B1 B2 is a proper subgroup of B1 whose order divides 3 by Lagrange's theorem, hence B1 B2 = 1 and we conclude that H has (at least) 8 elements of order 3. Now the Sylow 2-subgroups of H have order 4, and every element of a Sylow 2-subgroup has order a power of 2. If H had more than one Sylow 2-subgroup, then H would have at least 5 elements of order a power of 2, consequently H would have at least 5 + 8 = 13 elements, which is not possible because | H| = 12. Therefore H has a unique subgroup C of order 4, which must be normal in H. Since | H/C| = 3 and | C| = 4, we see that H/C and C are abelian, and we conclude that H is solvable. This completes the proof.
Now any element of Gal(K/) must send to ±, to ±, and to ±. It follows that every nonidentity element of Gal(K/) has order 2, and that Gal(K/) is elementary abelian of order 1,2,4 or 8.
We shall use the following result: if a and b are products of distinct prime numbers and () = (), then a = b. To see this, write = r + s where r, s . Then a = r2 + 2rs + s2b. Clearly s 0 and rs = 0, consequently r = 0 and a = s2b which establishes the result.
It follows immediately that there are at least 8 subfields between and K, namely () where c, d, e are 0 or 1. Now if | Gal(K/)|4, then there would be at most 5 subgroups of Gal(K/), consequently there would be at most 5 fields between K and . This is a contradiction, so we must have | Gal(K/)| = 8 and therefore Gal(K/) (/2)3.