The Sylow Theorems
In this chapter, we will prove the Sylow theorems. These are difficult results, but fundamental to our understanding of the structure of finite groups. In particular, we will show that if \(p^n\) is the largest power of a prime p dividing the order of a f
- PDF / 255,112 Bytes
- 18 Pages / 439.37 x 666.142 pts Page_size
- 38 Downloads / 212 Views
The Sylow Theorems
In this chapter, we will prove the Sylow theorems. These are difficult results, but fundamental to our understanding of the structure of finite groups. In particular, we will show that if p n is the largest power of a prime p dividing the order of a finite group G, then G has at least one subgroup of order p n . Furthermore, we will discover that any two such subgroups are conjugate to each other, and determine a restriction upon the number of such subgroups. We will then explore various applications of these theorems, and conclude the chapter by classifying all groups of order smaller than 16.
7.1 Normalizers and Centralizers We are very familiar with the centre of a group, which consists of all elements that commute with everything. Let us generalize. Definition 7.1. Let G be a group, a ∈ G and H a subgroup of G. Then the centralizer of a is the set of all elements of G that commute with a. We write C(a) = {g ∈ G : ag = ga}. Also, the centralizer of H is C(H ) = {g ∈ G : gh = hg for all h ∈ H }. Example 7.1. If a ∈ Z (G), then C(a) = G. If H ≤ Z (G), then C(H ) = G. In particular, C(e) = G, so we cannot assume that centralizers are necessarily abelian. Example 7.2. Let G = D8 . Then we find that C(R270 ) = R90 , C(R180 ) = G and C(F1 ) = {R0 , R180 , F1 , F2 }.
© Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_7
115
116
7 The Sylow Theorems
Theorem 7.1. Let G be a group, a ∈ G and H a subgroup of G. Then 1. C(H ) = h∈H C(h); 2. C(a) and C(H ) are both subgroups of G; 3. if H is a normal subgroup of G, then so is C(H ); 4. Z (G) is a subgroup of both C(a) and C(H ); and 5. C(a) = C(a). Proof. (1) This follows from the definition. (2) Clearly ae = a = ea, so e ∈ C(a). Suppose that b, c ∈ C(a). Then bca = bac = abc, so bc ∈ C(a). Also, ba = ab, so b−1 (ba)b−1 = b−1 (ab)b−1 . Thus, ab−1 = b−1 a, so b−1 ∈ C(a). Hence, C(a) ≤ G. Furthermore, combining this fact with (1) and Exercise 3.37, we see that C(H ) ≤ G. (3) See Exercise 4.4. (4) Central elements commute with everything hence, in particular, they commute with a and elements of H . (5) If b ∈ C(a), then since a ∈ a, we see that b commutes with a. Thus, b ∈ C(a). Conversely, if b ∈ C(a), then ab = ba. Therefore, a ∈ C(b). As C(b) ≤ G by (2), we see that a i ∈ C(b) for all integers i. That is, a i b = ba i , for all i ∈ Z. In other words, b ∈ C(a). Suppose we have a subgroup H of G that is not normal. Of course, H is a normal subgroup of H . Furthermore, it is easy to see that H is normal in H Z (G). How big a subgroup of G can we find in which H is a normal subgroup? This is where normalizers come in. Definition 7.2. Let G be a group and H a subgroup. Then the normalizer of H is the set N (H ) = {a ∈ G : a −1 H a = H }. If K is another subgroup of G, then we write N K (H ) = N (H ) ∩ K , and call it the normalizer of H in K . Remember, if a ∈ C(H ), then a −1 ha = h, for all h ∈ H . But if a ∈
Data Loading...
