{"id":1537,"date":"2020-09-28T22:25:02","date_gmt":"2020-09-28T16:55:02","guid":{"rendered":"https:\/\/soulofmathematics.com\/?page_id=1537"},"modified":"2020-10-11T11:38:23","modified_gmt":"2020-10-11T06:08:23","slug":"sylows-theorem","status":"publish","type":"page","link":"https:\/\/soulofmathematics.com\/index.php\/sylows-theorem\/","title":{"rendered":"SYLOW’S THEOREM"},"content":{"rendered":"\n

Sylow theorems<\/strong> are a collection of theorems named after the Norwegian mathematician Peter Ludwig Sylow (1872) that give detailed information about the number of subgroups of fixed order that a given finite group contains. The Sylow theorems form a fundamental part of finite group theory and have very important applications in the classification of finite simple groups.<\/p>\n\n\n\n

Sylow\u2019s theorems and their proofs.<\/h2>\n\n\n\n

Definitions.<\/strong> Let G be a group, and let p be a prime number.<\/p>\n\n\n\n

\u2022 A group of order pk<\/sup> for some k \u2265 1 is called a p-group<\/strong>. A subgroup of order pk<\/sup> for some k \u2265 1 is called a p-subgroup<\/strong>.
\u2022 If |G| = p\u03b1<\/sup> m where p does not divide m, then a subgroup<\/strong> of order p\u03b1<\/sup> is called a Sylow p-subgroup<\/strong> of G.<\/p>\n\n\n\n

Notation.<\/strong> Sy<\/sub>lp<\/sub>(G) = the set of Sylow p-subgroups<\/strong> of G<\/em> <\/p>\n\n\n\n

np<\/sub>(G) = the # of Sylow p-subgroups<\/strong> of G = |Sy<\/sub>lp<\/sub>(G)|<\/em><\/p>\n\n\n\n

Diving Into Details<\/h3>\n\n\n\n
\n

Let G be a group of order p\u03b1<\/sup> m, where p is a prime, m \u2265 1, and p does not divide m. Then:<\/p>\n\n\n\n

  1. Sy<\/sub>lp<\/sub>(G) not equal to \u2205, i.e. Sylow p-subgroups exist<\/em>.<\/li>
  2. All Sylow p-subgroups are conjugate in G, i.e., if P1<\/sub> and P2<\/sub> are both Sylow p-subgroups, then there is some g \u2208 G such that P1<\/sub> = g P1<\/sub>g \u22121<\/sup>. In particular, np<\/sub>(G) = (G : NG<\/sub>(P)).<\/em><\/li>
  3. Any p-subgroup of G is contained in a Sylow p-subgroup.<\/em><\/li>
  4. np<\/sub>(G) \u2261 1 mod p.<\/em><\/li><\/ol>\n<\/div><\/div>\n\n\n\n

    First Sylow Theorem:<\/strong> There is a subgroup of G whose order is pe<\/sup>.<\/p>\n\n\n\n

    Proof:<\/em> We let g be the set of all subsets of G of order pe<\/sup>. One of these subsets is the subgroup we are looking for, but instead of finding it directly we will show that one of these subsets has a stabilizer of order pe<\/sup>. The stabilizer will be the required subgroup.<\/p>\n\n\n\n

    Lemma.<\/strong> The number of subsets of order pe<\/sup> in a set of n = pe<\/sup>m elements
    (p not dividing m) is the binomial coefficient<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Moreover N is not divisible by p.<\/p>\n\n\n\n

    Proof. It is a standard fact that the number of subsets of order pe<\/sup>: is this binomial coefficient. To see that N is not divisible by p, note that every time p divides a term (n – k) in the numerator of N, it also divides the term (pe<\/sup> – k) of the denominator exactly the same number of times: If we write k in the form k = pi<\/sup> l, where p does not divide l, then i < e. Therefore (n – k) and (pe<\/sup> – k) are both divisible by pi but not divisible by p1+i<\/sup>.<\/p>\n\n\n\n

    We decompose g into orbits for the operation of left multiplication, obtaining the formula.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Since p does not divide N, some orbit has an order which is not divisible by p, say the orbit of the subset U.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Second Sylow Theorem:<\/strong> Let K be a subgroup of G whose order is divisible by p, and let H be a Sylow p-subgroup of G. There is a conjugate subgroup H’ = gHg– 1<\/sup> such that K n H’ is a Sylow subgroup of K.<\/p>\n\n\n\n

    Proof:<\/em> We are given a subgroup K and a Sylow subgroup H of G, and we are to show that for some conjugate subgroup H I of H, the intersection K n H I is a Sylow subgroup of K. Let S denote the set of left cosets G\/H. The facts that we need about this set are that G operates transitively, that is, the set forms a single orbit, and that H is the stabilizer of one of its elements, namely of s = 1H. So the stabilizer of as is the conjugate subgroup aHa-1<\/sup>. We restrict the operation of G to K and decompose S into K-orbits. Since H is a Sylow subgroup, the order of S is prime to p. So there is some K-orbit O whose order is prime to p. Say that O is the K-orbit of the element as. Let H’ denote the stabilizer aHa-1<\/sup> of as for the operation of G. Then the stabilizer of as for the restricted operation of K is obviously H’ n K, and the index [K:H’nK] is |O|, which is prime to p. Also, since it is a conjugate of H, H’ is ap-group. Therefore H’ n K is a p -group. It follows that H’ n K is a Sylow subgroup of K.<\/p>\n\n\n\n

    Third Sylow Theorem:<\/strong> Let |G| = n, and n = pe<\/sup>m. Let s be the number of Sylow p-subgroups. Then s divides m and is congruent 1 (modulo p): s|m, and s = ap + 1 for some integer a \u2265 O.<\/p>\n\n\n\n

    Proof:<\/em> The Sylow subgroups of G are all conjugate to a given one, say to H. So the number of Sylow subgroups is s = [G:N], where N is the normalizer of H. Since H\u2282N, [G:N] divides [G:H] = m. To show s \u2261 1 (modulo p), we decompose the set {H1, \u2026 , Hs} of Sylow subgroups into orbits for the operation of conjugation by H = H1. An orbit consists of a single group Hi if and only if H is contained in the normalizer M of Hi. If so, then H and Hi are both Sylow subgroups of M, and Hi is normal in M. There is only one H orbit of order 1, namely {H}. The other orbits have orders divisible by p because their orders divide |H|, by the Counting Formula. This shows that s \u2261 1 (modulo p).<\/p>\n\n\n