1.0SOUL OF MATHEMATICShttps://soulofmathematics.comRajarshi Deyhttps://soulofmathematics.com/index.php/author/rajarshidey1729gmail-com/rich600338<blockquote class="wp-embedded-content" data-secret="tlIJ40TKOg"><a href="https://soulofmathematics.com/index.php/sylows-theorem/">SYLOW’S THEOREM</a></blockquote><iframe sandbox="allow-scripts" security="restricted" src="https://soulofmathematics.com/index.php/sylows-theorem/embed/#?secret=tlIJ40TKOg" width="600" height="338" title="“SYLOW’S THEOREM” — SOUL OF MATHEMATICS" data-secret="tlIJ40TKOg" frameborder="0" marginwidth="0" marginheight="0" scrolling="no" class="wp-embedded-content"></iframe><script type="text/javascript"> /* <![CDATA[ */ /*! This file is auto-generated */ !function(d,l){"use strict";l.querySelector&&d.addEventListener&&"undefined"!=typeof URL&&(d.wp=d.wp||{},d.wp.receiveEmbedMessage||(d.wp.receiveEmbedMessage=function(e){var t=e.data;if((t||t.secret||t.message||t.value)&&!/[^a-zA-Z0-9]/.test(t.secret)){for(var s,r,n,a=l.querySelectorAll('iframe[data-secret="'+t.secret+'"]'),o=l.querySelectorAll('blockquote[data-secret="'+t.secret+'"]'),c=new RegExp("^https?:$","i"),i=0;i<o.length;i++)o[i].style.display="none";for(i=0;i<a.length;i++)s=a[i],e.source===s.contentWindow&&(s.removeAttribute("style"),"height"===t.message?(1e3<(r=parseInt(t.value,10))?r=1e3:~~r<200&&(r=200),s.height=r):"link"===t.message&&(r=new URL(s.getAttribute("src")),n=new URL(t.value),c.test(n.protocol))&&n.host===r.host&&l.activeElement===s&&(d.top.location.href=t.value))}},d.addEventListener("message",d.wp.receiveEmbedMessage,!1),l.addEventListener("DOMContentLoaded",function(){for(var e,t,s=l.querySelectorAll("iframe.wp-embedded-content"),r=0;r<s.length;r++)(t=(e=s[r]).getAttribute("data-secret"))||(t=Math.random().toString(36).substring(2,12),e.src+="#?secret="+t,e.setAttribute("data-secret",t)),e.contentWindow.postMessage({message:"ready",secret:t},"*")},!1)))}(window,document); /* ]]> */ </script> https://i1.wp.com/soulofmathematics.com/wp-content/uploads/2020/09/Sylow_2.jpeg?fit=262%2C326&ssl=1262326Sylow theorems 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. Sylow’s theorems and their proofs. Definitions. Let G be a group, and let p be a prime number. • A group of order pk for some k ≥ 1 is called a p-group. A subgroup of order pk for some k ≥ 1 is called a p-subgroup.• If |G| = pα m where p does not divide m, then a subgroup of order pα is called a Sylow p-subgroup of G. Notation. Sylp(G) = the set of Sylow p-subgroups of G np(G) = the # of Sylow p-subgroups of G = |Sylp(G)| Diving Into Details First Sylow Theorem: There is a subgroup of G whose order is pe. Proof: We let g be the set of all subsets of G of order pe. 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. The stabilizer will be the required subgroup. Lemma. The number of subsets of order pe in a set of n = pem elements(p not dividing m) is the binomial coefficient Moreover N is not divisible by p. Proof. It is a standard fact that the number of subsets of order pe: 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 – k) of the denominator exactly the same number of times: If we write k in the form k = pi l, where p does not divide l, then i < e. Therefore (n – k) and (pe – k) are both divisible by pi but not divisible by p1+i. We decompose g into orbits for the operation of left multiplication, obtaining the formula. Since p does not divide N, some orbit has an order which is not divisible by p, say the orbit of the subset U. Second Sylow Theorem: 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 such that K n H’ is a Sylow subgroup of K. Proof: 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. 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 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. Third Sylow Theorem: Let |G| = n, and n = pem. 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 ≥ O. Proof: 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⊂N, [G:N] divides [G:H] = m. To show s ≡ 1 (modulo p), we decompose the set {H1, … , 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 ≡ 1 (modulo p).