{"id":31004,"date":"2025-06-21T21:41:00","date_gmt":"2025-06-21T21:41:00","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=31004"},"modified":"2025-06-21T21:41:01","modified_gmt":"2025-06-21T21:41:01","slug":"find-all-the-sylow-2-subgroups-of-s_3-and-show-that-they-are-conjugate","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-all-the-sylow-2-subgroups-of-s_3-and-show-that-they-are-conjugate\/","title":{"rendered":"Find all the Sylow 2-subgroups of S_3 and show that they are conjugate"},"content":{"rendered":"\n<p>Find all the Sylow 2-subgroups of S_3 and show that they are conjugate.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To find all the Sylow 2-subgroups of the symmetric group S3S_3 and show that they are conjugate, we proceed as follows:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Order of the Group<\/h3>\n\n\n\n<p>The symmetric group S3S_3 has order \u2223S3\u2223=3!=6|S_3| = 3! = 6<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Apply Sylow&#8217;s Theorems<\/h3>\n\n\n\n<p>We want to find the Sylow 2-subgroups. The highest power of 2 dividing 6 is 21=22^1 = 2, so any Sylow 2-subgroup of S3S_3 has order 2.<\/p>\n\n\n\n<p>By Sylow&#8217;s theorems:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The number n2n_2 of Sylow 2-subgroups divides 3 (since 6\/2 = 3) and<\/li>\n\n\n\n<li>n2\u22611mod\u2009\u20092n_2 \\equiv 1 \\mod 2<\/li>\n<\/ul>\n\n\n\n<p>So n2n_2 divides 3 and is congruent to 1 mod 2. The possible values are 1 and 3.<\/p>\n\n\n\n<p>We will find all subgroups of order 2 in S3S_3.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: List Elements of S3S_3<\/h3>\n\n\n\n<p>The elements of S3S_3 are: id,&nbsp;(12),&nbsp;(13),&nbsp;(23),&nbsp;(123),&nbsp;(132)\\text{id},\\ (12),\\ (13),\\ (23),\\ (123),\\ (132)<\/p>\n\n\n\n<p>The three transpositions (12),(13),(23)(12), (13), (23) each generate a subgroup of order 2:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H1={id,(12)}H_1 = \\{ \\text{id}, (12) \\}<\/li>\n\n\n\n<li>H2={id,(13)}H_2 = \\{ \\text{id}, (13) \\}<\/li>\n\n\n\n<li>H3={id,(23)}H_3 = \\{ \\text{id}, (23) \\}<\/li>\n<\/ul>\n\n\n\n<p>These are all the subgroups of order 2, so n2=3n_2 = 3. Thus, there are three Sylow 2-subgroups in S3S_3.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Show They Are Conjugate<\/h3>\n\n\n\n<p>Two subgroups HH and KK are conjugate in a group GG if there exists g\u2208Gg \\in G such that gHg\u22121=KgHg^{-1} = K.<\/p>\n\n\n\n<p>Let us take H1={id,(12)}H_1 = \\{ \\text{id}, (12) \\}. We can find elements of S3S_3 that conjugate it to the other two subgroups.<\/p>\n\n\n\n<p>Let g=(123)\u2208S3g = (123) \\in S_3. Then (123)(12)(132)=(23)(123)(12)(132) = (23)<\/p>\n\n\n\n<p>So (123)H1(132)={id,(23)}=H3(123)H_1(132) = \\{ \\text{id}, (23) \\} = H_3<\/p>\n\n\n\n<p>Similarly, (132)(12)(123)=(13)(132)(12)(123) = (13), so (132)H1(123)={id,(13)}=H2(132)H_1(123) = \\{ \\text{id}, (13) \\} = H_2<\/p>\n\n\n\n<p>This shows that all three Sylow 2-subgroups are conjugate to each other in S3S_3.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>There are <strong>three Sylow 2-subgroups<\/strong> of S3S_3: {id,(12)},&nbsp;{id,(13)},&nbsp;{id,(23)}\\{ \\text{id}, (12) \\},\\ \\{ \\text{id}, (13) \\},\\ \\{ \\text{id}, (23) \\}<\/p>\n\n\n\n<p>They are all conjugate to each other under elements of S3S_3, which satisfies Sylow&#8217;s conjugacy condition.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Find all the Sylow 2-subgroups of S_3 and show that they are conjugate. The correct answer and explanation is: To find all the Sylow 2-subgroups of the symmetric group S3S_3 and show that they are conjugate, we proceed as follows: Step 1: Order of the Group The symmetric group S3S_3 has order \u2223S3\u2223=3!=6|S_3| = 3! [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-31004","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31004","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=31004"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31004\/revisions"}],"predecessor-version":[{"id":31010,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31004\/revisions\/31010"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=31004"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=31004"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=31004"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}