<?xml version="1.0" encoding="UTF-8"?><rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:atom="http://www.w3.org/2005/Atom" version="2.0" xmlns:itunes="http://www.itunes.com/dtds/podcast-1.0.dtd" xmlns:googleplay="http://www.google.com/schemas/play-podcasts/1.0"><channel><title><![CDATA[Pseudovariety]]></title><description><![CDATA[A pseudovariety of mathematics and general (mostly mathematics) which purports to talk more than just about associative algebra.]]></description><link>https://pseudovariety.com</link><image><url>https://substackcdn.com/image/fetch/$s_!Lxrf!,w_256,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2F9ad4f314-8f59-468a-8bf4-e1a46d66a1f0_382x382.png</url><title>Pseudovariety</title><link>https://pseudovariety.com</link></image><generator>Substack</generator><lastBuildDate>Fri, 17 Apr 2026 05:16:57 GMT</lastBuildDate><atom:link href="https://pseudovariety.com/feed" rel="self" type="application/rss+xml"/><copyright><![CDATA[Peter Alspaugh]]></copyright><language><![CDATA[en]]></language><webMaster><![CDATA[pseudovariety@substack.com]]></webMaster><itunes:owner><itunes:email><![CDATA[pseudovariety@substack.com]]></itunes:email><itunes:name><![CDATA[Peter Alspaugh]]></itunes:name></itunes:owner><itunes:author><![CDATA[Peter Alspaugh]]></itunes:author><googleplay:owner><![CDATA[pseudovariety@substack.com]]></googleplay:owner><googleplay:email><![CDATA[pseudovariety@substack.com]]></googleplay:email><googleplay:author><![CDATA[Peter Alspaugh]]></googleplay:author><itunes:block><![CDATA[Yes]]></itunes:block><item><title><![CDATA[2. String Rewriting Systems]]></title><description><![CDATA[Definitions of string rewriting systems.]]></description><link>https://pseudovariety.com/p/2-string-rewriting-systems</link><guid isPermaLink="false">https://pseudovariety.com/p/2-string-rewriting-systems</guid><dc:creator><![CDATA[Peter Alspaugh]]></dc:creator><pubDate>Fri, 04 Jul 2025 23:50:11 GMT</pubDate><enclosure url="https://substackcdn.com/image/fetch/$s_!kexk!,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png" length="0" type="image/jpeg"/><content:encoded><![CDATA[<p>A <em>string rewriting system</em>, also known as an abstract reduction system or abstract rewriting system,<em> </em>is a pair consisting of an alphabet A and a set of rewriting rules R, where R is a relation</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\mathcal{R}\\subseteq A^*\\times A^*.&quot;,&quot;id&quot;:&quot;QPHMOHIPCW&quot;}" data-component-name="LatexBlockToDOM"></div><p>Here, A* denotes the free monoid generated by A, or the set of all strings over A equipped with concatenation of strings as its multiplication. It is not hard to see that concatenation is associative and the identity element is the empty word consisting of no letters, often denoted as</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\varepsilon&quot;,&quot;id&quot;:&quot;IYQFYUSXLN&quot;}" data-component-name="LatexBlockToDOM"></div><p>or &#8220;&#8220; for Python purposes. Note that rewriting rules are ordered pairs, meaning any rewriting happens in a specific order. We always assume that a word may rewrite to itself with precisely 0 applications of any rewriting rule.</p><p>Any string rewriting system defines a monoid presentation (or semigroup presentation, but we will focus on monoids) via the congruence closure of its set of rewriting rules. This provides a way to think about equivalence of words under the congruence as rewriting. Here, if two strings in A* are equal, such as a=b, then for any x and y in A* we have that xay=xby (by definition of congruence). We can say that xay rewrites to xby, denoted xay &#8594; xby or vice versa.</p><p>Now, let&#8217;s say that we have a monoid presentation and we would like to determine if two words are equal under the given presentation (or <em>solve the word problem</em> for that monoid). We could give a computer our presentation and tell it to determine if those words are equal by rewriting one into the other with the relations in the presentation like so: xaybz &#8594; xbybz &#8594; xbyaz (letting us conclude that xaybz=xbyaz). However, we will inevitably run into problems. For one, equalities in the presentation to not prescribe a direction for us to rewrite. Also, there are infinitely many ways that the rewritings could happen, or none at all (yet infinitely many ways we must rule out). </p><p>We then must make some specifications for the rewriting rules defined by a presentation. The following definition will be our most important.</p><h4>Definition: A rewriting system (A,R) is <em>Noetherian</em> if there does not exist any infinite sequence of words in A* such that</h4><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;x_1\\rightarrow x_2\\rightarrow x_3\\rightarrow\\cdots\\text{   (for }x_i\\in A^*,\\text{ }i\\in\\mathbb{Z^+)}&quot;,&quot;id&quot;:&quot;ATFXXNBSIP&quot;}" data-component-name="LatexBlockToDOM"></div><h4>and</h4><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\forall i\\in\\mathbb{Z}^+,\\text{ }x_i\\neq x_{i+1}.&quot;,&quot;id&quot;:&quot;JUQHLBWRZQ&quot;}" data-component-name="LatexBlockToDOM"></div><p>That is, every rewriting of a word must terminate at some point. Any finite rewriting system (meaning one with a finite alphabet and finite set of rewriting rules) is Noetherian if it contains no &#8220;loops&#8220;, where there exist words x and y such that x&#8594;y and y&#8594;x using the rewriting rules in the system.</p><h4>Definition: A word which does not contain as a subword the left hand side of any rewriting rule is called <em>irreducible</em>. That is, it cannot be written by the given rewriting system. If x and y are two words such that x&#8594;y and y is irreducible, then we say that y is a normal form for x.</h4><p>Using just this definition a congruence class may contain several normal forms, or none at all! Luckily, we have the following theorems.</p><h4>Theorem: Every word (and thus congruence class) has a normal form under a Noetherian rewriting system.</h4><p>Proof: Let x be a word in A* and consider a sequence of rewritings of x which results in a word y. Since x has no normal form, y cannot be irreducible. Then y may be further rewritten, and hence x may be rewritten ad infinitum, so there exists an infinite sequence of rewritings of x (where each resulting word is not equal to the last). This contradicts the assumption that the rewriting system is Noetherian. &#9633;</p><h4>Definition: A rewriting system is <em>confluent</em> if for any word x in A*, if x rewrites to two words y and z, then there exists some word w such that y and z both rewrite to w. A rewriting system is <em>locally confluent</em> if for any x in A*, if x rewrites to both y and z with just one application of any rewriting rule, then y and z both rewrite to some w in A*.</h4><p>In the below diagram, I depict confluence and local confluence. Arrows with stars represent rewritings that may use zero or multiple rules, whereas arrows with no stars use precisely one rewriting rule. This notation is typical, but I neglect it in this series.</p><div class="captioned-image-container"><figure><a class="image-link image2 is-viewable-img" target="_blank" href="https://substackcdn.com/image/fetch/$s_!kexk!,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png" data-component-name="Image2ToDOM"><div class="image2-inset"><picture><source type="image/webp" srcset="https://substackcdn.com/image/fetch/$s_!kexk!,w_424,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 424w, https://substackcdn.com/image/fetch/$s_!kexk!,w_848,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 848w, https://substackcdn.com/image/fetch/$s_!kexk!,w_1272,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 1272w, https://substackcdn.com/image/fetch/$s_!kexk!,w_1456,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 1456w" sizes="100vw"><img src="https://substackcdn.com/image/fetch/$s_!kexk!,w_1456,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png" width="903" height="470" data-attrs="{&quot;src&quot;:&quot;https://substack-post-media.s3.amazonaws.com/public/images/d3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png&quot;,&quot;srcNoWatermark&quot;:null,&quot;fullscreen&quot;:null,&quot;imageSize&quot;:null,&quot;height&quot;:470,&quot;width&quot;:903,&quot;resizeWidth&quot;:null,&quot;bytes&quot;:29273,&quot;alt&quot;:null,&quot;title&quot;:null,&quot;type&quot;:&quot;image/png&quot;,&quot;href&quot;:null,&quot;belowTheFold&quot;:true,&quot;topImage&quot;:false,&quot;internalRedirect&quot;:&quot;https://pseudovariety.com/i/167555724?img=https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png&quot;,&quot;isProcessing&quot;:false,&quot;align&quot;:null,&quot;offset&quot;:false}" class="sizing-normal" alt="" srcset="https://substackcdn.com/image/fetch/$s_!kexk!,w_424,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 424w, https://substackcdn.com/image/fetch/$s_!kexk!,w_848,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 848w, https://substackcdn.com/image/fetch/$s_!kexk!,w_1272,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 1272w, https://substackcdn.com/image/fetch/$s_!kexk!,w_1456,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fd3649b40-99b2-48c5-9a9b-5ed93cfab040_903x470.png 1456w" sizes="100vw" loading="lazy"></picture><div class="image-link-expand"><div class="pencraft pc-display-flex pc-gap-8 pc-reset"><button tabindex="0" type="button" class="pencraft pc-reset pencraft icon-container restack-image"><svg role="img" width="20" height="20" viewBox="0 0 20 20" fill="none" stroke-width="1.5" stroke="var(--color-fg-primary)" stroke-linecap="round" stroke-linejoin="round" xmlns="http://www.w3.org/2000/svg"><g><title></title><path d="M2.53001 7.81595C3.49179 4.73911 6.43281 2.5 9.91173 2.5C13.1684 2.5 15.9537 4.46214 17.0852 7.23684L17.6179 8.67647M17.6179 8.67647L18.5002 4.26471M17.6179 8.67647L13.6473 6.91176M17.4995 12.1841C16.5378 15.2609 13.5967 17.5 10.1178 17.5C6.86118 17.5 4.07589 15.5379 2.94432 12.7632L2.41165 11.3235M2.41165 11.3235L1.5293 15.7353M2.41165 11.3235L6.38224 13.0882"></path></g></svg></button><button tabindex="0" type="button" class="pencraft pc-reset pencraft icon-container view-image"><svg xmlns="http://www.w3.org/2000/svg" width="20" height="20" viewBox="0 0 24 24" fill="none" stroke="currentColor" stroke-width="2" stroke-linecap="round" stroke-linejoin="round" class="lucide lucide-maximize2 lucide-maximize-2"><polyline points="15 3 21 3 21 9"></polyline><polyline points="9 21 3 21 3 15"></polyline><line x1="21" x2="14" y1="3" y2="10"></line><line x1="3" x2="10" y1="21" y2="14"></line></svg></button></div></div></div></a></figure></div><h4>Theorem: A Noetherian rewriting system is confluent iff it is locally confluent.</h4><p>I do not give the proof of the theorem, but it should be accessible to the reader and can be found in Book &amp; Otto (1993).</p>]]></content:encoded></item><item><title><![CDATA[1. Semigroup Presentations]]></title><description><![CDATA[Recall the definitions of congruences, congruence closures, and presentations.]]></description><link>https://pseudovariety.com/p/1-semigroup-presentations</link><guid isPermaLink="false">https://pseudovariety.com/p/1-semigroup-presentations</guid><dc:creator><![CDATA[Peter Alspaugh]]></dc:creator><pubDate>Fri, 04 Jul 2025 22:58:49 GMT</pubDate><enclosure url="https://substackcdn.com/image/fetch/$s_!Lxrf!,w_256,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2F9ad4f314-8f59-468a-8bf4-e1a46d66a1f0_382x382.png" length="0" type="image/jpeg"/><content:encoded><![CDATA[<p>First, let us recall some basic algebraic definitions. A <em>semigroup</em> is a set which is closed under an associative binary operation, a <em>monoid</em> is a semigroup which contains an identity element, and a <em>group</em> is a monoid in which every element has a two-sided inverse.</p><p>A <em>semigroup homomorphism </em>is a map f between semigroups S and T such that for all r and s in S, f(rs)=f(r)f(s). A <em>congruence</em> is an equivalence relation on S such that for any two equivalence classes [r] and [s], if q is in [r] and t is in [s], then qt is in [rs]. That is, it is an equivalence relation that is compatible with the multiplication on S, so multiplication of equivalence classes is well-defined. Fundamentally, a partition of S into preimages of each element under a homomorphism defines a congruence (this generalizes the link between normal subgroups and homomorphisms in Group Theory).</p><p>Let </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\mathcal{R}\\subseteq S^2&quot;,&quot;id&quot;:&quot;SUBZZTLYDW&quot;}" data-component-name="LatexBlockToDOM"></div><p>be a relation. The equivalence closure of R is the smallest equivalence relation on S containing R and the <em>congruence closure</em> of R is the smallest congruence on S containing R.</p><p>A <em>presentation</em> for a semigroup (resp. monoid, group, etc.) is a pair consisting of a set of generators X and a relation on the free semigroup (resp. monoid, group, etc.) generated by X (which is written F with a subscript X). It is usually written</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\langle X\\text{ }|\\text{ }\\mathcal{R}\\rangle.&quot;,&quot;id&quot;:&quot;UTOPZZDSHL&quot;}" data-component-name="LatexBlockToDOM"></div><p>The relation provided defines which words over X are equal to each other in the semigroup. Relations are often written using equalities rather than ordered pairs. Every presentation defines a semigroup S such that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;S\\cong F_X/\\overline{\\mathcal{R}}&quot;,&quot;id&quot;:&quot;SIINWCAYOJ&quot;}" data-component-name="LatexBlockToDOM"></div><p>where R with an overset bar is the congruence closure of R.</p><p>Let us consider an example. We define a monoid with the following presentation (since it is a monoid, we take the free monoid generated by X rather than the free semigroup, so there is no need to specify the existence of the identity or include relations for it):</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\langle a \\text{ }|\\text{ } a^5=1 \\rangle.&quot;,&quot;id&quot;:&quot;GIKXYKEDDC&quot;}" data-component-name="LatexBlockToDOM"></div><p>This presentation defines the cyclic group of integers modulo 5.</p><p>Providing a presentation is a common way to define a semigroup, and perhaps the only way for a large or even infinite one. However, presentations can be impractical to work with computationally. Consider the following presentation for the <em>monoid</em> M:</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\langle a,b \\text{ }|\\text{ } (ab)^2a^2=aba,\\text{ }b^2(ab)^2=a^2,\\text{ }b^2a=a,\\text{ }b^2ab=1,\\text{ }a^2=a \\rangle&quot;,&quot;id&quot;:&quot;WZKBEIMEUJ&quot;}" data-component-name="LatexBlockToDOM"></div><p>Could you tell that this actually defines the trivial group? if so, it probably took you a second. If you need proof, observe that </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;b^2a=a\\text{ and }b^2ab=1\\text{ }\\Rightarrow\\text{ }ab=1\\Rightarrow b=a^{-1}.&quot;,&quot;id&quot;:&quot;EUCIQQVDTZ&quot;}" data-component-name="LatexBlockToDOM"></div><p>This means that a is contained in a subgroup of M with identity 1. Then, since a is idempotent, it must be the identity of M. Then b is also the identity, so the group is trivial.</p><p>Luckily, there are methods to make these computations easier. The focus of this series will be the Knuth-Bendix algorithm, which can extend a set of relations such that they can be used to rewrite any word into a unique <em>normal form </em>that represents the congruence class of that word defined by the presentation. However, this algorithm is not guaranteed to terminate, as when it does, it in effect solves the <em>word problem, </em>the undecidable problem of determining if two given words are equal under the presentation, for the semigroup it defines. In the subsequent posts in this series, I cover the basics of string rewriting systems before moving on to the Knuth-Bendix algorithm, for which I also provide a Python implementation.</p>]]></content:encoded></item><item><title><![CDATA[3. Identity Elements and Monoids, Inverses and Groups]]></title><description><![CDATA[Building up to Group Theory.]]></description><link>https://pseudovariety.com/p/3-identity-elements-and-monoids-inverses</link><guid isPermaLink="false">https://pseudovariety.com/p/3-identity-elements-and-monoids-inverses</guid><dc:creator><![CDATA[Peter Alspaugh]]></dc:creator><pubDate>Thu, 03 Jul 2025 21:32:13 GMT</pubDate><enclosure url="https://substackcdn.com/image/fetch/$s_!Lr_z!,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png" length="0" type="image/jpeg"/><content:encoded><![CDATA[<p>Let S be a semigroup and suppose there is an element 1 in S such that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\forall a\\in S,\\text{ }1\\cdot a=a\\cdot 1=a.&quot;,&quot;id&quot;:&quot;QHCLSONVFF&quot;}" data-component-name="LatexBlockToDOM"></div><p>Then we say that 1 is the identity element of S. I write 1 for notational purposes only (meaning 1 isn&#8217;t necessarily the number 1&#8212;just a symbol representing the identity element). If S contains an identity element, then we say that S is a <em>monoid.</em></p><h4>Fact: The identity element of a monoid is unique.</h4><p>Proof: Exercise.</p><p>Let a be in a monoid M with identity 1. We say that an element b is an inverse of a if ab=ba=1 and we write</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;b=a^{-1}.&quot;,&quot;id&quot;:&quot;EBVJFANECM&quot;}" data-component-name="LatexBlockToDOM"></div><h4>Fact: The inverse of an element in a monoid is unique.</h4><p>Proof: Exercise.</p><p>An idempotent is an element a in a semigroup (or monoid) such that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a^2=a.&quot;,&quot;id&quot;:&quot;WVMQXMYYDU&quot;}" data-component-name="LatexBlockToDOM"></div><p>The identity element of a monoid is an idempotent. I now give the fundamental definition for this series.</p><h4>Definition: A <em>group</em> is a monoid such that every element has an inverse.</h4><p>That is, a group is a monoid G such that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\forall a\\in G,\\text{ }\\exists a^{-1}\\in G\\text{ s.t. }aa^{-1}=a^{-1}a=1.&quot;,&quot;id&quot;:&quot;DPVBOQVJCM&quot;}" data-component-name="LatexBlockToDOM"></div><h4>Theorem: The only idempotent in a group G is the identity.</h4><p>Proof: Suppose to the contrary that e is an idempotent in G. Since G is a group e has an inverse. That is, </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\exists e^{-1}\\in G\\text{ s.t. } ee^{-1}=e^{-1}e=1.&quot;,&quot;id&quot;:&quot;NONAHFRRWS&quot;}" data-component-name="LatexBlockToDOM"></div><p>Since 1 is the identity, we also have that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;e^2=e=1\\cdot e.&quot;,&quot;id&quot;:&quot;ZDMZHPCNQO&quot;}" data-component-name="LatexBlockToDOM"></div><p>Then by left multiplication of the inverse of e on both sides of the =, we obtain</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;e=e^2\\cdot e^{-1}=1\\cdot e\\cdot e^{-1}=1&quot;,&quot;id&quot;:&quot;IMAWVVEEKN&quot;}" data-component-name="LatexBlockToDOM"></div><p>so e=1. &#9633;</p><h4>Theorem: Let M be a finite monoid with a unique idempotent. Then M is a group.</h4><p>Proof: Since the identity 1 is always idempotent, we have that the unique idempotent must be the identity. Consider the set P of all powers of some a in M</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\{a^n\\in M: n\\in\\mathbb{Z}^{+}\\}.&quot;,&quot;id&quot;:&quot;LYOGRDWNWY&quot;}" data-component-name="LatexBlockToDOM"></div><p>P is a subsemigroup of M since it is closed under multiplication.  (in fact it is a subsemigroup of M (the &#8220;monogenic&#8220; subsemigroup of M generated by a) since it is closed under multiplication. Hence P must be finite, so there must exist some positive integers n and m (with n less than or equal to m) such that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a^n=a^m.&quot;,&quot;id&quot;:&quot;WWFIZXOBKO&quot;}" data-component-name="LatexBlockToDOM"></div><p>I claim that there exists some k which is greater than or equal to n and less than or equal to m such that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(a^k)^2=a^k.&quot;,&quot;id&quot;:&quot;XFGOUYPFZU&quot;}" data-component-name="LatexBlockToDOM"></div><p>That is, a to the power of k is idempotent. Why? Suppose to the contrary that there is no such k. Let k be between n and m as mentioned before. By assumption, 2k mod m-n is not equal to k mod m-n. Let</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\ell= (2k \\mod m-n)-(k\\mod m-n)&quot;,&quot;id&quot;:&quot;JMTUBZPEDF&quot;}" data-component-name="LatexBlockToDOM"></div><p>Then </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(k+\\ell)\\mod (m-n)=2(k+\\ell)\\mod (m-n)&quot;,&quot;id&quot;:&quot;UCMIFFTOLG&quot;}" data-component-name="LatexBlockToDOM"></div><p>so a to the power of k+l is idempotent. The below figure illustrates this argument more clearly.</p><div class="captioned-image-container"><figure><a class="image-link image2 is-viewable-img" target="_blank" href="https://substackcdn.com/image/fetch/$s_!Lr_z!,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png" data-component-name="Image2ToDOM"><div class="image2-inset"><picture><source type="image/webp" srcset="https://substackcdn.com/image/fetch/$s_!Lr_z!,w_424,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 424w, https://substackcdn.com/image/fetch/$s_!Lr_z!,w_848,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 848w, https://substackcdn.com/image/fetch/$s_!Lr_z!,w_1272,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 1272w, https://substackcdn.com/image/fetch/$s_!Lr_z!,w_1456,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 1456w" sizes="100vw"><img src="https://substackcdn.com/image/fetch/$s_!Lr_z!,w_1456,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png" width="946" height="402" data-attrs="{&quot;src&quot;:&quot;https://substack-post-media.s3.amazonaws.com/public/images/e0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png&quot;,&quot;srcNoWatermark&quot;:null,&quot;fullscreen&quot;:null,&quot;imageSize&quot;:null,&quot;height&quot;:402,&quot;width&quot;:946,&quot;resizeWidth&quot;:null,&quot;bytes&quot;:18713,&quot;alt&quot;:null,&quot;title&quot;:null,&quot;type&quot;:&quot;image/png&quot;,&quot;href&quot;:null,&quot;belowTheFold&quot;:true,&quot;topImage&quot;:false,&quot;internalRedirect&quot;:&quot;https://pseudovariety.com/i/167475213?img=https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png&quot;,&quot;isProcessing&quot;:false,&quot;align&quot;:null,&quot;offset&quot;:false}" class="sizing-normal" alt="" srcset="https://substackcdn.com/image/fetch/$s_!Lr_z!,w_424,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 424w, https://substackcdn.com/image/fetch/$s_!Lr_z!,w_848,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 848w, https://substackcdn.com/image/fetch/$s_!Lr_z!,w_1272,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 1272w, https://substackcdn.com/image/fetch/$s_!Lr_z!,w_1456,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fe0425148-e083-4fb8-8953-c65e622f5f8e_946x402.png 1456w" sizes="100vw" loading="lazy"></picture><div class="image-link-expand"><div class="pencraft pc-display-flex pc-gap-8 pc-reset"><button tabindex="0" type="button" class="pencraft pc-reset pencraft icon-container restack-image"><svg role="img" width="20" height="20" viewBox="0 0 20 20" fill="none" stroke-width="1.5" stroke="var(--color-fg-primary)" stroke-linecap="round" stroke-linejoin="round" xmlns="http://www.w3.org/2000/svg"><g><title></title><path d="M2.53001 7.81595C3.49179 4.73911 6.43281 2.5 9.91173 2.5C13.1684 2.5 15.9537 4.46214 17.0852 7.23684L17.6179 8.67647M17.6179 8.67647L18.5002 4.26471M17.6179 8.67647L13.6473 6.91176M17.4995 12.1841C16.5378 15.2609 13.5967 17.5 10.1178 17.5C6.86118 17.5 4.07589 15.5379 2.94432 12.7632L2.41165 11.3235M2.41165 11.3235L1.5293 15.7353M2.41165 11.3235L6.38224 13.0882"></path></g></svg></button><button tabindex="0" type="button" class="pencraft pc-reset pencraft icon-container view-image"><svg xmlns="http://www.w3.org/2000/svg" width="20" height="20" viewBox="0 0 24 24" fill="none" stroke="currentColor" stroke-width="2" stroke-linecap="round" stroke-linejoin="round" class="lucide lucide-maximize2 lucide-maximize-2"><polyline points="15 3 21 3 21 9"></polyline><polyline points="9 21 3 21 3 15"></polyline><line x1="21" x2="14" y1="3" y2="10"></line><line x1="3" x2="10" y1="21" y2="14"></line></svg></button></div></div></div></a></figure></div><p>First, observe that </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(a^6)^2=a^{12}=a^{10}\\cdot a^2=a^4\\cdot a^2=a^6&quot;,&quot;id&quot;:&quot;HJFNOVENYO&quot;}" data-component-name="LatexBlockToDOM"></div><p>so a to the power of 6 is an idempotent. Now choose an arbitrary k, lets say k=5. From the diagram, we see that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(a^5)^2=a^{10}=a^4&quot;,&quot;id&quot;:&quot;HNDOHEYPDX&quot;}" data-component-name="LatexBlockToDOM"></div><p>so we have fallen &#8220;1 a short&#8220; of reaching around the loop to get back to a to the power of five by just multiplying by 5. We then would try a to the power of 6 next, since 6 is one greater than 5. This must work because we now walk 2 a&#8217;s further around the circle when squaring a to the power of 6, but we also must stretch an additional one a around to get to a to the power of 6. Try again with k=7 and we see that when walking a distance of 7 around the circle, we go one too far, so we should try k=7-1=6, which we know works.</p><p>Now that that is settled, return to the question at hand: showing that every element has an inverse. Luckily, now that this fact has been established, this is easy. For every a in M, there is some power k of a which is idempotent. We then get the following implication</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a^k=1\\Rightarrow a\\cdot a^{k-1}=a^{k-1}\\cdot a=1&quot;,&quot;id&quot;:&quot;DTFSWSCRFZ&quot;}" data-component-name="LatexBlockToDOM"></div><p>so the inverse of a is a to the power of k-1. Then every element a in M has an inverse, so M is a group! &#9633;</p><p>I now give an important characterization of groups.</p><h4>Theorem: A semigroup G is a group if and only if for any a and b in G, there exist x and y in G such that ax=b and ya=b.</h4><p>Proof: Exercise.</p><p>We say that a semigroup S is cancellative if it has the following property:</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\forall a,b,c\\in S,\\text{ }ca=cb\\text{ or }ac=bc\\Rightarrow a=b. &quot;,&quot;id&quot;:&quot;COLCBMINSY&quot;}" data-component-name="LatexBlockToDOM"></div><p>That is, you can strike the same element off of both sides of an equals sign and preserve equality. All groups are cancellative, since multiplication on both sides of an equals sign by the inverse of c and reducing will result in a=b.</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;ca=cb\\Rightarrow c^{-1}ca=c^{-1}cb\\Rightarrow 1\\cdot a=1\\cdot b\\Rightarrow a=b.&quot;,&quot;id&quot;:&quot;NUKZFVEVJX&quot;}" data-component-name="LatexBlockToDOM"></div><p>We then have the following theorem.</p><h4>Theorem: A finite cancellative semigroup S is a group.</h4><p>Proof: Let a be in S. I use the following notation</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;aS=\\{as:s\\in S\\}.&quot;,&quot;id&quot;:&quot;KLTKSHKAHD&quot;}" data-component-name="LatexBlockToDOM"></div><p>Sa is defined similarly with multiplication on the right. By the contrapositive of the cancellation property, we have that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a\\neq b\\Rightarrow ca\\neq cb&quot;,&quot;id&quot;:&quot;XMPLZEHESR&quot;}" data-component-name="LatexBlockToDOM"></div><p>so |aS| is greater than or equal to |S|. Since S is closed, we have that |aS|=|S|. Similarly, |Sa|=|S|. Then for any b in S, there exists some x and y in S such that ax=b and ya=b. &#9633;</p>]]></content:encoded></item><item><title><![CDATA[2. Binary Operations, Closure, and Semigroups]]></title><description><![CDATA[The basic definitions of algebra.]]></description><link>https://pseudovariety.com/p/2-binary-operations-and-closure</link><guid isPermaLink="false">https://pseudovariety.com/p/2-binary-operations-and-closure</guid><dc:creator><![CDATA[Peter Alspaugh]]></dc:creator><pubDate>Thu, 03 Jul 2025 03:38:09 GMT</pubDate><enclosure url="https://substackcdn.com/image/fetch/$s_!Lxrf!,w_256,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2F9ad4f314-8f59-468a-8bf4-e1a46d66a1f0_382x382.png" length="0" type="image/jpeg"/><content:encoded><![CDATA[<p>A function</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;* : A\\times A\\to A&quot;,&quot;id&quot;:&quot;FPGWHMUXSD&quot;}" data-component-name="LatexBlockToDOM"></div><p>is called a binary operation on A and a subset B of A is said to be closed under the binary operation if for any a and b in B &#8727;(a,b) is in B. We usually write a&#8727;b  or even ab (and we say &#8220;a times b&#8220; for a generic operation) for &#8727;(a,b). A magma is a set A equipped with a binary operation &#8727;, written (A,&#8727;) or simply A when &#8727; is understood. </p><p>An operation on A is said to be associative if for any a, b, and c in A,</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(ab)c=a(bc).&quot;,&quot;id&quot;:&quot;WSQXETOSBT&quot;}" data-component-name="LatexBlockToDOM"></div><p>In this case, we call A a <em>semigroup</em>. if B is a subset of A which is closed under the same operation as A, then we call B a <em>subsemigroup</em> of A.</p><p>It is important to note that associativity in no way implies commutativity, the property where for any a and b in A, ab=ba. I now give some examples of semigroups. It is left to the reader to prove that they are indeed semigroups.</p><ul><li><p>The set of real numbers under addition or multiplication</p></li><li><p>The set of integers under addition/multiplication is a subsemigroup of the real numbers under addition/multiplication</p></li><li><p>The set of even integers under addition or multiplication is a subsemigroup of the set of integers under addition/multiplication</p></li><li><p>The set of all functions from a set to itself</p></li><li><p>The set of all bijections from a set to itself is a subsemigroup of the set of all functions from a set to itself</p></li><li><p>Square matrices under matrix multiplication or addition</p></li></ul><p>Notably, the set of odd integers is NOT a semigroup, since 1+1=2, which is not odd. Finite semigroups may be described with a multiplication table where each entry represents the product of its row label (on the left) by its column label (on the right). A multiplication table can be checked for associativity via <a href="https://en.wikipedia.org/wiki/Light%27s_associativity_test">Light&#8217;s test</a>.</p><h4>Exercise: Consider the set S composed of rock, paper, and scissors defined by the below multiplication table. Is S a semigroup? Is S commutative?</h4><div class="captioned-image-container"><figure><a class="image-link image2" target="_blank" href="https://substackcdn.com/image/fetch/$s_!VOcb!,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png" data-component-name="Image2ToDOM"><div class="image2-inset"><picture><source type="image/webp" srcset="https://substackcdn.com/image/fetch/$s_!VOcb!,w_424,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 424w, https://substackcdn.com/image/fetch/$s_!VOcb!,w_848,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 848w, https://substackcdn.com/image/fetch/$s_!VOcb!,w_1272,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 1272w, https://substackcdn.com/image/fetch/$s_!VOcb!,w_1456,c_limit,f_webp,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 1456w" sizes="100vw"><img src="https://substackcdn.com/image/fetch/$s_!VOcb!,w_1456,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png" width="482" height="174" data-attrs="{&quot;src&quot;:&quot;https://substack-post-media.s3.amazonaws.com/public/images/bb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png&quot;,&quot;srcNoWatermark&quot;:null,&quot;fullscreen&quot;:null,&quot;imageSize&quot;:null,&quot;height&quot;:174,&quot;width&quot;:482,&quot;resizeWidth&quot;:null,&quot;bytes&quot;:12784,&quot;alt&quot;:null,&quot;title&quot;:null,&quot;type&quot;:&quot;image/png&quot;,&quot;href&quot;:null,&quot;belowTheFold&quot;:true,&quot;topImage&quot;:false,&quot;internalRedirect&quot;:&quot;https://pseudovariety.com/i/167408351?img=https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png&quot;,&quot;isProcessing&quot;:false,&quot;align&quot;:null,&quot;offset&quot;:false}" class="sizing-normal" alt="" srcset="https://substackcdn.com/image/fetch/$s_!VOcb!,w_424,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 424w, https://substackcdn.com/image/fetch/$s_!VOcb!,w_848,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 848w, https://substackcdn.com/image/fetch/$s_!VOcb!,w_1272,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 1272w, https://substackcdn.com/image/fetch/$s_!VOcb!,w_1456,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2Fbb641023-2cf0-48f3-ae38-24ebb4e80821_482x174.png 1456w" sizes="100vw" loading="lazy"></picture><div></div></div></a></figure></div><p>We can also use the Cartesian product to construct a new semigroup from two existing semigroups. </p><h4>Proposition: Let S and T be semigroups (written multiplicatively). Then the cartesian product of S and T is a semigroup with the following multiplication (s&#8217;s are in S and t&#8217;s in T):</h4><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(s_1,t_1)(s_2,t_2)=(s_1s_2,t_1t_2).&quot;,&quot;id&quot;:&quot;YHCXVREHPN&quot;}" data-component-name="LatexBlockToDOM"></div><p>Proof: It is evident from the definition that the Cartesian product is closed under multiplication and that multiplication is defined for all pairs. It then remains to check that multiplication is associative. We compute:</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;((s_1,t_1)(s_2,t_2))(s_3,t_3)=(s_1s_2,t_1t_2)(s_3,t_3)=((s_1s_2)s_3,(t_1t_2)t_3)&quot;,&quot;id&quot;:&quot;KOSQKXHYOG&quot;}" data-component-name="LatexBlockToDOM"></div><p>and </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(s_1,t_1)((s_2,t_2)(s_3,t_3))=(s_1,t_1)(s_2s_3,t_2t_3)=(s_1(s_2s_3),t_1(t_2t_3)).&quot;,&quot;id&quot;:&quot;UBXDUGDYND&quot;}" data-component-name="LatexBlockToDOM"></div><p>Since multiplication in S and T is associative, we have that</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;((s_1s_2)s_3,(t_1t_2)t_3)=(s_1(s_2s_3),t_1(t_2t_3))&quot;,&quot;id&quot;:&quot;ATNVYRSKLJ&quot;}" data-component-name="LatexBlockToDOM"></div><p>so multiplication in the cartesian product is associative. &#9633;</p>]]></content:encoded></item><item><title><![CDATA[1. Preliminaries]]></title><description><![CDATA[Let's be clear on the basics.]]></description><link>https://pseudovariety.com/p/preliminaries</link><guid isPermaLink="false">https://pseudovariety.com/p/preliminaries</guid><dc:creator><![CDATA[Peter Alspaugh]]></dc:creator><pubDate>Wed, 02 Jul 2025 22:07:10 GMT</pubDate><enclosure url="https://substackcdn.com/image/fetch/$s_!Lxrf!,w_256,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2F9ad4f314-8f59-468a-8bf4-e1a46d66a1f0_382x382.png" length="0" type="image/jpeg"/><content:encoded><![CDATA[<p>This section is intended to be a quick review. if you are not already comfortable with the material, then it may be wise to review it before reading any further.</p><h2>Sets</h2><p>A set is a well-defined collection of objects that can be described precisely, perhaps by a shared property. For two sets A and B, we say that A is a <em>subset</em> of B and write</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;A\\subseteq B&quot;,&quot;id&quot;:&quot;UUZACZVMQT&quot;}" data-component-name="LatexBlockToDOM"></div><p>if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a\\in A\\Rightarrow a\\in B.&quot;,&quot;id&quot;:&quot;XLHTGREKAG&quot;}" data-component-name="LatexBlockToDOM"></div><p>If A is a subset of B, but</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\exists b\\in B \\text{ s.t. } b\\notin A&quot;,&quot;id&quot;:&quot;RQQESNGZWZ&quot;}" data-component-name="LatexBlockToDOM"></div><p>then we write </p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;A\\subsetneq B&quot;,&quot;id&quot;:&quot;COUXQFZLQB&quot;}" data-component-name="LatexBlockToDOM"></div><p>and say that A is a <em>proper</em> subset of B.</p><h2>Cartesian Product</h2><p>For two sets A and B, we denote the <em>Cartesian product</em> of A and B to be</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;A\\times B :=\\{(a,b):a\\in A,\\text{ }b\\in B\\}.&quot;,&quot;id&quot;:&quot;EJIELDRLWD&quot;}" data-component-name="LatexBlockToDOM"></div><p>For a finite collection of sets (such that i is bounded by 1 and n), we may write</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;A_1\\times A_2\\times\\cdots\\times A_n=\\{(a_1,a_2,\\dots, a_n): a_i\\in A_i,\\text{ }1\\leq i\\leq n\\}.&quot;,&quot;id&quot;:&quot;LOHEWPJAVY&quot;}" data-component-name="LatexBlockToDOM"></div><p>For an arbitrary collection of sets indexed by a set I, we may write</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\prod_{i\\in I} A_i=\\{(a_i)_{i\\in I}: a_i\\in A_i\\}.&quot;,&quot;id&quot;:&quot;ZYWLQGYAHE&quot;}" data-component-name="LatexBlockToDOM"></div><p></p><h2>Relations</h2><p>A relation (between sets A and B) is a subset of the Cartesian product (of sets A and B). For a relation</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\mathcal{R}\\subseteq A\\times B&quot;,&quot;id&quot;:&quot;TCCAHDIYBV&quot;}" data-component-name="LatexBlockToDOM"></div><p>if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(a,b)\\in\\mathcal{R}&quot;,&quot;id&quot;:&quot;ETBYVKUEND&quot;}" data-component-name="LatexBlockToDOM"></div><p>then we may for convenience write</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a\\mathcal{R}b&quot;,&quot;id&quot;:&quot;HXQLFPJLPC&quot;}" data-component-name="LatexBlockToDOM"></div><p>and say that a and b are related under R. One of the most commonly encountered types of relations is that of a function. A function f with domain A and codomain B, written</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;f:A\\to B&quot;,&quot;id&quot;:&quot;YITLWBUMVP&quot;}" data-component-name="LatexBlockToDOM"></div><p>is a subset</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;f\\subseteq A\\times B\\text{ s.t. } \\forall a\\in A,\\text{ } (a,b_1),(a,b_2)\\in f\\Rightarrow b_1=b_2.&quot;,&quot;id&quot;:&quot;GDYRYVFIUF&quot;}" data-component-name="LatexBlockToDOM"></div><p>That is, any a in A cannot be related to two distinct elements in B. In secondary school algebra, one learns that a function (from the Real line to itself) must &#8220;pass the vertical line test&#8220; to be a function. This just simply means that any real number cannot be sent to two distinct real numbers under a function. We also use the following notation, which may also be used for relations in general.</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(a,b)\\in f\\Leftrightarrow f:a\\mapsto b\\Leftrightarrow f(a)=b&quot;,&quot;id&quot;:&quot;ZAVHYGJJSP&quot;}" data-component-name="LatexBlockToDOM"></div><p>Reframed in this notation, the definition of a function is familiar.</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a=b\\Rightarrow f(a)=f(b)&quot;,&quot;id&quot;:&quot;QNRZFUKYTA&quot;}" data-component-name="LatexBlockToDOM"></div><p>The range of f is the set of all b&#8217;s in B that are mapped onto by the function.</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\text{Range}f:=\\{b\\in B:\\text{ }\\exists a\\in A\\text{ s.t. }f(a)=b\\}&quot;,&quot;id&quot;:&quot;VFYGQKDOYR&quot;}" data-component-name="LatexBlockToDOM"></div><p>Just like functions, all relations can be composed. Formally, the composition of relations R and S is defined below. Let R be a relation from A to B and S be a relation from B to C.</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;S\\circ R:=\\{(a,c)\\subseteq A\\times C:\\text{ }\\exists b\\in B\\text{ s.t. }(a,b)\\in R\\text{ and }(b,c)\\in S\\}.&quot;,&quot;id&quot;:&quot;KPIYEFFUIA&quot;}" data-component-name="LatexBlockToDOM"></div><p>All this means is that whenever R(a)=b and S(b)=c then</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(S\\circ R)(a)=c.&quot;,&quot;id&quot;:&quot;XNLBZZPVJW&quot;}" data-component-name="LatexBlockToDOM"></div><p>You may notice that function and relation composition is written &#8220;backwards&#8220; relative to how we think about it. This is done to fit with the f(a)=b notation, however some authors write composition the &#8220;correct way&#8220;.</p><p>A function f from A to B is called one-to-one or injective if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a\\neq b\\Rightarrow f(a)\\neq f(b).&quot;,&quot;id&quot;:&quot;YEQSACGSRL&quot;}" data-component-name="LatexBlockToDOM"></div><p>It is called onto or surjective if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\forall b\\in B,\\text{ }\\exists a\\in A\\text{ s.t. } f(a)=b.&quot;,&quot;id&quot;:&quot;WUEIVPPTKS&quot;}" data-component-name="LatexBlockToDOM"></div><p>A function which is both one-to-one and onto is called a one-to-one correspondence or a bijection.</p><p>We denote the <em>cardinality </em>of a set A as |A| and we define</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;|A|\\leq |B|&quot;,&quot;id&quot;:&quot;CATVBQVGIV&quot;}" data-component-name="LatexBlockToDOM"></div><p>if there exists an injection from A to B or if there exists a surjection from B to A. Hence two sets have equal cardinality if there exists a bijection between them. If a set has finitely many elements n, then it is in bijection with the set of positive integers less than or equal to n, and we may say that it has cardinality n.</p><p>The identity map on a set A is the function</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;f:A\\to A\\text{ s.t. } a\\mapsto a\\text{ }\\forall a\\in A&quot;,&quot;id&quot;:&quot;FGDYAJWZKY&quot;}" data-component-name="LatexBlockToDOM"></div><p>and is denoted id (when A is understood). Let</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;f:A\\to B&quot;,&quot;id&quot;:&quot;UILWUOSNMG&quot;}" data-component-name="LatexBlockToDOM"></div><p>and</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;g:B\\to A&quot;,&quot;id&quot;:&quot;UVHHLUVXHK&quot;}" data-component-name="LatexBlockToDOM"></div><p>be functions. g is called a left (resp. right) inverse of f if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;g\\circ f=\\text{id} \\text{ (resp. }f\\circ g=\\text{id )}.&quot;,&quot;id&quot;:&quot;NIEYBYTPMF&quot;}" data-component-name="LatexBlockToDOM"></div><h4>Theorem: A function f has a left inverse iff it is injective and a right inverse iff it is onto.</h4><p>Proof: Exercise.</p><h3>Equivalence Relations</h3><p>Consider the following relation on a set A.</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;\\sim\\subseteq A\\times A&quot;,&quot;id&quot;:&quot;JBVQYRRWRL&quot;}" data-component-name="LatexBlockToDOM"></div><p>We say that ~ is <em>reflexive</em> if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a\\in A\\Rightarrow (a,a)\\in\\sim\\Leftrightarrow a\\sim a,&quot;,&quot;id&quot;:&quot;WSBFDUENGS&quot;}" data-component-name="LatexBlockToDOM"></div><p><em>symmetric</em> if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;a_1\\sim a_2\\Leftrightarrow (a_1,a_2)\\in\\sim\\Rightarrow (a_2,a_1)\\in\\sim \\Leftrightarrow a_2\\sim a_1,&quot;,&quot;id&quot;:&quot;UHQRJEMKFF&quot;}" data-component-name="LatexBlockToDOM"></div><p>and <em>transitive </em>if</p><div class="latex-rendered" data-attrs="{&quot;persistentExpression&quot;:&quot;(a_1,a_2),(a_2,a_3)\\in\\sim\\Rightarrow (a_1,a_3)\\in\\sim.&quot;,&quot;id&quot;:&quot;CKPZAOLQWD&quot;}" data-component-name="LatexBlockToDOM"></div><p>A curious reader is encouraged to consider a finite set A and find a way to represent reflexive and symmetric relations in terms of a binary matrix. How could one interpret matrix multiplication in this context?</p><p>A partition of a set A is a collection of disjoint subsets of A such that their union equals A.</p><h4>Theorem: Every equivalence relation defines a partition and every partition defines an equivalence relation.</h4><p>Otherwise stated, every equivalence relation <em>is</em> a partition and every partition <em>is </em>an equivalence relation.</p><p>Proof: (&#10233;) Let R be an equivalence relation and for all a&#8217;s in A, let the equivalence class of a be the set of all elements which are related to a and denote it [a]. From reflexivity, every a is contained in some equivalence class. Hence the union of all such classes contains A. We also see that by symmetry, if b is in [a], then a is in [b], so [a]=[b] must be true. Suppose that for two elements a and b in A, [a]&#8800;[b] and suppose to the contrary that there exists some c in the intersection of [a] and [b]. By transitivity, a and b must be related, so [a]=[b], which is a contradiction. Then each equivalence class is disjoint.</p><p>(&#10232;) Consider a partition of A into disjoint subsets. We construct a relation R such that for each such subset P and each a and b in P, R contains (a,b) and (b,a). Since every a in A is contained in some subset in the partition, (a,a) is in R, so R is reflexive. If (a,b) is in R, then a and b must be in the same subset P, so naturally a is in the same subset as b and therefore (b,a) is in R. Then R is symmetric. Now suppose that both (a,b) and (b,c) are in R. Then a ,b, and c must all be in the same subset, so (a,c) is in R. Then R is transitive and we may conclude that it is an equivalence relation. &#9633;</p><p>That sums up the preliminaries section. We may now move on to Algebra&#8230;.</p>]]></content:encoded></item><item><title><![CDATA[0. Outline for the Notes]]></title><description><![CDATA[What are we trying to do here?]]></description><link>https://pseudovariety.com/p/outline</link><guid isPermaLink="false">https://pseudovariety.com/p/outline</guid><dc:creator><![CDATA[Peter Alspaugh]]></dc:creator><pubDate>Wed, 02 Jul 2025 21:45:47 GMT</pubDate><enclosure url="https://substackcdn.com/image/fetch/$s_!Lxrf!,w_256,c_limit,f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2F9ad4f314-8f59-468a-8bf4-e1a46d66a1f0_382x382.png" length="0" type="image/jpeg"/><content:encoded><![CDATA[<p>This series will be focused on basic Group Theory, but with a slight twist. Semigroups will also be introduced as a way to give intuition about certain ideas in Group Theory, which is not common in Algebra textbooks. It is not intended to be a course on Semigroup Theory&#8212;or even Group Theory&#8212;but rather as a companion of sorts to a first course in Group Theory. Recommended reading for such a first course would be <em>A First Course in Abstract Algebra</em> by Fraleigh, which is commonly used, but is far from the only option.</p><p>Below is an outline of topics:</p><ol><li><p><a href="https://pseudovariety.com/p/preliminaries">preliminaries</a>: a reminder of na&#239;ve set theory, and definitions of relations, functions, and the cartesian product</p></li><li><p><a href="https://pseudovariety.com/p/2-binary-operations-and-closure">operations, closure, and semigroups</a></p></li><li><p><a href="https://pseudovariety.com/p/3-identity-elements-and-monoids-inverses">identity elements and monoids, inverses and groups</a></p></li><li><p>generators and Cayley graphs</p></li><li><p>Green&#8217;s relations, and the ax=b &amp; ya=b definition of a group</p></li><li><p>monogenic semigroups and cyclic groups</p></li><li><p>homomorphisms, congruences, and isomorphisms</p></li><li><p>the First Isomorphism Theorem</p></li><li><p>transformation semigroups and symmetric groups</p></li><li><p>actions on a set</p></li><li><p>important theorems: Cayley, Lagrange, Cauchy</p></li><li><p>free objects and presentations</p></li></ol><p>The outline may be changed during development to avoid circular logic and for the convenience of the reader, as I am not directly mimicking any particular textbook.</p>]]></content:encoded></item></channel></rss>