<?xml version="1.0"?>
<feed xmlns="http://www.w3.org/2005/Atom" xml:lang="en">
	<id>https://groupprops.subwiki.org/w/index.php?action=history&amp;feed=atom&amp;title=Simplicity_is_directed_union-closed</id>
	<title>Simplicity is directed union-closed - Revision history</title>
	<link rel="self" type="application/atom+xml" href="https://groupprops.subwiki.org/w/index.php?action=history&amp;feed=atom&amp;title=Simplicity_is_directed_union-closed"/>
	<link rel="alternate" type="text/html" href="https://groupprops.subwiki.org/w/index.php?title=Simplicity_is_directed_union-closed&amp;action=history"/>
	<updated>2026-09-08T06:59:20Z</updated>
	<subtitle>Revision history for this page on the wiki</subtitle>
	<generator>MediaWiki 1.41.2</generator>
	<entry>
		<id>https://groupprops.subwiki.org/w/index.php?title=Simplicity_is_directed_union-closed&amp;diff=12790&amp;oldid=prev</id>
		<title>Vipul: New page: {{group metaproperty satisfaction| property = simple group| metaproperty = directed union-closed group property}}  ==Statement==  ===Verbal statement===  A directed union of simple subgrou...</title>
		<link rel="alternate" type="text/html" href="https://groupprops.subwiki.org/w/index.php?title=Simplicity_is_directed_union-closed&amp;diff=12790&amp;oldid=prev"/>
		<updated>2008-09-07T17:33:07Z</updated>

		<summary type="html">&lt;p&gt;New page: {{group metaproperty satisfaction| property = simple group| metaproperty = directed union-closed group property}}  ==Statement==  ===Verbal statement===  A directed union of simple subgrou...&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;{{group metaproperty satisfaction|&lt;br /&gt;
property = simple group|&lt;br /&gt;
metaproperty = directed union-closed group property}}&lt;br /&gt;
&lt;br /&gt;
==Statement==&lt;br /&gt;
&lt;br /&gt;
===Verbal statement===&lt;br /&gt;
&lt;br /&gt;
A directed union of simple subgroups is simple.&lt;br /&gt;
&lt;br /&gt;
===Statement with symbols===&lt;br /&gt;
&lt;br /&gt;
Suppose &amp;lt;math&amp;gt;G&amp;lt;/math&amp;gt; is a group, &amp;lt;math&amp;gt;I&amp;lt;/math&amp;gt; is a nonempty directed set, and &amp;lt;math&amp;gt;H_i, i \in I&amp;lt;/math&amp;gt; is a collection of subgroups such that &amp;lt;math&amp;gt;i &amp;lt; j \implies H_i \le H_j&amp;lt;/math&amp;gt;. Then, if all the &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt;s are simple, so is their union.&lt;br /&gt;
&lt;br /&gt;
==Definitions used==&lt;br /&gt;
&lt;br /&gt;
===Directed set===&lt;br /&gt;
&lt;br /&gt;
A directed set is a partially ordered set &amp;lt;math&amp;gt;(I, \le)&amp;lt;/math&amp;gt; such that if &amp;lt;math&amp;gt;i,j \in I&amp;lt;/math&amp;gt;, there exists &amp;lt;math&amp;gt;k \in I&amp;lt;/math&amp;gt;, such that &amp;lt;math&amp;gt;i \le k, j \le k&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
===Simple group===&lt;br /&gt;
&lt;br /&gt;
A nontrivial group is simple if it has no proper nontrivial [[normal subgroup]].&lt;br /&gt;
&lt;br /&gt;
==Facts used==&lt;br /&gt;
&lt;br /&gt;
# [[Directed union of subgroups is subgroup]]&lt;br /&gt;
# [[Normality satisfies transfer condition]]: The intersection of a normal subgroup with another subgroup is normal in that subgroup.&lt;br /&gt;
&lt;br /&gt;
==Proof==&lt;br /&gt;
&lt;br /&gt;
&amp;#039;&amp;#039;&amp;#039;Given&amp;#039;&amp;#039;&amp;#039;: A group &amp;lt;math&amp;gt;G&amp;lt;/math&amp;gt;, a collection of simple subgroups &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt; indexed by a directed set &amp;lt;math&amp;gt;I&amp;lt;/math&amp;gt;, such that &amp;lt;math&amp;gt;i &amp;lt; j \implies H_i \le H_j&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&amp;#039;&amp;#039;&amp;#039;To prove&amp;#039;&amp;#039;&amp;#039;: The union of &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt; is a simple subgroup.&lt;br /&gt;
&lt;br /&gt;
&amp;#039;&amp;#039;&amp;#039;Proof&amp;#039;&amp;#039;&amp;#039;: Suppose &amp;lt;math&amp;gt;H&amp;lt;/math&amp;gt; is the union of the &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt;s. By fact (1), &amp;lt;math&amp;gt;H&amp;lt;/math&amp;gt; is a subgroup, so it suffices to show that &amp;lt;math&amp;gt;H&amp;lt;/math&amp;gt; is simple. We do this by showing that any normal subgroup &amp;lt;math&amp;gt;N&amp;lt;/math&amp;gt; of &amp;lt;math&amp;gt;H&amp;lt;/math&amp;gt; is either trivial or equal to &amp;lt;math&amp;gt;H&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
For each &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt;, consider &amp;lt;math&amp;gt;N \cap H_i&amp;lt;/math&amp;gt;. By fact (2), this is a normal subgroup of &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt;, hence is either trivial or equals the whole of &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt;. &lt;br /&gt;
Suppose &amp;lt;math&amp;gt;N \cap H_i = H_i&amp;lt;/math&amp;gt;. Then, &amp;lt;math&amp;gt;N&amp;lt;/math&amp;gt; is a normal subgroup of &amp;lt;math&amp;gt;H&amp;lt;/math&amp;gt; containing &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Now, consider any subgroup &amp;lt;math&amp;gt;H_j, j \ne i&amp;lt;/math&amp;gt;. By the definition of directed set, there exists &amp;lt;math&amp;gt;k \in I&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;i,j &amp;lt; k&amp;lt;/math&amp;gt;. So, &amp;lt;math&amp;gt;H_k&amp;lt;/math&amp;gt; contains both &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;H_j&amp;lt;/math&amp;gt;. Thus, the intersection &amp;lt;math&amp;gt;N \cap H_k&amp;lt;/math&amp;gt; is nontrivial (since the intersection contains &amp;lt;math&amp;gt;H_i&amp;lt;/math&amp;gt;). By the simplicity of &amp;lt;math&amp;gt;H_k&amp;lt;/math&amp;gt;, and fact (2) again, &amp;lt;math&amp;gt;N \cap H_k = H_k&amp;lt;/math&amp;gt;, so &amp;lt;math&amp;gt;H_k \le N&amp;lt;/math&amp;gt;. In particular, &amp;lt;math&amp;gt;H_j \le N&amp;lt;/math&amp;gt;. Thus, &amp;#039;&amp;#039;every&amp;#039;&amp;#039; &amp;lt;math&amp;gt;H_j, j \in I&amp;lt;/math&amp;gt;, is contained in &amp;lt;math&amp;gt;N&amp;lt;/math&amp;gt;, Thus, &amp;lt;math&amp;gt;N = H&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Thus, if &amp;lt;math&amp;gt;N \cap H_i = H_i&amp;lt;/math&amp;gt; for any &amp;lt;math&amp;gt;i \in I&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;N = H&amp;lt;/math&amp;gt;. This leaves the case that &amp;lt;math&amp;gt;N \cap H_i&amp;lt;/math&amp;gt; is trivial for every &amp;lt;math&amp;gt;i \in I&amp;lt;/math&amp;gt;, forcing &amp;lt;math&amp;gt;N \cap (\bigcup_i) H_i&amp;lt;/math&amp;gt; to be trivial, and thus forcing &amp;lt;math&amp;gt;N \cap H = N&amp;lt;/math&amp;gt; to be trivial.&lt;/div&gt;</summary>
		<author><name>Vipul</name></author>
	</entry>
</feed>