Intersection of subgroups is subgroup: Difference between revisions

Revision as of 12:30, 22 March 2008

This article gives the statement, and possibly proof, of a basic fact in group theory.
View a complete list of basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

Statement

Verbal statement

The intersection of any arbitrary collection of subgroups of a group is again a subgroup.

Symbolic statement

Let $H_{i}$ be an arbitrary collection of subgroups of a group $G$ indexed by $i \in I$ . Then, $⋂_{i \in I} H_{i}$ is again a subgroup of $G$ .

Proof

Given: Let $H_{i}$ be an arbitrary collection of subgroups of a group $G$ indexed by $i \in I$ . Let us denote $H = ⋂_{i \in I} H_{i}$ .

To prove: We need to show that $H$ is a subgroup. In other words, we need to show the following:

$e \in H$
If $g \in H$ then $g^{- 1} \in H$
If $g, h \in H$ then $g h \in H$

Proof: Let's prove these one by one:

Since $e \in H_{i}$ for every $i$ , $e \in H$

Take $g \in H$ . Then $g \in H_{i}$ for every $i \in I$ . Since each $H_{i}$ is a subgroup, $g^{- 1} \in H_{i}$ for each $i \in I$ . Thus, $g^{- 1} \in H$ .
Take $g, h \in H$ . Then $g, h \in H_{i}$ for every $i$ , so $g h \in H_{i}$ for every $i \in I$ . Thus $g h \in H$ .

@@ Line 12: / Line 12: @@
 ==Proof==
-Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Let us denote <math>H = \bigcap_{i \in I} H_i</math>. We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:
+''Given'': Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Let us denote <math>H = \bigcap_{i \in I} H_i</math>.
+''To prove'': We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:
 # <math>e \in H</math>
@@ Line 18: / Line 20: @@
 # If <math>g,h \in H</math> then <math>gh \in H</math>
-Let's prove these one by one:
+''Proof'': Let's prove these one by one:
 * Since <math>e \in H_i</math> for every <math>i</math>, <math>e \in H</math>
 # Take <math>g \in H</math>. Then <math>g \in H_i</math> for every <math>i \in I</math>. Since each <math>H_i</math> is a subgroup, <math>g^{-1} \in H_i</math> for each <math>i \in I</math>. Thus, <math>g^{-1} \in H</math>.
 # Take <math>g,h \in H</math>. Then <math>g,h \in H_i</math> for every <math>i</math>, so <math>gh \in H_i</math> for every <math>i \in I</math>. Thus <math>gh \in H</math>.