Normality satisfies transfer condition: Difference between revisions

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This article gives the statement, and possibly proof, of a subgroup property satisfying a subgroup metaproperty
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This article gives the statement, and possibly proof, of a basic fact in group theory.
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Statement

Verbal statement

If a subgroup is normal in the group, its intersection with any other subgroup is normal in that subgroup.

Symbolic statement

Let $H \leq G$ be a normal subgroup and let $K$ be any subgroup of $G$ . Then, $H \cap K ◃ K$ .

Property-theoretic statement

The subgroup property of being normal satisfies the transfer condition.

Definitions used

Normal subgroup

A subgroup $H$ of a group $G$ is said to be normal if for any $g \in G$ and $h \in H$ , $g h g^{- 1} \in H$ .

Transfer condition

A subgroup property $p$ is said to satisfy transfer condition if whenever $H, K$ are subgroups of $G$ and $H$ has property $p$ in $G$ , $H \cap K$ has property $p$ in $K$ .

Generalizations

Stronger metaproperties satisfied by normality

Normality satisfies inverse image condition

Proof

Hands-on proof

Given: A group $G$ , a normal subgroup $H ◃ G$ and a subgroup $K \leq G$

To prove: $H \cap K ◃ K$ . In other words, we need to prove that given any $g \in K$ and $h \in H \cap K$ , $g h g^{- 1} \in H \cap K$ .

Proof: Since $h \in H \cap K$ , we in particular have $h \in H$ . Since $H ◃ G$ (viz $H$ is normal in $G$ ), $g h g^{- 1} \in H$ .

But we also have that $g \in K$ and $h \in K$ . Since $K$ is a subgroup, $g h g^{- 1} \in K$ .

Combining these two facts, $g h g^{- 1} \in H \cap K$ .

References

Textbook references

Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 88, Exercise 24
Topics in Algebra by I. N. Herstein, ^{More info}, Page 53, Problem 5

@@ Line 37: / Line 37: @@
 ===Hands-on proof===
-Suppose <math>H \triangleleft G</math> and <math>K \le G</math>. We need to prove that <math>H \cap K \triangleleft K</math>. In other words, we need to prove that given any <math>g \in K</math> and <math>h \in H \cap K</math>, <math>ghg^{-1} \in H \cap K</math>.
+''Given'': A group <math>G</math>, a normal subgroup <math>H \triangleleft G</math> and a subgroup <math>K \le G</math>
-Here's how the proof proceeds. Since <math>h \in H \cap K</math>, we in particular have <math>h \in H</math>. Since <math>H \triangleleft G</math> (viz <math>H</math> is normal in <math>G</math>), <math>ghg^{-1} \in H</math>.
+''To prove'': <math>H \cap K \triangleleft K</math>. In other words, we need to prove that given any <math>g \in K</math> and <math>h \in H \cap K</math>, <math>ghg^{-1} \in H \cap K</math>.
+''Proof'': Since <math>h \in H \cap K</math>, we in particular have <math>h \in H</math>. Since <math>H \triangleleft G</math> (viz <math>H</math> is normal in <math>G</math>), <math>ghg^{-1} \in H</math>.
 But we also have that <math>g \in K</math> and <math>h \in K</math>. Since <math>K</math> is a subgroup, <math>ghg^{-1} \in K</math>.
 Combining these two facts, <math>ghg^{-1} \in H \cap K</math>.
+==References==
+===Textbook references===
+* {{booklink-stated|DummitFoote}}, Page 88, Exercise 24
+* {{booklink-stated|Herstein}}, Page 53, Problem 5