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Why is the intersection of two normal subgroups a normal subgroup?

Contents

  • Why is the intersection of two normal subgroups a normal subgroup?
    • Intersection of two subgroups is a subgroup
    • Intersection of two normal subgroups is normal

Let \((G, \cdot)\) be a group and \(X \lhd G\) and \(Y \lhd G\) be two normal subgroups.

I will show this in two steps:

  1. Show that $X \cap Y$ is a group
  2. Show that $X \cap Y$ is a normal group of $(G, \cdot)$

Intersection of two subgroups is a subgroup

Theorem: \((X \cap Y) \leq G\)

Proof:

\(X \cap Y\) is not empty: \(e_G \in X \land e_G \in Y \Rightarrow e_G \in (X \cap Y)\)

\(X \cap Y\) has inverse elements. Let \(a \in (X \cap Y)\). As a is in \(X\) and \(X\) is a group, \(a^{-1} \in X\). The same is true for \(Y\). So: \(\forall a \in (X \cap Y) \exists a^{-1} \in (X \cap Y): a \cdot a^{-1} = a^{-1} \cdot a = e_G\)

\(\forall a,b \in (X \cap Y): a \cdot b^{-1} \in (X \cap Y)\), because both, \(a\) and \(b^{-1}\) are in \(X\). As \(X\) is a group, the result has to be in \(X\). Same argumentation for \(Y\). Then the result is in \(X\) and \(Y \blacksquare\)

Intersection of two normal subgroups is normal

First the definition of a normal subgroup:

Let $N \leq G$ be a subgroup of $G$. $N \lhd G :\Leftrightarrow \forall n \in N \forall g \in G: g \cdot n \cdot g^{-1} \in N$

Theorem: \((X \cap Y) \lhd G\)

Proof:

\(X \cap Y\) is a subgroup of \(G\) as I have proved above.

\(\forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in X\) and \(\forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in Y\)

\(\Rightarrow \forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in (X \cap Y)\)

\(\Rightarrow (X \cap Y) \lhd G \blacksquare\)


Published

Aug 25, 2013
by Martin Thoma

Category

Mathematics

Tags

  • Algebra 6
  • mathematics 59

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