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:
- Show that \(X \cap Y\) is a group
- 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:
\(N \lhd G :\Leftrightarrow \forall n \in N \forall g \in G: g \cdot n \cdot g^{-1} \in N\)
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 \(\forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in Y\)
\(\Rightarrow \forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in (X \cap Y)\)
\(\Rightarrow (X \cap Y) \lhd G \blacksquare\)