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

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

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