Prove that the union of two normal subgroups is also a normal subgroup I know that the statement "the union of two normal subgroups is also a normal subgroup" is false.

Is there a counterexample to show this?

I can prove that the intersection is normal, but I can't disprove this.
$3\mathbb{Z}$ and $2\mathbb{Z}$ are subgroups of $(\mathbb{Z}, +)$ but the union has elements $2$ and $3$. But $2 + 3 = 5 \notin 3\mathbb{Z} \cup 2\mathbb{Z}$. $(\mathbb{Z}, +)$ is cyclic, and hence abelian, and hence all subgroups of $(\mathbb{Z}, +)$ are normal. But the union of these 2 subgroups is not even a subgroup.
Consider $(\mathbb{Z}, +)$, the group of integers under addition. We can prove that $(2\mathbb{Z}, +)$ and $(3\mathbb{Z},+)$ are normal subgroups of $(\mathbb{Z}, +)$. Consider $2\mathbb{Z}\cup 3\mathbb{Z}= \{x\in \mathbb{Z}\mid  x\in 2\mathbb{Z}\ or \ x\in3\mathbb{Z}\}$. We know that $2,3\in 2\mathbb{Z}\cup 3\mathbb{Z}$, but $2+3=5 \notin 2\mathbb{Z}\cup 3\mathbb{Z}$ because $5 \notin 2\mathbb{Z}$ and $5\notin 3\mathbb{Z}$. Thus, $2\mathbb{Z}\cup 3\mathbb{Z}$ is not a subgroup of $\mathbb{Z}$. Consequently, $2\mathbb{Z}\cup 3\mathbb{Z}$ is not normal subgroup of $\mathbb{Z}$. Therefore, the union of any two normal subgroups need not be a normal subgroup of a group.