To prove that the intersection of two normal subgroups is a normal subgroup, let $G$ be a group, and let $N_1$ and $N_2$ be normal subgroups of $G$. The goal is to show that $N_1 \cap N_2$ is a normal subgroup of $G$.

Proof:

1. Subgroup Property:

First, $N_1 \cap N_2$ is a subgroup of $G$:

• The identity element $e \in G$ is in $N_1$ and $N_2$ since both $N_1$ and $N_2$ are subgroups, so $e \in N_1 \cap N_2$.
• Closure: If $x, y \in N_1 \cap N_2$, then $x, y \in N_1$ and $x, y \in N_2$. Since $N_1$ and $N_2$ are subgroups, $xy \in N_1$ and $xy \in N_2$. Therefore, $xy \in N_1 \cap N_2$.
• Inverses: If $x \in N_1 \cap N_2$, then $x \in N_1$ and $x \in N_2$. Since $N_1$ and $N_2$ are subgroups, $x^{-1} \in N_1$ and $x^{-1} \in N_2$. Therefore, $x^{-1} \in N_1 \cap N_2$.

Thus, $N_1 \cap N_2$ is a subgroup of $G$.

2. Normality:

Now, we prove that $N_1 \cap N_2$ is normal in $G$. To do this, we need to show that $g(N_1 \cap N_2)g^{-1} \subseteq N_1 \cap N_2$ for all $g \in G$.

Let $x \in N_1 \cap N_2$. Then:

• $x \in N_1$, so $gxg^{-1} \in N_1$ because $N_1$ is normal in $G$.
• $x \in N_2$, so $gxg^{-1} \in N_2$ because $N_2$ is normal in $G$.

Thus, $gxg^{-1} \in N_1 \cap N_2$, and therefore $g(N_1 \cap N_2)g^{-1} \subseteq N_1 \cap N_2$.

3. Conclusion:

Since $N_1 \cap N_2$ satisfies both the subgroup and normality properties, it follows that $N_1 \cap N_2$ is a normal subgroup of $G$.

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Related Questions:

1. Why are normal subgroups important in group theory?
2. What happens if $N_1$ or $N_2$ is not normal? Does their intersection remain normal?
3. Can this proof be generalized to the intersection of more than two normal subgroups?
4. How does the concept of normality relate to quotient groups?
5. Can you provide examples of groups where the intersection of normal subgroups is not trivial?

Tip:

Remember that the normality condition $gNg^{-1} = N$ for all $g \in G$ ensures that conjugation respects the subgroup structure, making it compatible with quotient constructions.