To solve question 1:

Problem:

Let $G$ be a group, $H$ and $N$ be subgroups of $G$. If $N$ is a normal subgroup of $G$, prove that $H \cap N$ is a normal subgroup of $H$.

Solution:

1. Normal subgroup definition: $N$ is a normal subgroup of $G$, so for any $g \in G$ and $n \in N$, we have: $g n g^{-1} \in N.$

2. Intersection $H \cap N$: The subgroup $H \cap N$ consists of all elements that are in both $H$ and $N$: $H \cap N = \{x \in G : x \in H \text{ and } x \in N\}.$

3. Goal: Prove that $H \cap N$ is a normal subgroup of $H$. For this, we need to show: $\forall h \in H, \forall x \in H \cap N, \; h x h^{-1} \in H \cap N.$

4. Verification: Let $h \in H$ and $x \in H \cap N$. Since $x \in H \cap N$, we have:

   • $x \in H$,
   • $x \in N$.

   Since $h \in H$ and $N$ is normal in $G$, we know: $h x h^{-1} \in N.$

   Additionally, since $h \in H$ and $x \in H$, the group $H$ being closed under conjugation implies: $h x h^{-1} \in H.$

   Combining these, we have: $h x h^{-1} \in H \cap N.$

5. Conclusion: Since $h x h^{-1} \in H \cap N$ for all $h \in H$ and $x \in H \cap N$, $H \cap N$ is a normal subgroup of $H$.

If you have further questions, feel free to ask!

Related Questions:

1. Can you provide examples of groups $G$, $H$, and $N$ satisfying the problem's conditions?
2. How can we extend this result if $H$ is not a subgroup of $G$?
3. Why is the normality of $N$ in $G$ critical for this proof?
4. What is the significance of normal subgroups in group theory?
5. How does this result generalize to intersections of multiple normal subgroups?

Tip:

Always verify the subgroup properties (closure, associativity, identity, and inverses) before proving normality.