## The set of normal subgroups of a fixed group is closed under arbitrary intersections

Let $G$ be a group.

1. Let $H,K \leq G$ be normal subgroups. Prove that $H \cap N \leq G$ is normal.
2. Let $\{H_i\}_{i \in I}$ be a family of subgroups $H_i \leq G$, all normal. Show that $\bigcap H_i \leq G$ is normal.

1. Using a lemma to a previous exercise, for all $g \in G$ we have $g(H \cap K)g^{-1} = gHg^{-1} \cap gKg^{-1} = H \cap K$. Thus $H \cap K$ is normal in $G$.
2. Using a lemma to a previous exercise, for all $g \in G$ we have $g(\bigcap H_i)g^{-1} = \bigcap gH_ig^{-1} = \bigcap H_i$. By Theorem 6 in the text, $\bigcap H_i$ is normal in $G$.

Let $\varphi$ be an automorphism of a group $G$ and let $H,K \leq G$. We want to show that $\varphi[H \cap K] = \varphi[H] \cap \varphi[K]$. $(\subseteq)$ Suppose $x \in \varphi[H \cap K]$. Then $x = \varphi(y)$, where $y \in H \cap K$. In particular, $x \in \varphi[H]$ and $x \in \varphi[K]$, so $x \in \varphi[H] \cap \varphi[K]$. $(\supseteq)$ Now let $x \in \varphi[H] \cap \varphi[K]$. Say $x = \varphi(h)$ and $x = \varphi(k)$, with $h \in H$ and $k \in K$. Since $\varphi$ is an automorphism, it is injective, and we have $h = k \in H \cap K$. Thus $x \in \varphi[H \cap K]$, and we have $\varphi[H \cap K] = \varphi[H] \cap \varphi[K]$.
In particular, conjugation by a fixed element $g$ is an automorphism.