## The join of an arbitrary collection of normal subgroups is normal

Let $G$ be a group. Prove that the join of any nonempty collection $\mathcal{H} = \{ H_i \}_{i \in I}$ of normal subgroups of $G$ is normal.

Lemma: Let $G$ be a group, $g \in G$, and $\{ A_i \}_{i \in I}$ a nonempty collection of subgroups of $G$. Then $g(\bigcup A_i)g^{-1} = \bigcup g A_i g^{-1}$. Proof: $\subseteq)$ Let $x \in g(\bigcup A_i)g^{-1}$. Then $x = gyg^{-1}$ for some $y \in \bigcup A_i$. Thus there exists $k \in I$ with $y \in A_k$, so that $x \in g A_k g^{-1}$. Hence $x \in \bigcup g A_i g^{-1}$. $\supseteq$ Let $x \in \bigcup g A_i g^{-1}$. Then there exists $k \in I$ with $x \in g A_k g^{-1}$, so that $x = gyg^{-1}$ for some $y \in A_k$. But $y \in \bigcup A_i$, so that $x \in g (\bigcup A_i) g^{-1}$. $\square$

Recall that the join of $\mathcal{H}$ is $\langle \bigcup \mathcal{H} \rangle$. Now let $g \in G$; we saw in a previous exercise that $g \langle \bigcup \mathcal{H} \rangle g^{-1} = g \langle \bigcup H_i \rangle g^{-1} = \langle g(\bigcup H_i)g^{-1} \rangle$ $= \langle \bigcup gH_ig^{-1} \rangle$ $= \langle \bigcup H_i \rangle$ $= \langle \bigcup \mathcal{H} \rangle$. Hence $\langle \bigcup \mathcal{H} \rangle$ is normal in $G$.