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.

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