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Let $G$ be a set and let $A \subseteq G$. Then the normal subgroup of $G$ generated by the set $A$ is $$\langle A \rangle ^N= \{ g_1a_1^{i_1}g_1^{-1} \dots g_ma_m^{i_m}g_m^{-1} \mid m \geq 0, a_j \in A, g_j \in G, i_j \in \{1,-1\}, \forall 1\leq j\leq m\}.$$

Show that $\langle A \rangle ^N$ is the intersection of all the normal subgroups of $G$ that contain $A$.

I showed that $\langle A \rangle ^N$ is minimal by the following argument: Suppose that $M$ is a normal subgroup of $H$ containing $A$. Then $N$ must contain $ga^ig^{-1}$ for all $g\in G$, $a\in A$ and $i\in \{1,-1\}$. in order to be normal and contain $A$, and $M$ must contain all products of such elements in order to be a subgroup of $G$. So $M$ must contain $\langle A \rangle^N$.

However, I am not sure what to do with this. How can I use this to conclude that $\displaystyle \langle A \rangle^N = \bigcap_{A\subseteq N \trianglelefteq G}N$? Note that I have not shown before that $\bigcap_{A\subseteq N \trianglelefteq G}N$ is minimal.

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  • $\begingroup$ "the intersection of all the normal subgroups of $G$ that contain $A$" is just another way of saying "the smallest normal subgroup of $G$ that contains $A$" $\endgroup$ – Kenny Lau Sep 19 '18 at 3:54
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First, if $K\trianglelefteq G$, then $\displaystyle\bigcap_{A\subseteq N \trianglelefteq G}N\subseteq K$ since $K$ is actually one of the sets in the intersection. In particular $\displaystyle\bigcap_{A\subseteq N \trianglelefteq G}N \subseteq\langle A \rangle^N$. Also, since $\langle A \rangle^N$ is minimal, we have that $\langle A \rangle^N \subseteq \displaystyle\bigcap_{A\subseteq N \trianglelefteq G}N$. So you have your result. (Note that you have to show that the intersection of normal subgroups is a normal subgroup)

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