- #1

Mr Davis 97

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## Homework Statement

Let ##G## be a group and let ##A \subseteq G## be a set. The normal subgroup of ##G## generated by ##A##, denoted ##\langle A \rangle ^N##, is the set of all products of conjugates of elements of ##A## and inverses of elements of ##A##. In symbols,

$$\langle A \rangle ^N= \{ g_1a_1^{i_1}g_1^{-1} \dots g_ma_m^{i_m}g_m^{-1} \mid m \geq 0, a_j \in A, g_j \in G, i_j \in \{1,-1\}, \forall 1\leq j\leq m\}.$$

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

## Homework Equations

## The Attempt at a Solution

I amount really sure how to start here. I don't see anything that I can really grab on to to start the proof. Maybe we show that one is a subset of the other, and vice versa. But even then I still don't see how to show that an arbitrary element of ##\langle A \rangle ^N## is in the intersection of all the normal subgroups of ##G## that contain ##A##.