User:Dfeuer/Definition:Normal Subset of Group

Definition 1
Let $(G,\circ)$ be a group.

Let $N \subseteq G$.

Then $N$ is a normal subset of $G$ iff:
 * $\forall g \in G: g \circ N \circ g^{-1} = N$

Definition 2
Let $(G,\circ)$ be a group.

Let $N \subseteq G$.

Then $N$ is a normal subset of $G$ iff:
 * $\forall g \in G: g \circ N \circ g^{-1} \subseteq N$

Definition 3
Let $(G,\circ)$ be a group.

Let $N \subseteq G$.

Then $N$ is a normal subset of $G$ iff:
 * $\forall g \in G: g \circ N \subseteq N \circ g$

Definition 1 implies Definition 2
Every set is a subset of itself, so the result follows.

Definition 3 implies Definition 2
Suppose that
 * $\forall g \in G: g \circ N \subseteq N \circ g$

Let $g \in G$.

Let $n \in N$

Then $g \circ n \in N \circ g$

Thus for some $m \in N$: $g \circ n = m \circ g$

Then $g \circ n \circ g^{-1} = m$

Since this holds for each $n \in N$:
 * $g \circ N \circ g^{-1} \subset N$.

Since this holds for each $g \in G$:
 * $\forall g \in G: g \circ N \circ g^{-1} \subseteq N$

Definition 2 implies Definition 1
Let $(G,\circ)$ be a group

Let $N$ be a normal subset of $(G,\circ)$ by definition 2.

Then for each $g \in G$:
 * $g \circ N \circ g^{-1} \subseteq N$.

Let $x \in N$.

Then $g^{-1} \circ x \circ g \in N$.

So for some $n \in N$, $g^{-1} \circ x \circ g = n$

Thus $x \circ g = g \circ n$.

Then $x = g \circ n \circ g^{-1}$.

Thus so $x \in g \circ N \circ g^{-1}$.

Since this holds for all $x \in N$:
 * $N \subseteq g \circ N \circ g^{-1}$

Since we also have $g \circ N \circ g^{-1} \subseteq N$:


 * $N = g \circ N \circ g^{-1}$

Thus $N$ is a normal subset of $(G,\circ)$ by definition 1