User:Dfeuer/Definition:Normal Subset of Group
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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.
$\blacksquare$
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 $g \in G$:
- $\forall g \in G: g \circ N \circ g^{-1} \subseteq N$
$\blacksquare$
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