Equivalences for Normal Subgroups

Lemma

For any $H \leq G$ (see Subgroup) and $g \in G$ we have:

g H = H g ⟺ g H g^{- 1} = H

(see Left and Right Cosets) In fact, the following are equivalent:

$g H = H g$ for any $g \in G$
$g H g^{- 1} = H$ for all $g \in G$
$g H g^{- 1} \subseteq H$ for all $g \in G$
$g h g^{- 1} \in H$ for all $g \in G$
$N_{G} (H) = G$ (see Normalizer)
$H ⊴ G$ (see Normal Subgroup) (comes directly from (5))

Proof

For the former, we'll do both directions:

$(\Rightarrow)$ : Suppose $g H = H g$ . Take any $h \in H$ where then:

g h \in g H = H g \Rightarrow g h \in H g

so then there is some $h^{'} \in H$ where $g h = h^{'} g$ . Thus:

g h g^{- 1} = h^{'} \in H

thus proving (3), namely $g H g^{- 1} \subseteq H$ .

We just need to prove that $H \subseteq g H g^{- 1}$ . So take any $h \in H$ . We want to show (want to find) some $h^{'} \in H$ such that $h = g h^{'} g^{- 1}$ . So then we'd want $h = g h^{'} g^{- 1} \Leftarrow g^{- 1} h g$ .

So choose $h^{'} := g^{- 1} h g$ . Now notice that $h g \in H g = g H$ . Thus then $h g = g \tilde{h}$ for some $\tilde{h} \in H$ . Then:

h^{'} = g^{- 1} h g = g^{- 1} (g \tilde{h}) = \tilde{h} \in H

as we needed to verify. Thus then:

\Rightarrow h = g (g^{- 1} h g) g^{- 1} \in g H g^{- 1} \Rightarrow H \subseteq g H g^{- 1}

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