Proof of Characterizations and Examples of Normal Subgroups

Throughout, products are regrouped by associativity, part of Groups and Abelian Groups (i), without comment.

We begin with (ii). If every conjugate $\{gng^{-1}:n\in N\}$ equals $N$, then in particular $gng^{-1}\in N$ for all $g\in G$ and $n\in N$, so $N$ is normal by Normal Subgroups (i). Conversely, let $N$ be normal and fix $g\in G$. The inclusion $\{gng^{-1}:n\in N\}\subseteq N$ is Normal Subgroups (i). For the reverse inclusion let $n\in N$ and write

$$n=g(g^{-1}ng)g^{-1}.$$

Here $g^{-1}ng=g^{-1}n(g^{-1})^{-1}$ by Cancellation Laws and Basic Inverse Identities in a Group (iii), and this element lies in $N$ by Normal Subgroups (i) applied to $g^{-1}$. So $n$ is the conjugate by $g$ of an element of $N$.

Now (i), with left and right cosets as in Cosets and the Index of a Subgroup (i) and Cosets and the Index of a Subgroup (ii). Let $N$ be normal and $g\in G$. For $n\in N$,

$$gn=(gng^{-1})g\in Ng\qquad\text{and}\qquad ng=g(g^{-1}ng)\in gN,$$

since $gng^{-1}\in N$ by Normal Subgroups (i) for $g$, and $g^{-1}ng=g^{-1}n(g^{-1})^{-1}\in N$ by Normal Subgroups (i) for $g^{-1}$ together with Cancellation Laws and Basic Inverse Identities in a Group (iii). Hence $gN\subseteq Ng$ and $Ng\subseteq gN$, so $gN=Ng$.

Conversely, suppose $gN=Ng$ for every $g\in G$, and let $g\in G$ and $n\in N$. Then $gn\in gN=Ng$, so $gn=n'g$ for some $n'\in N$, and by the inverse law of Groups and Abelian Groups (i)

$$gng^{-1}=n'gg^{-1}=n'\in N.$$

Thus $N$ is normal by Normal Subgroups (i).

For (iii), if $G$ is abelian (Groups and Abelian Groups (ii)), then $gng^{-1}=ngg^{-1}=n\in N$ for all $g\in G$ and $n\in N$, so $N$ is normal by Normal Subgroups (i).

Finally, (iv). The sets $\{e\}$ and $G$ are subgroups by Subgroup Criterion and Basic Examples (iv). For every $g\in G$ we have $geg^{-1}=gg^{-1}=e$, and $gng^{-1}\in G$ for $n\in G$ since $G$ is closed under its operation. Hence both subgroups are normal by Normal Subgroups (i).