[proofplan]
We first translate the set containment $gNg^{-1} \subset N$ into its elementwise form, which gives the equivalence of conditions 2 and 4. The only nontrivial point is that the one-sided containment in condition 2 automatically gives equality, because the same containment can be applied to $g^{-1}$. Once conjugation equality $gNg^{-1}=N$ is obtained, multiplying on the right by $g$ gives the coset equality $gN=Ng$, and conversely multiplying $gN=Ng$ on the right by $g^{-1}$ gives conjugation equality.
[/proofplan]
[step:Unpack conjugation containment into its elementwise form]
For each $g \in G$, the set $gNg^{-1}$ is defined by
\begin{align*}
gNg^{-1} = \{gng^{-1} : n \in N\}.
\end{align*}
Therefore $gNg^{-1} \subset N$ holds if and only if every element of the form $gng^{-1}$ with $n \in N$ belongs to $N$. Quantifying over $g \in G$, condition 2 is exactly condition 4.
[/step]
[step:Upgrade one-sided conjugation containment to conjugation equality]
Assume condition 2. Fix $g \in G$. The hypothesis gives
\begin{align*}
gNg^{-1} \subset N.
\end{align*}
Applying the same hypothesis to $g^{-1} \in G$ gives
\begin{align*}
g^{-1}Ng \subset N.
\end{align*}
We prove the reverse containment $N \subset gNg^{-1}$. Let $n \in N$. Since $g^{-1}ng \in g^{-1}Ng$, the containment $g^{-1}Ng \subset N$ gives $g^{-1}ng \in N$. Define $m := g^{-1}ng \in N$. Then
\begin{align*}
gmg^{-1} = g(g^{-1}ng)g^{-1} = n.
\end{align*}
Thus $n \in gNg^{-1}$. Since $n \in N$ was arbitrary, $N \subset gNg^{-1}$. Combining both containments gives
\begin{align*}
gNg^{-1} = N.
\end{align*}
Conversely, if $gNg^{-1}=N$ for every $g \in G$, then $gNg^{-1} \subset N$ for every $g \in G$. Hence condition 2 is equivalent to
\begin{align*}
gNg^{-1} = N \quad \text{for every } g \in G.
\end{align*}
[guided]
The point of this step is to show that the apparently weaker one-sided condition is already symmetric. Assume condition 2, so for every element $h \in G$ we have
\begin{align*}
hNh^{-1} \subset N.
\end{align*}
Fix $g \in G$. Taking $h=g$ gives
\begin{align*}
gNg^{-1} \subset N.
\end{align*}
To prove equality, we still need $N \subset gNg^{-1}$.
How do we get the reverse containment? We use the same hypothesis with the inverse element $g^{-1}$. Since $G$ is a group, $g^{-1} \in G$, so condition 2 applies to $h=g^{-1}$ and gives
\begin{align*}
g^{-1}Ng \subset N.
\end{align*}
Now let $n \in N$. The element $g^{-1}ng$ belongs to $g^{-1}Ng$ by definition of that set, so the displayed containment implies $g^{-1}ng \in N$. Define
\begin{align*}
m := g^{-1}ng.
\end{align*}
Then $m \in N$, and direct multiplication in the group gives
\begin{align*}
gmg^{-1} = g(g^{-1}ng)g^{-1} = n.
\end{align*}
Thus $n$ has the form $gmg^{-1}$ for some $m \in N$, so $n \in gNg^{-1}$. Since $n \in N$ was arbitrary, $N \subset gNg^{-1}$. Together with $gNg^{-1} \subset N$, this proves
\begin{align*}
gNg^{-1} = N.
\end{align*}
The converse direction is immediate from set equality: if $gNg^{-1}=N$, then in particular $gNg^{-1} \subset N$. Therefore condition 2 is equivalent to conjugation equality for every $g \in G$.
[/guided]
[/step]
[step:Translate conjugation equality into equality of left and right cosets]
We show that, for each fixed $g \in G$,
\begin{align*}
gNg^{-1} = N
\end{align*}
is equivalent to
\begin{align*}
gN = Ng.
\end{align*}
Assume first that $gNg^{-1}=N$. Multiplying both sides on the right by $g$ gives
\begin{align*}
(gNg^{-1})g = Ng.
\end{align*}
Associativity in $G$ gives $(gNg^{-1})g = gN$, because every element on the left has the form $(gng^{-1})g = gn$ with $n \in N$. Hence $gN=Ng$.
Conversely, assume $gN=Ng$. Multiplying both sides on the right by $g^{-1}$ gives
\begin{align*}
(gN)g^{-1} = (Ng)g^{-1}.
\end{align*}
Again by associativity, $(gN)g^{-1}=gNg^{-1}$ and $(Ng)g^{-1}=N$. Therefore
\begin{align*}
gNg^{-1}=N.
\end{align*}
Thus condition 3 is equivalent to conjugation equality for every $g \in G$.
[/step]
[step:Identify conjugation equality with normality and conclude the equivalence]
By the standard definition of a [normal subgroup](/page/Normal%20Subgroup), $N \trianglelefteq G$ means that conjugation by every element of $G$ preserves $N$ as a set:
\begin{align*}
gNg^{-1}=N \quad \text{for every } g \in G.
\end{align*}
The previous steps proved that this conjugation equality is equivalent to condition 2, equivalent to condition 3, and equivalent to condition 4. Therefore conditions 1, 2, 3, and 4 are all equivalent.
[/step]
