Questions on Proofs - Equivalent Conditions of Normal Subgroup - Fraleigh p. 141 Theorem 14.13 

(1.) Why did Fraleigh shirk the proof for $(2) \implies (1)$?
  By dint of Arthur's comment, $(2) \iff \color{crimson}{gHg^{-1} \subseteq H} \quad \wedge \quad gHg^{-1} \supseteq H \implies \color{crimson}{gHg^{-1} \subseteq H} \iff (1)$

(2.) In $(1) \implies (2)$, how does $\{ghg^{-1} : h \in H \} \subseteq H$?
(3.) I know left cosets $\neq$ right cosets. The same $H$ appears on both sides $gH = Hg$ in:
 $gH = Hg \iff gh_1 = h_2g$, hence why isn't $h_1 = h_2$ always?
I read $gH = Hg$ is a set equality and not an equality elementwise. But I'm confounded. 
To boot, I know $gh_1 = h_2g \iff G$ Abelian $\iff \color{magenta}{g^{-1}}gh_1 = \color{magenta}{g^{-1}}gh_2 \iff h_1 = h_2$.
$G$ can be nonAbelian hence if it is nonabelian, the previous line muffs. 
 A: Let $f_g:G\mapsto G$ defined by $f_g(x)=gxg{-1}$. Note that $f_g$ is an bijection from $G$ to $G$(actually it is an isomorphism from $G$ to $G$) Thus,restriction of $f_g$ on any subset $S$ of $G$ is also a bijection from $S$ to its image.
Now,if $G$ is finite your question is very trivial since if $gNg^{-1}\subseteq N$ then we know that $|gNg^{-1}|=|N|\implies gNg^{-1}=N $.(their order equal since it is an bijection)
For general case,Since $f_g$ is bijection so it has inverse namely $f_{g^{-1}} $ or vice versa.Thus,restriction of $f_g$ on a subset $S$,$f_g:S\mapsto Im(S)$ is invertable.
Now, it is given that $f_g(N)\subseteq N$ for all $g\in G$.So,
$$f_{g^{-1}}(N)\subseteq N$$
$$f_g(f_{g^{-1}}(N))\subseteq f_g(N)$$
$$N\subseteq f_g(N)\implies N=f_g(N)$$ Which is showing that $gNg^{-1}\subseteq N\implies gNg^{-1}=N$ we are done.
A: 2) We have to prove that if $ghg^{-1}\in H$ for each $g\in G$ and $h\in H$ then $\{ghg^{-1}:h\in H\}\subseteq H$ for each $g\in G$.
Take $x\in \{ghg^{-1}:h\in H\}$. We must prove $x \in H$. $x\in \{ghg^{-1}:h\in H\}$ means there exists $h\in H$ such that $x=ghg^{-1}$. By assumption $ghg^{-1}\in H$, thus $x\in H$.
3) Equality $gH=Hg$ means $\{gh:h\in H\}=\{hg:h\in H\}$.
In other words, for each $h\in H$ there exists $h^\prime\in H$ such that $gh=h^\prime g$, but, in general, we have $h\neq h^\prime$. On the other hand, if $G$ is abelian, then you have $h=h^\prime$.
4) If $gHg^{-1}\subseteq H$ for each $g\in G$ then also $g^{-1}Hg=g^{-1}H(g^{-1})^{-1}\subseteq H$ thus you obtain $H\subseteq gHg^{-1}$.
5) The equivalence of the following stataments:


*

*$gHg^{-1}\subseteq H$ for each $g\in G$;

*$gHg^{-1}=H$ for each $g\in G$;


follows from the presence of the for all $g\in G$. Without this for all $g\in G$ the equivalence will be no longer true.
