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(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.

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    $\begingroup$ Are you trying to construct a Solutions Manual to Fraleigh's book? So many questions.... $\endgroup$ – Gerry Myerson Feb 4 '14 at 9:08
  • $\begingroup$ (1): Because it is trivial. (2): That is precisely the assumption (this actually also answers (4)). $\endgroup$ – Tobias Kildetoft Feb 4 '14 at 9:08
  • $\begingroup$ (1) says that $gHg^{-1} \subseteq H$. (2) implies this trivially. $\endgroup$ – Arthur Feb 4 '14 at 9:09
  • $\begingroup$ (3) Why would anyone think $gh_1=h_2g$ implies $h_1=h_2$? Examine your assumptions and prejudices; in the noncommutative world we can't cancel something on the left with something elsewhere on the right. (3) Rewriting $gH=Hg$ yields $gHg^{-1}=H$, and (ostensibly) weakening this yields $gHg^{-1}\subseteq H$. Sometimes "moves" in math are like "moves" in chess: you have to learn how to see multiple steps ahead, like in this case where you need to see two steps ahead. (For example, one can solve $2x+1=0$ immediately mentally by seeing two steps ahead, "intuition" or not.) $\endgroup$ – anon Feb 4 '14 at 9:22
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    $\begingroup$ You are asking me about your feelings? I have no idea. Other than my guess that you are prejudiced by your abelian preconceptions, or somesuch. Just because conjugation by $g$ fixes $H$ setwise does not mean it fixes $H$ pointwise: it can move things around inside $H$, without anything spilling out, and that's all that $gH=Hg$ means. $\endgroup$ – anon Feb 21 '14 at 7:33
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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.

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  • $\begingroup$ Apologies, but how does this answer my questions? $\endgroup$ – Group Theory Apr 28 '14 at 13:51
  • $\begingroup$ @FrankMuer: You edited your question $6$ hours ago and my answer was given almost one moth ago. (I do not even remember previous version of the question). When I have time I will try to edit my answer. $\endgroup$ – mesel Apr 28 '14 at 20:07
  • $\begingroup$ Thanks. But my previous questions weren't about proofs. $\endgroup$ – Group Theory May 4 '14 at 16:06
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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:

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

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  • $\begingroup$ I deleted the redundant X. Please notify if my edit's bungled. $\endgroup$ – Group Theory Apr 28 '14 at 13:51

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