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This is the excerpt from Herstein's book. I have read these paragraphs and understood almost everything but there are some moments which I cannot comprehend.

1) Using homomorphism theorem they showed that $\bar{H}\cong H/K$. But what does it give?

2) Why this association is well-defined?

3) Where did they get that association is one-to-one?

4) I read the proof entirely and understood almost everything. But can anyone point out to the moment where was showed that there is a one-to-one mapping from the set of all subgroups of $\bar{G}$ onto the set of all subgroups of $G$ which contain $K$?

I would be very thankful if somebody will answer my questions.

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  • $\begingroup$ 1) You need to show that the homomorphism $\phi\upharpoonright H$ is onto. $\endgroup$ – Gödel Dec 25 '17 at 16:19
  • $\begingroup$ @Gödel, what is the meaning of your notation? $\endgroup$ – ZFR Dec 25 '17 at 16:25
  • $\begingroup$ 2), 3) Note that in the proof, you begin with a arbitrary subgroup of $\hat{H}$ and you associate it with $H$, where $H$ has a particular form and it contains the kernel. Next, you choose a arbitrary subgruop in $H$ with the property that contains $K$ and you shows that it has the same form that the last one. This is a ono-to-one correspondence, so is well-defined. $\endgroup$ – Gödel Dec 25 '17 at 16:28
  • $\begingroup$ $\phi\upharpoonright H$ means the homomorphism $\phi$ restric to $H$. $\endgroup$ – Gödel Dec 25 '17 at 16:30
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1) It allows to make the proof of the lemma for $G/K$ instead of $\overline G$ since, as they're isomorphic, subgroups of the former correspond bijectively to subgroups of the latter.

2) Simply because, with the previous notations, $H=\phi^{-1}\bigl(\overline{H}\mkern1mu\bigr)$.

3) It's surjective because, as $\phi$ is onto, and we consider subgroups $H$ of $G$ which contain $\ker\phi $, we have $\;\phi^{-1}\bigl(\phi(H)\bigr)=H$.

4) Injectivity is in the paragraph which begins with ‘Suppose, conversely’: the author shows that, for any subgroup $L$ such that $K\subset L\subset G$, $\;\phi^{-1}\bigl(\phi(L)\bigr)=L$.

To maket more explicit:

We've defined two maps, from the set $\overline{\mathcal S}$ of subgroups of $\overline S$ to the set $\mathcal S_K$ of subgroups of $S$ which contain $K$, and the reverse way: $$\begin{aligned} f\colon\overline{\mathcal S}&\longrightarrow\mathcal S_K&\qquad\mathcal g\mkern1mu\colon S_K, &\longrightarrow\overline{\mathcal S},\\ \overline H&\longmapsto \phi^{-1}(\overline H),&\qquad L&\longmapsto \phi(L), \end{aligned}$$ and we've proved that $g\circ f=\operatorname{id}_{\overline{\mathcal S\mathstrut}}$, $\;f\circ g= \operatorname{id}_{\mathcal S_K}$, so that $f$ and $g$ are bijective and inverse of each other.

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  • $\begingroup$ Dear, Bernard. Thanks for post but I did not undestand none of your answers. 1) How from $\overline{H}\simeq H/K$ you conclude that association is onto? 2) Why $H=\phi^{-1}(\overline{H})$? How it follows that association is well-defined? 3) Why $\phi(\phi^{-1}(\overline{H}))=\overline{H}$? How do you derive that association is one-to-one? 4) The same situation. I cannot understand your answer. $\endgroup$ – ZFR Dec 25 '17 at 17:12
  • $\begingroup$ @RFZ: I have somewhat mixed two points of view for my answer to your first question, and see it made it unclear. I've changed for a (hopefully) clearer explanation. For 2), that's the way the author defines $H$ – as the inverse image of $\overline H$. Are you familiar with direct and inverse images of subsets? For question 3) and 4), due to the mixed p.o.v., the explanations were swapped. I've updated my answer, please see it it's clearer. $\endgroup$ – Bernard Dec 25 '17 at 18:01
  • $\begingroup$ @RFZ (continuation): Please keep in mind that, essentially, Herstein proves that the inverse image and direct image maps, when restricted to the set of subgroups of $\overline G$ and the set of subgroups of $G$ containing $\ker\phi$, are inverse bijections. $\endgroup$ – Bernard Dec 25 '17 at 18:03
  • $\begingroup$ Thanks for edit! But let me ask some additional questions: 2) If I get you right, then for each subgroup $\overline{H}$ of $\overline{G}$ we correspond it's inverse image in $G$ namely $H:=\phi^{-1}(\overline{H})$ which is a subgroup of $G$. Then this association is well-defined since we are dealing just with inverse images. Right? 3) I proved the $\phi^{-1}(\phi(H))=H$ considering two inclusions but I've never used that $H$ contains $\text{Ker} \phi$. How did you realize that this relations holds? Could you describe it in detail. 4) The same for $\phi^{-1}(\phi(L))=L$. How did you derive it? $\endgroup$ – ZFR Dec 26 '17 at 8:26
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    $\begingroup$ @RFZ: I've just found your last comments, sorry for answering late. I agree with what you wrote in your last comment, but not with the previous one: if you ‘take two subgroups $L_1$ and $L_2$ such that $L_1=L_2$’, you really take a single subgroup, and that proves nothind. For the injectivity of $f$, you have to take two subgroups $\overline L_i$ in $\overline{\mathcal S}$, and show that if $\phi^{-1}(\overline L_1)=\phi^{-1}(\overline L_2)$, then $\overline L_1=\overline L_2$. $\endgroup$ – Bernard Dec 28 '17 at 11:02

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