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I'm reading the proof of maximal normal subgroup in this answer: https://math.stackexchange.com/a/161593/435467.

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The question is when $\frac{A}{H} = \frac{G}{H}$, how could we know that $\frac{G}{H}$ is a simple group?

Ari Royce Hidayat
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  • Why do you guys downvote my question?? – Ari Royce Hidayat Mar 15 '20 at 11:50
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    I'm voting to close this question as off-topic because adding a question and then an answer to fill a "gap" in another answer is nonsense. This whole ordeal could, and should, have simply been a comment under the original question. – AnalysisStudent0414 Mar 15 '20 at 15:25
  • Let us assume there exist $A$ so that $\frac{A}{H}\lhd \frac{G}{H}$. Then by definion, $H$ must be normal in $A$. Because $H$ is maximal, we get $H=A$ and therefore $\frac{A}{H}={1}$. – Ari Royce Hidayat Mar 15 '20 at 15:42
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    @AnalysisStudent0414 I post a question looking for an answer, I found another one I like, post it here. If it is a better answer, why not? Why is it nonsense? – Ari Royce Hidayat Mar 15 '20 at 17:19
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    @AnalysisStudent Maybe I'm missing something here, but the original post is ancient (and has comments by the OP here but no comments by the answerer). Moreover, this question here was answered by a third party. I cannot see any other way of resolving the issue, and it all seems honest and above board. (There is an answer to this question by the OP, but it was the second answer.) – user1729 Mar 15 '20 at 22:23
  • @user1729 OP had answered the question. Then she deleted it, and accepted the other answer. – AnalysisStudent0414 Mar 16 '20 at 10:50
  • @AnalysisStudent0414 Yes, but as I said the other answer was given first. There was a one-hour pause before the OP gave their own answer, and I assume they worked out the answer during this time, which is good, no? (Timeline can be found here.) – user1729 Mar 16 '20 at 11:23
  • @AnalysisStudent0414 Why are you still discussing this, while you seem to be very eager to downvote a question? No worry any more though, I've been blocked to post another question. That's ok, but I just don't get it. The site told me that my low level questions are impacting its quality, which is that means downvotes from you all decide it. What I don't get is what kind of quality we are talking about? Is it about the quality of helping other people to learn, or is it about a community of 1st rate mathematicians solving 1st rate of mathematical problems? – Ari Royce Hidayat Mar 16 '20 at 12:08
  • @user1729 yeah that's even stranger to me because the solutions provided are nearly identical. – AnalysisStudent0414 Mar 16 '20 at 12:15
  • @AnalysisStudent0414 May be, but it's not clear in the accepted answer that the maximality of H would force it to become $H/H ⊲ G/H$ -- which is crystal clear that $G/H$ is a simple group. Yes, the accepted answer mentions that in the beginning, but then it continues with "Consequently..." which is for me is kind of making it obscure again. – Ari Royce Hidayat Mar 16 '20 at 12:54
  • @AnalysisStudent0414 For what its worth, I feel that the OPs answer is clearer (but of course the solutions are nearly identical - the question is low-level and requires only a short answer). – user1729 Mar 16 '20 at 13:11

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Note that $A/H < G/H$ is normal in $G/H$. This means that $H\trianglelefteq A \triangleleft G$, but the maximality of $H$ forces $H=A$. Consequently, the only normal subgroups of $G/H$ are $H/H$ and $G/H$. And if $G/H=A/H$, you must have $A=G$. Again, assuming $N/H\triangleleft G/H$ forces $H=N$.