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I'm preparing for a midterm and one of the suggested practice questions was Chapter 3.1 #32 in Dummit and Foote (p.88):

32. Which subgroups of $D_8$ are normal? (You can use the lattice.)

The textbook's advice on how to check if $N$ is a normal subgroup includes:

  • If you have a set of generators of $N$, you can check all conjugates of the generators lie in $N$.
  • If you have a set of generators of $G$, you can check that the generators for $G$ normalize $N$.

I have some questions below about technical details in my following attempt. I want to clarify my understanding of the textbook's advice, so going through this analysis is useful for me even if it's not the recommended method (which I would like to hear in addition if there is one.)

Using the lattice of $D_8$, I'm examining each subgroup one by one.

Consider $\langle s,r^2\rangle$.

First question: When checking that all conjugates of $s$ and $r^2$ live in $\langle s,r^2\rangle$, does it suffice to check the generators of $G$?

If so, we verify that

  • $rsr^-1 \in \langle s,r^2\rangle$
  • $sss^-1 \in \langle s,r^2\rangle$

Or, do we need to actually conjugate by every single element in $D_8$, including for example the following element?

  • $r^3sr^{-3} \in \langle s,r^2\rangle$

If there are other approaches that could be useful, I'd be glad to hear them. But note that since I'm preparing for an exam, hopefully other approaches can be generalized for other groups and not just work for $D_8$.

(I also noticed that the center of $D_8$ is $\langle r\rangle$, and that the center is a normal subgroup, but I don't know how to exploit this fact further for this question.)

Thank you!

Arturo Magidin
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    Use \langle and \rangle for delimiters, not < and > – Arturo Magidin Sep 27 '24 at 15:15
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  • @ArturoMagidin thank you for the link. Sounds like I don't need to check any subgroups of index 2 since I know they will be normal. All the groups on the first row of the $D_8$ lattice had index 2 when I checked. So, is it just a coincidence that all subgroups of index 2 were on the first row of the lattice? Or in general, should I expect that groups on the same row on the lattice have the same order (and thus index?) – Flying Spaghetti Sep 27 '24 at 15:36
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    Any subgroup of index two will be at the top of the lattice, because there can be no subgroup strictly between it and the whole group. But no, lattices do not usually have subgroups of the same order on the same height or vice versa, as it can get too complicated. The lattice for the cyclic group of order $10$ is usually represented as a symmetrical diamond, with the groups of order $2$ and $5$ symmetrically placed. – Arturo Magidin Sep 27 '24 at 15:39
  • @ArturoMagidin I've posted my proof here and would appreciate any feedback, thank you! https://math.stackexchange.com/questions/4977263/review-proof-of-identifying-normal-subgroups-of-d-8 – Flying Spaghetti Sep 27 '24 at 16:38
  • Nope. This is not a "here's my proof, give me feedback" site. Read the description of the "solution-verification" tag next time before posting. And I'm not your tutor. – Arturo Magidin Sep 27 '24 at 16:47
  • Certainly, I will read the tag. But personally, I don't think there's anything wrong with asking for feedback from a specific person I have engaged with online. Such a response would be rude at any workplace I've been at; I've presented query building techniques to audiences of 100+ with Q&A and a simple "no" would've sufficed - I would've never responded with "I'm not your tutor" to any audience member. – Flying Spaghetti Sep 27 '24 at 16:58
  • You seem confused. I am not someone giving a presentation to you in which you ask me a question. Rather, it is as if you are someone giving a presentation, and then singling out someone in the room and asking them to reply to your question, simply because they've spoken before about something else you said; worse, you are calling them on their cell phone and asking them to go somewhere else to look at your work and to comment on it. – Arturo Magidin Sep 27 '24 at 17:02

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