Determining elements in Aut(D_8)

  • #1
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Homework Statement


Find all of the elements in ##\operatorname{Aut}(D_8)##.

Homework Equations




The Attempt at a Solution


I am trying to come with a repeatable procedure for finding automorphisms of groups with presentations. I'm taking ##D_8## as a good example.

Here is how I would like to start. Let ##\psi\in \operatorname{Aut}(D_8)##. Then by the universal mapping property for presentations, ##\psi## is completely determined by ##r'=\psi(r)## and ##s'=\psi(s)## which must satisfy ##(r')^4 = (s')^2 = r's'r's'=1##. Since ##\psi## is an automorphism, it must preserve order, so the possible automorphisms are given by ##\psi(r) \in \{r,r^{-1}\}## and ##\psi(s) \in \{r^2,s,sr,sr^2,sr^3\}##. But all of these may not be automorphisms, since we haven't checked if the relations are satisfied. Note we can exclude the case when ##\psi(r)=r^{\pm1}## and ##\psi(s) = r^2##, since this does not satisfy the relation ##r's'r's'=1##. So we are left with ##8## possible automorphisms. Note that ##\operatorname{im}(\psi)## contains the subgroup ##\langle r' \rangle## of order ##6## and at least one element ##s'## not in this group, so ##|\operatorname{im}(\psi)| \ge 5## and by Lagrange ##|\operatorname{im}(\psi)|## divides ##8##. Hence ##\operatorname{im}(\psi) = D_8##, so ##\psi## is surjective (and hence injective), so ##\psi## is a bijection. All that's left to show is that all of these ##8## possibilities are indeed homomorphisms...

This is where I get stuck. Do I need to show that the ##r's'r's'=1## for every choice of ##r'## and ##s'## to show that the relations hold and that we get homomorphisms for all of these? This seems like a lot of work.
 

Answers and Replies

  • #3
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Yes, but I don't think I completely understood what I was doing. Here I want to explicitly indicate which steps show that what I've found are bijections and homomorphisms. What I see is that, assuming we have an automorphism, we get 10 possibilities based on order considerations. We then cut this down to ##8## possibilities, and I showed that these 8 are bijections. But nowhere here did I show that all of these ##8## are actually homomorphisms, which would complete the proof.
 
  • #4
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Note we can exclude the case when ##\psi(r)=r^{\pm1}## and ##\psi(s) = r^2##, since this does not satisfy the relation ##r's'r's'=1##.
I find ##\operatorname{im}\psi \subseteq \langle r\rangle## implies that ##\psi## is no automorphism quicker to see.

What you have shown are all necessary conditions, derived from the properties of the group and ##\psi##. So sufficiency has still to be proven somehow, i.e. that all these combinations are actually isomrphisms. You already ruled out two combinations, so what makes you sure there aren't others which were impossible? I think that the isomorphism ##D_{2n} \cong \mathbb{Z}_n \rtimes_\sigma \mathbb{Z}_2## should be of help here, i.e. will probably get you a lower bound for ##|\operatorname{Aut}(D_{2n})|\,.##

I suppose that ##\operatorname{Aut}(D_{2n}) \cong \mathbb{Z}_n \bowtie \mathbb{Z}_n^*##, i.e. I suspect that at least one of the factors is normal (the inner autormphisms), but I'm not sure which is which. I'm still convinced of
Couldn't we generalize this [your argument] for ##\phi(r)=r\; , \;\phi(s)=sr^l## to answer the question which automorphisms are outer and which are inner instead of the counting argument?
 

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