##\phi(R_{180})##, if ##\phi:D_n\to D_n## is an automorphism

  • #1
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Homework Statement


Determine ##\phi(R_{180})##, if ##\phi:D_n\to D_n## is an automorphism where ##n## is even so let ##n=2k##.
The solutions manual showed that since the center of ##D_n## is ##\{R_0, R_{180}\}## and ##R_{180}## is not the identity then it can only be that ##\phi(R_{180})=R_{180}##. This made some sense since to preserve structure ##\phi## must map centers to centers. My solution below is indeed sloppy. Although, I want to know if the procedure is acceptable in the general case, for instance when we want to know the image of some rotation not in the center of ##D_n##. Thanks!
2. Relevant propositions
($) Let ##H## be the subgroup of all rotations in ##D_n## and ##\phi:D_n\to D_n## be an automorphism, then ##\phi(H)=H##

The Attempt at a Solution


Let ##n=2k## such that ##k\in\Bbb{N}##. By ($), if ##H## is a subgroup of rotations, ##\phi(H)=H##; hence ##\phi(R_{180})\in H##. Also, note that if ##R_{360}=R_0=R_2k##, then ##R_{180}=R_k##. Picking an integer ##m\in\Bbb{Z}_{2k}## where ##gcd(m,2k)=1## so that ##\Bbb{Z}_{2k}=\langle m\rangle##; that is ##\forall s\in\Bbb{Z}_{2k}\exists t\in\Bbb{Z}## such that ##s\equiv mt \pmod {2k}##. Suppose ##k\equiv mv\pmod {2k}##. Then ##\phi(R_{180})=\phi(R_k)=\phi(R_{mv})=\phi(R_{m}^{v})=\phi(R_m)^v##.
 

Answers and Replies

  • #2
andrewkirk
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Let ##n=2k## such that ##k\in\Bbb{N}##. ....
if ##R_{360}=R_0=R_2k##, then ##R_{180}=R_k##. Picking an integer ##m\in\Bbb{Z}_{2k}## where ##gcd(m,2k)=1##.....
I can't follow this. ##k## is defined as half the degree of the dihedral group, ie ##n/2##, and then is said to equal 180, which is a number that is entirely dependent on the measure used for angle, and bears no relation to the group's degree. Did you mean the second ##k## to be a different value, not the same as that which the first ##k## represents? If so, a different variable name needs to be used. Even if not, I can't see any reason for using a variable that represents the number of degrees in a rotation, and then using that as the base for a modular integer field.

Perhaps you didn't mean what the statement ##R_{360}=R_{2k}## means, which is simply ##k=180##. If not, what was intended by that statement?
 
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  • #3
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It does not seem to me to follow that ϕ(H)=Hϕ(H)=H\phi(H)=H.
An argument I have in mind: Since ##\phi## must take ##R_{360/n}## to an element of order ##n## and reflections in ##D_n## have order 2, then it can only be that ##\phi(R_{360/n})## is a rotation. Thus, for any rotation ##R_{\frac{360k}{n}}## (##k \in\Bbb{N}##), can be denoted ##(R_{360/n})^k##, has an image ##\phi(R_{\frac{360k}{n}})=\phi((R_{360/n})^k)=\phi(R_{360/n})^k=R_{i}^k \in H## where ##R_i## is also a generator in ##H##. Would this be a valid one?
 
  • #4
andrewkirk
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An argument I have in mind: Since ##\phi## must take ##R_{360/n}## to an element of order ##n## and reflections in ##D_n## have order 2, then it can only be that ##\phi(R_{360/n})## is a rotation. Thus, for any rotation ##R_{\frac{360k}{n}}## (##k \in\Bbb{N}##), can be denoted ##(R_{360/n})^k##, has an image ##\phi(R_{\frac{360k}{n}})=\phi((R_{360/n})^k)=\phi(R_{360/n})^k=R_{i}^k \in H## where ##R_i## is also a generator in ##H##. Would this be a valid one?
Ah, I see you replied before I realised that what I wrote may have misinterpreted the OP and changed it. On looking at your OP again I realised (or thought I did) that that claim which I challenged was a given theorem, rather than something you were asked to prove. Is that correct? If so then there's no need to prove it, but it can be used to prove the question that was set.

As for the argument in your reply - that looks fine.
 
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  • #5
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I realised (or thought I did) that that claim which I challenged was a given theorem, rather than something you were asked to prove. Is that correct?
Yes :)
As for the argument in your reply - that looks fine.
Thanks. How about my sloppy looking solution to a general way of finding the image of a rotation under the automorphism ##\phi##? My solution was written in the OP(attempt at a solution).
 
  • #6
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Perhaps you didn't mean what the statement R360=R2kR360=R2kR_{360}=R_{2k} means, which is simply k=180k=180k=180.
Yes, I'm really sorry for this. It was a typo and I pass the time limit for changes in the OP.
 
  • #7
andrewkirk
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Thanks. How about my sloppy looking solution to a general way of finding the image of a rotation under the automorphism ϕ\phi? My solution was written in the OP(attempt at a solution).
I'm afraid I couldn't work out what you were trying to prove. I understand the original proof but I don't understand what generalisation of it you are trying to prove.
 
  • #8
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I couldn't work out what you were trying to prove.
It wasn't a proof :smile:. I was writing a general procedure in determining the image of rotation with respect to where the automorphism ##\phi## maps the generator of ##H##.
 
  • #9
andrewkirk
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It wasn't a proof :smile:. I was writing a general procedure in determining the image of rotation with respect to where the automorphism ##\phi## maps the generator of ##H##.
I'm afraid I can't follow the procedure. In the last line it looks like you are working with the image of ##R_{180}##, rather than of a general rotational element ##R_{360j/n}## for ##j\in\{0,1,...,n-1\}##.

For any case other than ##j\in \{0,n/2\}## there will be at least two possible values of ##\phi(R_{360j/n})##, being ##R_{360j/n}## and ##R_{360(n-j)/n}##, since both the identity and the 'reversal' map ##\phi(R_\theta)=R_{-\theta}## are automorphisms of ##D_n##. So we cannot determine the image of ##\phi(R_{360j/n})## without knowing more about ##\phi##.
 
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