1st Isomorphism thm for dihedral gps

  • #1
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Homework Statement


Prove that ##D_\infty/\langle R^n \rangle\cong D_{2n}##, where ##D_\infty=\langle R,S \mid S^2=e, SRS=R^{-1}\rangle##.

Homework Equations




The Attempt at a Solution


Pick ##g:\{R,S\} \to D_{2n}## such that ##g(R) = r## and ##g(S) = s##. We note that ##g(S)^2 = 1## and ##g(S)g(R)g(S)g(R)=1## since it is given in the presentation of ##D_{2n}## that ##s^2=1## and that ##srsr=1##. By the UMP for presentations there exists a unique homomorphism ##f: D_{\infty} \to D_{2n}## such that ##f(R)=r## and ##f(S)=s##.

This map ##f## is surjective because any word formed by the generators of ##D_{2n}## can be mapped to by the corresponding word in ##D_{\infty}## with ##r## replaced by ##R## and ##s## replaced by ##S##.

Let ##\Gamma## be an arbitrary word in the kernel of ##f## so that ##f(\Gamma) = 1##. Suppose that ##S## is a part of the word ##\Gamma##. Then ##f(\Gamma) \not = 1## since there will necessarily be an ##s## on the RHS. Hence only ##R## can be a part of this word, and the word must be an integral power of ##R^n## in order to map to ##1##, since otherwise the word will map to an element ##r^p## with ##p\in [0,n-1)##, which can never be ##1## (since the order of ##r## is ##n##). Hence ##\Gamma \in \langle R^n \rangle##. So ##\ker(f) = \langle R^n \rangle##

By the first isomorphism theorem, ##D_\infty/\langle R^n \rangle\cong D_{2n}##.
 

Answers and Replies

  • #2
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I don't see why ##f(\Gamma) \neq 1##.

We can always bring ##\Gamma \in \operatorname{ker}(f)## into the form ##\Gamma = SR^k\; , \;k\in\mathbb{Z}\; , \;## or ##\Gamma=R^k##. The second case is easy. Now in the first case ##f(\Gamma)=1=sr^k## and thus ##s=r^k##, i.e. ##1=r^{2k}##. Since ##r^{2k}\in D_{2n}##, we now have ##n\,|\,k## or ##\Gamma = S(R^n)^l##. But how do you conclude from ##s=r^{n\cdot l}## that ##S\in \langle R^n\rangle\,?##

Theoretically there can be an ##s## on the RHS, e.g. ##\Gamma = SRSR## maps to ##1=srsr##. Sure, this is ##1##, but it shows that in principle more complex words ##\Gamma## can be mapped on ##1## even if they contain an ##S##.

Edit: Got it. ##s=r^k## is not possible and therefore ##\Gamma \neq SR^k## and ##\Gamma = R^k## is the only left possibility.
 
  • #3
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I don't see why ##f(\Gamma) \neq 1##.

We can always bring ##\Gamma \in \operatorname{ker}(f)## into the form ##\Gamma = SR^k\; , \;k\in\mathbb{Z}\; , \;## or ##\Gamma=R^k##. The second case is easy. Now in the first case ##f(\Gamma)=1=sr^k## and thus ##s=r^k##, i.e. ##1=r^{2k}##. Since ##r^{2k}\in D_{2n}##, we now have ##n\,|\,k## or ##\Gamma = S(R^n)^l##. But how do you conclude from ##s=r^{n\cdot l}## that ##S\in \langle R^n\rangle\,?##

Theoretically there can be an ##s## on the RHS, e.g. ##\Gamma = SRSR## maps to ##1=srsr##. Sure, this is ##1##, but it shows that in principle more complex words ##\Gamma## can be mapped on ##1## even if they contain an ##S##.

Edit: Got it. ##s=r^k## is not possible and therefore ##\Gamma \neq SR^k## and ##\Gamma = R^k## is the only left possibility.
I'm a little confused. I see that we have two possibilities. If ##\Gamma = R^k## then we get that ##k## must be a multiple of ##n## becuase we must have that ##r^k=1##. But in the case ##\Gamma = SR^k## we get that ##r^{2k}=1##. How does this lead us to conclude that there is no such ##k## for which this holds, and hence no elements of form ##SR^k## are in the kernel?
 
  • #4
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I'm a little confused. I see that we have two possibilities. If ##\Gamma = R^k## then we get that ##k## must be a multiple of ##n## becuase we must have that ##r^k=1##. But in the case ##\Gamma = SR^k## we get that ##r^{2k}=1##. How does this lead us to conclude that there is no such ##k## for which this holds, and hence no elements of form ##SR^k## are in the kernel?
No, that's what my "edit" says. It took me a moment to see (likely due to the current time here).

If ##\Gamma = SR^k## then ##s=r^k##. But ##s\notin \langle r\rangle \subseteq D_{2n}##, so this is not possible.
 
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  • #5
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No, that's what my "edit" says. It took me a moment to see (likely due to the current time here).

If ##\Gamma = SR^k## then ##s=r^k##. But ##s\notin \langle r\rangle \subseteq D_{2n}##, so this is not possible.
One last thing. How do I know that the elements of ##D_{\infty}## always look like either ##SR^k## or ##R^k##? i.e how can we tell from the presentation that ##D_{\infty} = \{R^{-2},R^{-1}, \dots, 1,R,R^2, \dots, SR^{-1}, S, SR^1,SR^2, \dots\}##
 
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  • #6
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We have ##S^2=1## so no higher powers as ##1## are possible for ##S##. Then we have ##SRS=R^{-1}## which is ##SR=R^{-1}S^{-1}=R^{-1}S## so we can always change the order such that ##S## will end up on one side of ##R## to the expense of powers of ##R##, which we are not interested in. We can allow integers, so it doesn't matter if it changes.
 
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