1st Isomorphism thm for dihedral gps

[Mr Davis 97]

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, $D_\infty/\langle R^n \rangle\cong D_{2n}$.

 

[fresh_42]

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.

 

[Mr Davis 97]

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 $n$ because 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?

 

[fresh_42]

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 $n$ because 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.

 

[Mr Davis 97]

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 $D_{\infty} = \{R^{-2},R^{-1}, \dots, 1,R,R^2, \dots, SR^{-1}, S, SR^1,SR^2, \dots\}$

 

[fresh_42]

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.