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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}##.