I have to show that, given $k|n,$ and $\langle{r^k}\rangle$ is a normal subgroup of the Dihedral group $D_{2n}$ then $D_{2n}/\langle{r^k}\rangle\cong D_{2k}$

Given this, I know that I need to show that a particular homomorphism is a bijection. The natural one I was thinking was $$\phi:D_{2k}\rightarrow D_{2n}/\langle{r^k}\rangle$$ $$r^is^j\mapsto r^is^j\langle{r^k}\rangle$$ First I have to show that this is a homomorphism, which is easy, i think, since $$\phi(r^{i_1}s^{j_1}r^{i_2}s^{j_2})=r^{i_1}s^{j_1}r^{i_2}s^{j_2}\langle{r^k}\rangle=r^{i_1}s^{j_1}\langle{r^k}\rangle r^{i_2}s^{j_2}\langle{r^k}\rangle=\phi(r^{i_1}s^{j_1})\phi(r^{i_2}s^{j_2})$$ Now that I have the homomorphism, I need to show bijectivity. So given $\phi(r^{i_1}s^{j_1})=\phi(r^{i_2}s^{j_2})$ $$\phi(r^{i_1}s^{j_1})=\phi(r^{i_2}s^{j_2})\Rightarrow r^{i_1}s^{j_1}\langle{r^k}\rangle=r^{i_2}s^{j_2}\langle{r^k}\rangle$$ and it's here I'm getting stuck...drawing a blank

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    $\begingroup$ You may find this link useful. $\endgroup$ – vasmous Oct 3 '15 at 17:39

I think that is the "hard way" to show isomorphism. An easier way is to use the Fundamental Isomorphism Theorem, and display a homomorphism:

$\psi: D_{2n} \to D_{2k}$

with $\text{ker }\psi = \langle r^k\rangle$.

I suggest taking $\psi(r^j) = r^{j\text{ mod }k}$, and $\psi(s) = s$.

To prove $\psi$ is a homomorphism, it will suffice to show $\psi(r^n) = 1 = \psi(s^2)$, and $\psi(s)\psi(r) = \psi(r)^{-1}\psi(s)$. You will need to use the fact that $k|n$, here.


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