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Let m and n be positive integers. I am asked to Prove that $D_{mn}/ \langle r^m \rangle \cong D_m$.

i belive what i want to do is use the First isomorphism theorem by asking what homomorphism would result in the $\ker \phi =\langle r^m \rangle $ and then looking for the homomorphism.

My question is how do i show that $\langle r^m \rangle$ is a normal subgroup of $D_{mn} $ ?

i managed to do so in a specific example by showing every right coset was equal to a left coset so the subgroup was normal but id like to do it better...( well and for a subgroup of order m.)

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You don't have to show it's normal. Just find a morphism $\phi:D_{mn}\to D_{m}$ which is surjective with $Ker(\phi)=\left\langle r^{m}\right\rangle$. Then $D_{mn}/Ker(\phi)\simeq Im(\phi)$ by the First Isomorphism Theorem.

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  • $\begingroup$ Thats easy to state but i dont really understand whats going on so it gives the impression of it being magic instead of math... $\endgroup$ – Faust Oct 14 '17 at 4:27
  • $\begingroup$ It just follows from the Theorem, which guarantees that the kernel of the morphism is a normal subgroup. For a proof of the statement, I'm sure you can find it in your course textbook. $\endgroup$ – Sir_Math_Cat Oct 14 '17 at 4:30
  • $\begingroup$ I looked through it and still feels odd how do u show that the $\ker \phi $ is the same as $N$ do you show containment both ways? $\endgroup$ – Faust Oct 14 '17 at 4:43
  • $\begingroup$ @Faust Showing that the kernel is equal to $\langle r^m \rangle$ would indeed involve showing containment both ways, as does showing equality between any sets. But the kernel of a group homomorphism is always normal, so once you show that $\langle r^m \rangle$ is the kernel you don't need to worry about anything else. What exactly is bothering you about it? $\endgroup$ – wgrenard Oct 14 '17 at 4:49
  • $\begingroup$ So basically i show that the $\ker \phi = \langle r^m \rangle $ by finding a homomorphism so that $\ker \phi = \langle r^m \rangle $ and i can show its the same by showing containment both ways which implys that $ \langle r^m \rangle $ is a normal subgroup of $D_{mn}$ ? $\endgroup$ – Faust Oct 14 '17 at 4:55

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