# Normal subgroup of a homomorphism.

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.)

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.
• 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? Oct 14, 2017 at 4:43
• @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? Oct 14, 2017 at 4:49
• 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 $D_{mn}$ ? Oct 14, 2017 at 4:55