Show that: $\frac{D_n}{\langle a\rangle}\simeq\mathbb{Z_2}$

Question

Show that: $$\frac{D_n}{\langle a\rangle}\simeq\mathbb{Z_2}$$ where $D_n$ is dihedral group and $a$ is generator of order $n$.

Answer 1

Let $D_n$ be the dihedral group of the regular $n$-sided polygon, and let $\langle a \rangle$ be the cyclic subgroup of $D_n$ generated by the rotation $a$. Then, the subgroup $\langle a \rangle$ has index 2 in $D_n$, and hence is a normal subgroup of $D_n$. The quotient group $D_n / \langle a \rangle$ has order 2. This group is isomorphic to $\mathbb{Z}_2$ because there is only one group of order 2 (up to isomorphism), i.e. $\mathbb{Z}_2$.

Answer 2

The quotient group is the group of cosets. When there are only two cosets, the quotient group is a group with only two elements. There is (up to isomorphism) only one group with only two elements.

Answer 3

Let $D_n=\langle a,b:a^n=1, b^2=1, b^{-1}ab=a^{-1}\rangle$ and write multiplicatively $\mathbb{Z}_2\cong C_2=\{1,x\}$.

Define a map $\phi: D_n \rightarrow C_2$ by $$\phi(a)=1$$ $$\phi(b)=x.$$

You can check that $\phi$ yields a surjective homomorphism and $\ker(\phi)=\langle a \rangle$.