Identify the isomorphism of the group $D_6 / \langle\rho ^3\rangle$.

Question

Identify the isomorphism of the group $D_6 / \langle \rho ^3\rangle$ among the following options. Choose one.

$\mathbb{Z}_3, D_3, \mathbb{Z}_2 \times \mathbb{Z}_3, \mathbb{Z}_6$

My thought is that since the index of $\langle \rho ^3\rangle $ is $2$, the number of cosets is $6$.

So I thought I only need to find a group of order $6$ but there are three of which the order is $6.$

Is there any other restrictions I need to consider?

Answer 1

A presentation for $D_6$ is

$$\langle \rho ,\sigma \mid \rho^6, \sigma^2, \sigma \rho=\rho^{-1}\sigma\rangle,\tag{1}$$

so that taking the quotient of $D_6$ by $\langle \rho^3\rangle$ amounts to killing $\rho^3$; that is, setting it equal to the identity in $(1)$, like so:

$$\langle \rho ,\sigma \mid \rho^3, \sigma^2, \sigma \rho=\rho^{-1}\sigma\rangle.\tag{2}$$

The relation $\sigma \rho=\rho^{-1}\sigma$ stays the same, since $\rho^2=\rho^5=\rho^{-1}$.

But $(2)$ is just a presentation of $D_3$. Hence the answer is $D_3$.

Answer 2

The order of $\Bbb Z_3$ is wrong. Next $\Bbb Z_3×\Bbb Z_2\cong\Bbb Z_6$. But $\Bbb Z_6$ is out, because of the fact that $\langle\rho^3\rangle=Z(D_6)$. It's well known that if a group modulo its center is cyclic, then the group is abelian.

We are left with $D_3$. As a check, it is well-known that $D_6\cong S_3\times\Bbb Z_2$. Finally, $S_3\cong D_3$.