Describe all homomorphic images of the dihedral group $D_3$ of order 6

Question

I saw a solution for $D_4$ where they just found all the normal subgroups and found the quotient groups. Is that sufficient here? How do I even do that? I know that one of them is the kernel of $D_3$, which is just all of the symmetries

Answer 1

Is that sufficient here?

Yes, this is the content of the first isomorphism theorem.

How do I even do that?

I think $D_3$ is small enough to do by hand (listing out subgroups and finding the normal ones, as suggested in the comments).

---

You did not list the elements of $D_3$ correctly if you are writing in terms of generators. (Part of it is my fault, I wrote the presentation of $D_4$ rather than $D_3$ in the comments.) Moreover, you have subgroups of size $4$, which does not divide $|D_3|=6$.

If you use the presentation $\langle r,s \mid r^3=s^2=(sr)^2=e \rangle$, then the elements can be listed as $\{e, r, r^2, s, sr, sr^2\}$. The subgroups are $\{e\}$, $\{e, s\}$, $\{e, sr\}$, $\{e, sr^2\}$, $\{e, s, s^2\}$, and $D_3$. You can check that only $\{e\}$, $\{e, s, s^2\}$, and $D_3$ are normal.

If you want to use the notation given in your link, then the subgroups are $\{id\}$, $\{id, m_1\}$, $\{id, m_2\}$, $\{id, m_3\}$, $\{id, r_1, r_2\}$, and $D_3$. Only $\{id\}$, $\{id, r_1, r_2\}$, and $D_3$ are normal.