Calculate quotients of $\mathbb S_3$ and $\mathbb D_4$ Problem
Calculate all the quotients by normal subgroups of $\mathbb S_3$ and $\mathbb D_4$,i.e., charactertize all the groups that can be obtained as quotients of the mentioned groups.
For the case $\mathbb S_3$, by Lagrange's theorem, all possible subgroups of $\mathbb S_3$ have to be of order $2$ or $3$. If $ord(S)=3$, then as $|G:S|=2$, so $S \lhd \mathbb S_3$. If the subgroup is of order $2$, then $S=\{id,(ab)\}$, it is easy to see that $S$ is not normal just by calculating $(bc)(ab)(bc)^{-1}$ for $c \neq a,b$. So the only normal subgroups are the ones of order $3$.I don't know how  I could "characterize" $\mathbb S_3/S$. I would appreciate help with that.
As for the case $\mathbb D_4$, I am having a hard time trying to figure out which are the normal subgroups (appart from the subgroup of rotations), I would appreciate suggestions to calculate the normal subgroups and then an idea on how could I characterize them. Thanks in advance.
 A: Your solution for $S_3$ looks good. 
For $D_4$, or more generally for
$$D_{n}=\langle r,s:r^n=s^2=1,srs=r^{-1}\rangle$$
note that $\langle r\rangle$ is a normal subgroup of index $2$ in $D_{n}$. If $A$ is any (non-trivial) normal subgroup of $D_{n}$, then there are two cases


*

*$A\subset \langle r\rangle$: These are easy to find since $\langle r\rangle$ is a cyclic groups.

*$A\not\subset \langle r\rangle$: Then $A\cap \langle r\rangle$ is a subgroup of index $2$ in $\langle r\rangle$ which is unique if $n$ even and non-existent if $n$ is odd.


Apply this for $n=4$, we see that


*

*$A$ can be: $1$, $\langle r^2\rangle\cong C_2$, $\langle r\rangle\cong C_4$.

*$A$ can be: $\langle r^2,s\rangle\cong C_2\times C_2$.

A: To find all the quotients of a group $G$, is equivalent to finding all of its normal subgroups. We always have the trivial quotients $G/G$ and $G/\{e\}$.
$D_4$ has non-trivial proper subgroups of orders $2$ and $4$. So let's find all of these groups, and see which ones are normal.
The subgroups of order $2$ are easiest to find, they are just $\{e,b\}$ for some element $b \in D_4$ of order $2$. We have $5$ of these ($1$ rotation, and $4$ reflections).
Suppose we write $r$ for a rotation of order $4$, and $s$ for a reflection. Then:
$D_4 = \{e,r,r^2,r^3,s,sr,sr^2,sr^3\}$
Let's handle all the reflections at once: let $H_i = \{e,sr^i\}$, for $i = 0,1,2,3$. We know that: $rs = sr^3 = sr^{-1}$.
So let's look at $rH_ir^{-1}$. Clearly, $rer^{-1} = rr^{-1} = e$, so the identity element is unaffected. So we may focus solely on $r(r^is)r^{-1}$.
Now $r(sr^i)r^{-1} = (rs)(r^ir^{-1}) = (sr^{-1})r^{i-1} = sr^{i-2}$. If this is to equal $sr^i$, we must have:
$i - 2 \equiv i\text{ (mod }4)$, which is never the case for $i = 0,1,2,3$. So none of these subgroups are normal.
Thus the only possible normal subgroup of order $2$ is: $\{e,r^2\}$. Note that: $sr^2 = sr^{-2} = (sr^{-1})r^{-1} = (rs)r^{-1} = r(sr^{-1}) = r(rs) = r^2s$, so $r^2$ commutes with $s$, and clearly commutes with $r$, so commutes with all of $D_4$. It follows, then, that for any $g \in D_4$ we have $gr^2g^{-1} = r^2gg^{-1} = r^2$, so this subgroup is normal.
Now any subgroup of order $4$ is of index $2$, and is thus necessarily normal. Since the product of two reflections is a rotation, these subgroups must either be:
$\{$identity,rotation,rotation,rotation$\}$, or:
$\{$identity,reflection,reflection,rotation$\}$
Clearly the first subgroup is $\{e,r,r^2,r^3\}$, so the only ones we haven't found are of the latter type. The rotation must be of order $2$, or else we would have a cyclic group which would be all rotations.
So we have at least one such group: $\{e,r^2,s,sr^2\}$. If we have another such subgroup, it must be of the form: $\{e,r^2,a,b\}$ and neither $a$ nor $b$ can be $s,r^2s$. Since we cannot choose:
$a = e,r,r^3,s,sr^2$, are only options are $a = sr$ or $sr^3$. No matter which we choose, the "other one" must be the $4$th element, and you can easily verify $\{e,r^2,sr,sr^3\}$ is a subgroup of $D_4$.
So that's all the normal subgroups. If our normal subgroup $N$ is of order $4$, then $D_4/N$ is of order $2$, and isomorphic to $C_2$.
That leaves us with just $D_4/\langle r^2\rangle$. Note that (setting $\langle r^2\rangle = K)\ $ $sr^iK$ is of order $2$ since $sr^i \not\in K$ and $(sr^i)^2 = e$. Also, $rK = r^3K$ is of order $2$ for the same reason. We conclude $D_4/K$ has no elements of order $4$, and is thus isomorphic to $V_4 \cong C_2 \times C_2$. So we obtain the following quotients of $D_4$:


*

*$D_4$

*$V_4$

*$C_2$

*$\{e\}$

A: Hint for $\mathbb D_4$
$\mathbb D_4 = \{1, r, r^2, r^3, m, mr, mr^2, mr^3\}$
You are looking for six groups in total. There are two trivial cases, one subgroup of order 2, and three subgroups of order 4.
Try to get the trivial, and the order 2, out of the way first.
A: The only non-trivial normal subgroups of $S_3$ is $A_3.$ Hence quotients are
$$S_3/S_3\cong1,\qquad S_3/A_3\cong\mathbb{Z}_2,\qquad S_3/\{\omega\}=6.$$
The subgroups of $D_8$ is given by 
Among these subgroups except $\langle s\rangle, \langle rs\rangle, \langle r^2s\rangle, \langle r^3s\rangle$ everything else is normal. In fact, the lattice of normal subgroups of $D_8$ is isomorphic to that of the quaternion group (this is a fun exercise do by your self). Lets focus on non-trivial, proper normal subgroups. Quotient of $D_8$ by any subgroup of order $4$ is isomorphic to $\mathbb{Z}_2.$ To quotient by $\langle r^2\rangle$ simply set $r^2=1$ in $D_8=\{1, r, r^2, r^3, s, rs, r^2s, r^3s\}.$ That gives us the quotient $$\langle r,s : r^2=s^2=1, rs=sr\rangle\cong \mathbb{Z}_2\times\mathbb{Z}_2.$$
