Show that $G/H\cong S_3$ 
If $G:=S_4$ and $H:=\{id,(12)(34),(13)(24),(14)(23)\}$ Show that $G/H$ has order $6$ and all of its elements have order less than or equal to $3$ (so by the classification of the groups of order 6 up to isomorphism, $G/H\cong S_3$)

$H$ is normal, since left- and right cosets are equal, There are $6$ cosets:
$\{id,(12)(34),(13)(24),(14)(23)\},\ \{(12),(34),(1423),(1324)\},\ \{(13),(24),(1234),(1423)\}$
$\{(123),(134),(243),(142)\},\ \{(14),(23),(1243),(1342)\},\ \{(132),(143),(234),(124)\}$
What does it mean: All elements of $G/H$ have order less than or equal to $3$?  Each element is a set of 4 permutations, what is the order here ?
$G/H\cong S_3$: How can I construct an isomorphism, since the cosets are disjoint, can I just pick an element from each group and compare them with $S_3$ ?
Thanks in advance.
 A: Here is a way to construct an homomorphism from $S_4$ to $S_3$, which has kernel $H$. Consider the three possible partitions of the set $\{1,2,3,4\}$ into two sets:
$$A=\{1,2\}\cup\{3,4\}; \quad B=\{1,3\}\cup \{2,4\}; \quad C=\{1,4\} \cup \{2,3\}.$$
Now you can see what does an element $p\in S_4$ to those partitions. For instance the element $p := (1234)$ does the following:
\begin{align*}
\{1,2\} &\overset{p}{\leadsto} \{2,3\}; &\{1,3\}\overset{p}{\leadsto}\{2,4\};\\\{1,4\}&\overset{p}{\leadsto} \{2,1\}; &\{2,3\}\overset{p}{\leadsto}\{3,4\}; \\
\{2,4\} &\overset{p}{\leadsto}\{3,1\}; &\{3,4\}\overset{p}{\leadsto}\{4,1\},
\end{align*}
so $$A \overset{p}{\leadsto}\{2,3\}\cup \{4,1\}; \quad B \overset{p}{\leadsto} \{2,4\}\cup \{1,3\}; \quad C \overset{p}{\leadsto} \{2,1\}\cup \{3,4\}.$$
This corresponds to the transposition $[ABC]\overset{p}{\leadsto}[CBA]$. Which is like the transposition $(13)$ in $S_3$. Now define the map $$\varphi:S_4\to S_3,$$
which sends an element of $S_4$ to the action on the partitions $A,B,C$. If $q,r \in S_4$ then the product $qr$ corresponds to the composition of the action of $q$ and of $r$, so the action of $qr$ is the result of the two actions, that is $\varphi(qr)=\varphi(q)\varphi(r)$ and so you have an homomorphism. Moreover you can see that the only elements which sends $[ABC]$ to $[ABC]$ (written in one line notation) are exactly the elements of $H$, so $H$ is the kernel of $\varphi$ and therefore $S_4/H\approx S_3$
A: $H$ is a conjugacy class. If a conjugacy class is a subgroup, then it is a normal subgroup. So, $H$ is normal in $S_4$.
There is no need to find an isomorphism. You can do it as follows which is easier.
$S_3 \cap H = \{ e \}$. Also, $|S_3H|=\frac{|H||S_3|}{|S_3 \cap H|}$. So $S_3H=G$.
According to the second isomorphism theorem:
$G/H \cong S_3H/H \cong S_3/ H\cap S_3 \cong S_3 $
