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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.

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    $\begingroup$ Since $H$ is normal, $G/H$ is a group, and for each $\sigma\in S_4$, $\sigma H$ is an element of $G/H$. The order of $\sigmaH$ is the least $k$ for which $\sigma^k\in H$. $\endgroup$ – Pedro Tamaroff Nov 1 '14 at 21:41
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    $\begingroup$ It's possible to construct a specific isomorphism, but it would not be easy. They want you to show that it's isomorphic to $S_3$ in an easier way. If the group had an element of order greater than $3$, then the group would be $\mathbb{Z}/6\mathbb{Z}$, but if the order of every element is $\leq 3$ then it must be $S_3$. $\endgroup$ – Matt Samuel Nov 1 '14 at 21:42
  • $\begingroup$ @Pedro Tamaroff Thanks but there are $24$ $\sigma$'s, so it is possible that $\sigma H=\sigma' H$ for $\sigma\neq\sigma'$ but must they have the same exponent to belong to $H$ ? $\endgroup$ – inequal Nov 1 '14 at 21:53
  • $\begingroup$ @Matt S Why it cannot be $\mathbb Z/6\mathbb Z$ ? I mean how can I show that no element has order $6$ ? Since $G/H$ has order $6$, any element, say $\sigma_1$ of a coset exponented with $6$ gives identity, but $id=((12)(34))^2$ for example and hence $\sigma_1^3=(12)(34)$ etc. Am I on the right track ? $\endgroup$ – inequal Nov 2 '14 at 11:18
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    $\begingroup$ A representative may have order six in $G$, but the order of its coset could still be less in $G/H$, if a lower power would belong to the subgroup $H$. When we pass to the quotient group $G/H$ the elements of $G$ lose their identity to the extent that any two belonging to the same coset of $H$ become equal. $\endgroup$ – Jyrki Lahtonen Nov 2 '14 at 11:56
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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 $S_4/H\approx S_3$

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