What is the meaning of the following quotient group? $S_4 / \{(1 \, 2)(3 \, 4), (1), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \}$
We let $H=\{(1 \, 2)(3 \, 4), (1), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \}$ be a subgroup. Then the above notation is the set of all cosets $\{H, aH, bH, cH, \ldots\}$ where $a, b, c, \ldots\in S_4$.
Now, we get stuck when we are asked to find the order of $S_4 / H$. Is it abelian, cyclic, simple?
We thought it was not abelian since $S_4$ isn't. What is the meaning of being cyclic in this case? etc. We are just massively stuck; I think we don't have a clear picture of quotient group...Can someone explain this problem and any concept behind it?
Thanks for the help!
 A: Some hints...
Order: $|S_4/H|=\frac{|S_4|}{|H|}$
Abelian: Try a 3-cycle.
Cyclic: $S_4/H$ is cyclic if it has a generator. A generator is some element of the form $gH$ with $g\in S_4$ that, only when composed with itself 6 times, is $H$.
In other words, $gH,g^2H,g^3H,g^4H,g^5H\neq H$, but $g^6H=H$. But then, $g,g^2,g^3,g^4,g^5\not\in H$. In particular, $g,g^2,g^3,g^4,g^5\neq e$. This tells us that $|g|> 5$. But what is the highest order of an element in $S_4$?
A: Another way to show $S_4/H \cong S_3$: 
$S_4$ acts on the conjugacy class $\{(12)(34), (13)(24), (14)(23) \}$ by conjugation and the kernel of the action is $H$.
A: Hint: Try listing all of the cosets in the quotient. You should find a group with $6$ elements. Can you show that this group is isomorphic to $S_3$?
A: If one has the language of homomorphisms and the first isomorphism available, then one way to understand $S_4/H$ is as the target of some group homomorphism $S_4\to\square$ with kernel $H$. 
As it happens, such a situation occurs 'in the wild': let $S_4$ act on $X=\{1,2,3,4\}$, and then consider the induced action on the space of all partitions of $X$ of the form $\{\{a,b\},\{c,d\}\}$; applying some permutation $\sigma$ to such a partition will yield $\{\{\sigma(a),\sigma(b)\},\{\sigma(c),\sigma(d)\}\}$. This yields a surjective group homomorphism $S_4\to S_3$ with kernel $H$. The kernel can be computed as the intersection of the stabilizers of these three partitions of given shape, e.g. as I did in this answer.
Since conjugating a permutation simply relabels the terms in its disjoint cycle decomposition, e.g.
$$\sigma(a_1~a_2~\cdots~a_r)\sigma^{-1}=(\sigma(a_1)~~\sigma(a_2)~~\cdots~~\sigma(a_r)),$$
this answer is actually equivalent to Serkan's answer.
