Let $G = S_4$ be a group, $N = \{1, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\}$ a normal subgroup of G. It's easy to see that $G/N$, the set of cosets is $G/N = \{a, b, c\}$, where
$$a = \{(1), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\},$$ $$b = \{(1, 3)(1, 2), (2, 3)(2, 4), (1, 2)(1, 4), (1, 4)(1, 3)\},$$ $$c = \{(1, 2)(1, 3), (2, 4)(2, 3), (1, 4)(1, 2), (1, 3)(1, 4)\}.$$
The question that I'm trying to solve is asking me to determine which known group $G/N$ is isomorphic to.
It's easy to check that the group of elements of $a, b, c$ is the alternating group $A_4$, but this isn't what the question is asking. Perhaps an intuitive question to ask would be "how should I view the elements of $G/N$?
$\;\left|S_4/N\right|=6\implies S_4/N=C_6\;\;or\;\;S_3\;$, because these two are the only groups of order six up to isomorphism.
Yet the first option is possible iff $\;[S_4,S_4]=A_4\le N\;$ , so it actually is the second one.