A group action proof without group actions? I am currently teaching an undergraduate abstract algebra course out of Saracino, Abstract Algebra: A First Course. Exercise 13.13 asks the following:

Let $K$ be the subgroup $\{ e, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\}$ in $S_4$. Prove that $K$ is normal in $S_4$ and that $S_4 / K \cong S_3$.

The solution that naturally occurs to me is to consider the natural group action on the set consisting of unordered disjoint pairs of pairs of elements of {1, 2, 3, 4}, to prove that it is surjective, and to prove that the kernel of the action is $K$. However, Saracino has not yet introduced group actions in his book! 
It is possible to translate this proof out of the language of group actions, but this strikes me as a little bit clumsy. Is there any other proof he may have plausibly had in mind?
 A: $\sigma(1\ 2)(3\ 4)\sigma^{-1}=(\sigma(1)\ \sigma(2))(\sigma(3)\ \sigma(4))\in K$ for any $\sigma\in S_4$, and so on. For the second part note that $S_4/K$ has six elements and no element of order six.
A: How about just picking a bijection between the set $\{1,2,3\}$ and the set of unordered disjoint pairs of pairs of elements of $\{1,2,3,4\}$. However you intend to prove surjectivity and that the kernel is $K$ with respect to permutations of the set of unordered disjoint pairs of pairs can be directly transferred to proving the same with respect to permutations of the set $\{1,2,3\}$. This could actually be a good thing to do pedagogically as well, because it edges the students away from thinking of $S_3$ symbolically and thinking of it instead as represented by the permutation group of any 3 element set.
A: I think  the best way is counting the elements of the form $(a,b)(c,d)$
$$=\dfrac{1}{2}.{4\choose2}{2\choose2}=3$$ which means that your group is uniqe as a result it is normal.
To see that $S_4/V_4=S_3$ show that $S_4/V_4$ is not abelian. 
Note: we divide by $2$ as $(a,b)(c,d)=(c,d)(a,b)$.
