A difficulty in understanding a statement in the method for finding all normal subgroups of the symmetric group $S_{4}$

Question

I was reading this question :

Normal subgroups of $S_4$

And I had a difficulty in understanding this statement:

"Now, any subgroup that contains all transpositions is the whole group."

Could anyone explain for me why this statement is true please?

Answer 1

The transpositions generate the symmetric group: each permutation is a product of transpositions. For example, a cycle $$(a_1\ a_2\ \cdots\ a_k)=(a_1\ a_2)(a_2\ a_3)\cdots(a_{k-1}\ a_k).$$

Answer 2

The answer to the linked question states that if a subgroup $H$ of $S_4$ contains all transpositions $(i\,j)$ for $i=1,2,3,4$, $j\ne i$ (there are $6$ such permutations because $(i\,j)=(j\,i)$), then $H=S_4$. It follows from the fact that any element (permutation) of the symmetric group $S_4$ (and more generaly $S_n$) is composition of a finite number of transpositions; in other words, transpositions of $S_4$ form a system of generators for $S_4$: explicitly $$\langle (1\,2),(1\,3),(1\,4),(2\,3),(2\,4),(3\,4)\rangle =S_4$$.