Every coset of $V_4$ in $S_4$ contains exactly one element from the set

Question

Consider the symmetric group $S_4$. The set $V_4=\{e,\left(1,2\right)\left(3,4\right),\left(1,3\right)\left(2,4\right),\left(1,4\right)\left(2,3\right)\}$ is a normal subgroup. The set $\{e,\left(1,2\right),\left(1,3\right),\left(2,3\right),\left(1,2,3\right),\left(1,3,2\right)\}$ is a full set of left coset representatives of $V_4$ in $S_4$, meaning that every coset of $V_4$ in $S_4$ contains exactly one element from this set. This set happens to form a subgroup of $S_4$, namely $S_3$ so we immediately see that the quotient $S_4/V_4$ can be identified with $S_3$

There are a lot of information in this and I am trying to understand the coset part.

My understanding so far:

1. $S_4$ contains 3 double transpositions, i.e. product of 2 disjoint 2 cycles and $V_4$ contains all of them. Conjugation preserves the length and the number of cycles in any element, hence $V_4$ is a normal subgroup of $S_4$. -- Is this correct?

2. $\left|S_4/V_4\right|=\left|S_4\right|/\left|V_4\right|=4!/4=6$. Since $S_4$ contains no element of order 6, it follows that $S_4/V_4$ does not contain any element of order 6, therefore $S_4/V_4$ is not abelian hence it is not cyclic. Since there are only 2 groups of order 6, cyclic group and $S_3$, $S_4/V_4$ is isomorphic to $S_3$. -- I need confirmation for this proof too.

I have been given 2 statements on this problem:

a) No two elements represent the same coset

b) Every coset of $V_4$ in $S_4$ contains exactly one element from the set

I have manually confirmed both statements, however I am sure how to prove these.

They do seem to be linked so once I understand one, I should be able to understand the other... But I don't know where to start.

Any help will be appreciated!