Given subgroup of $S_4$ how many cosets does it have?

Question

Given subgroup of $S_4$ how many cosets does it have?
Subgroup $D = \left\langle \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix}\right \rangle $

I am working on this problem. My thinking is that $|D|=4$ and $|S_4| = 4! = 24$. So size of $|Da|=4$, therefore there must be 6 cosets? I believe this falls from Lagrange's theorem, that $|G| = |G:H| |H| $,
thus $24 = |G:H| 4 $,
so $|G:H|=6$

Is this correct thinking or am I wildly off?

Answer 1

Your proof is fine.

You wrote "subgroups" instead of "cosets" though. You haven't said what $a$ is either.

As you have identified, just use the fact that

$$\lvert S_4\rvert=\lvert S_4: D\rvert\lvert D\rvert,$$

which can be rearranged to give

$$\begin{align} \lvert S_4: D\rvert&=\lvert S_4\rvert/\lvert D\rvert\\ &=4!/4\\ &=3!\\ &=6. \end{align}$$