I have managed to do part i and ii. In part ii, the number of left cosets of H is 4.
I am stuck with part iii. I am guessing I will have to use first isomorphism theorem but not sure how.
$V_4$ is the potential kernel as it is normal subgroup (part i)
But now i am not sure on what to do next.
Do I have to find homomorphism from $S_4$ to some group such that that group has a subgroup $S_3$? But then how do I show that this is actually isomorphic to S4/V4 ?
Thanks. I am confused about first isomorphism theorem.
Best Answer
Suppose $$ h_{1} V_{4} = h_{2} V_{4} ,$$ with $h_{1}, h_{2} \in H$. Then $h_{2} = h_{1} v$ for some $v \in V_{4}$. Therefore $$ 4 = 4 h_{2} = 4 h_{1} v = 4 v. $$ But the only element $v \in V_{4}$ that fixes $4$ is $v = e$, so $h_{2} = h_{1} e = h_{1}$.
Now you know that there are $6$ distinct cosets $h V_{4}$, for $h \in H$. Since $\lvert S_{4} : V_{4} \rvert = 6$, all cosets of $V_{4}$ are of this form, and thus $S_{4} = V_{4} H$. It follows (or you can check) that $V_{4} \cap H = \{ e \}$. Therefore $$ \frac{S_{4}}{V_{4}} = \frac{V_{4} H}{V_{4}} \cong \frac{H}{V_{4} \cap H} = \frac{H}{\{ e \}} \cong H. $$ Here I have used the second isomorphism theorem.