Let $K$ be the subgroup $K=\{e,(1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\}$ in $S_4$.
show that $K$ is normal in $S_4$.
show that $S_4/K \cong S_3$.
I know that I could try to prove this directly by the definition of normality, but that would be too long. I am thinking something like applying the First Isomorphism Theorem, like if I could some how make K a kernel of some surjective function. But I just don't know how to deal with functions of cycles.
Can anyone help please? Thanks
On
Hint:
For 1, note that $K$ contains all cycles of the form $(a,b)(c,d)$ in $S_4$
What can you say about conjugation of such a cycle? How can this help you?
For 2 $|S_4|/|K|=6$.
So $S_4/K\cong \mathbb{Z}_6\,\,\,\,\,$ or $\,\,\,\,S_4/K\cong S_3$.
Why can you eliminate the possibility of $\mathbb{Z}_6$?
Let $g \in S_4$ and $k \in K$. We need to show $g^{-1}kg \in K$. Recall that conjugating $k$ by $g$ gives a permutation that has the same cycle structure as $k$. Observe that $K$ contains all permutations of cycle structure $(i,j)(k,\ell)$ and $(i)(j)(k)(\ell)$, hence $g^{-1}kg \in K$ for all $g \in S_4$. Thus $K$ is normal in $S_4$.
Consider the quotient group $S_4/K$. We need to show it is isomorphic to $S_3$. The quotient group has order 6, and up to isomorphism there are only two groups of order 6, namely $S_3$ and $C_6$. But if a coset $gK$ in the quotient group has order 6 for some $g \in S_4$, then $g^6 \in K$, but $S_4$ does not have any nonidentity permutation whose sixth power is in $K$ and whose lesser powers are not in $K$.