Let $K$ be the commutator subgroup of $S_n$ and $t$ a transposition of $S_n$. Then $\langle t\rangle K$ is a normal subgroup of $S_n$ since $S_n/K$ is abelian.
I don't get it. How can I use the fact that $S_n/K$ is abelian to prove the $\langle t\rangle K$ is normal?
Best Answer
Let $g\in S_n$ and $x\in\langle t\rangle K$. We have to prove that $gxg^{-1}\in\langle t\rangle K$.
By definition we can write $x=t^ik$ for some $k\in K$. Since $S_n/K$ is abelian we have $gt^iK=t^igK$. So there is some $k_1\in K$ such that $gt^ik=t^igk_1$. And then:
$gxg^{-1}=gt^ikg^{-1}=t^i(gk_1g^{-1})\in \langle t\rangle K$
We used the fact that $gk_1g^{-1}\in K$. (which is true because $K$ is normal in $S_n$)