I'm going through a physicist viewpoint of Galois theory and a found a lecture by Takeuchi here. In the slide 47 he showed that $S_{n}$ is not solvable for $n\geq 5$. This is the proof
Let $G$ be a group of permutations of five objects or more that
include all cyclic permutations of three elements.
$(124)(142)=e$
$(135)(153)=e$
$(123)=(124)(135)(142)(153)$
Let $H$ be an invariant normal subgroup of $G$ such that $G/H$ is cyclic. And consider the homomorphism $f:G\rightarrow G/H$
$f[(124)]=x$
$f[(135)]=y$
$f[(123)]=f[(124)(135)(142)(153)]=xyx^{-1}y^{-1}=e$
Therefore $(123)\in H$. This is true for any cyclic permutation of three
elements. Therefore, $G$ is not solvable.
I don't understand particularly why the fact that $(123)$ is mapped to the identity implies $(123)\in H$. I would appreciate some help.
Best Answer
That map from $G$ onto $G/H$ is the map $g\mapsto gH$ and the identity element of $G/H$ is $H(=eH)$. So, $g(\in G)$ is maped into $eH$ if and only if $gH=eH$, which means that $g\in H$.