In this material, page 5, Section Construction of an outer automorphism of $S_6$.
$S_5$ acts by conjugation on six cyclic subgroups of cardinality 5 (its six Sylow 5-subgroups actually) induces homomorphism $j:S_5 \to S_6$.
Write $H=j(S_5)$, $H\triangleleft S_6$ and $[S_6:H]=[S_6:S_5]=6$, $S_6$ acts by left translation on $S_6/H$ induces $F:S_6\to S_6$.
Suppose $F((12))$ is a transposition, then there's a coset $xH$ s.t. $(12)xH=xH$, $x^{-1}(12)x \in H$ and is a transposition. It's proved before that $j$ preserves charity, so if $j(\sigma)=x^{-1}(12)x$, $\sigma$ is odd.
Then it claims without proof that $\sigma$ is a transposition and normalizes an Sylow $5$-subgroup $P$ (i.e. $\sigma P\sigma^{-1}=P$), and in the next step it assumes $P=\langle(12345)\rangle$ and $\sigma$ fixes $1$.
My question:
1.Why must $\sigma$ be a transposition?
What we know here is $\sigma$ is preimage of transposition, so it's odd.
For $\alpha=(24)(25)(23)=(2354)$, $\alpha$ is odd, $\alpha(12345)\alpha^{-1}=(2354)(12345)(2354)^{-1}=(13524)=(12345)^2$
2.Why can we assume $P=\langle(12345)\rangle$ and $\sigma$ fixes $1$ here?
Does this assumption really not lose any generality?
Thanks for your time and effort!
$1$. Look at its cycles decompositions. There are only $5$ elements, it can't be a disjoint product of $3$ transpositions.
EDIT: $\sigma$ is of order $2$ because it's the image of a monomorphism from an element of order $2$, so its cycles decomposition must be just transpositions. It's odd so the number of transpositions must be odd.
$2$. A group of $5$ elements must be generated by an element of order $5$, and in $S_{5}$ it must be a $5$-cycle. Choosing $1,2,3,4,5$ is just naming the elements to match the cycle. You can pick whatever $\sigma$ fixed and call it $1$, then follow it along the cycle and name the subsequent elements $2,3,4,5$.