Understanding a Sylow theory proof

[Mr Davis 97]

https://imgur.com/a/oSioYel

I am trying to understand this proof, but am tripped up on the part that says "Consider the action of $G$ on $\operatorname{Syl}_2(G)$ by conjugation." My question is, how is this a well-defined action if $\operatorname{Syl}_2(G)$ is not normal? Isn't this action by conjugation defined only on subgroups that are normal?

 

[fresh_42]

https://imgur.com/a/oSioYel

I am trying to understand this proof, but am tripped up on the part that says "Consider the action of $G$ on $\operatorname{Syl}_2(G)$ by conjugation." My question is, how is this a well-defined action if $\operatorname{Syl}_2(G)$ is not normal? Isn't this action by conjugation defined only on subgroups that are normal?

Given any operation $\varphi \, : \, U\times G \longrightarrow G$ with a subgroup $U\leqslant G$ we can define cosets $g.U=\varphi(g,U)=\{\,g.u\,|\,u\in U\,\}$ which in our case are the sets $gUg^{-1}$. This partitions $G$ into subsets of equal size.

Now the point is, that those sets are just that: sets. Since $U$ isn't normal, we simply cannot define a group structure on this set $G/U$ of sets. So as long as we are only counting elements, everything will be fine. If we want to consider homomorphisms $U \rightarrowtail G \twoheadrightarrow G/U$ we will have a problem if $U$ isn't normal. So normality directly corresponds to the fact, that $G/U$ is again a group or not.

 

[Infrared]

$Syl_2(G)$ is not a subgroup of $G$; rather, it is the set of all of the 2-Sylow subgroups of $G$. If $H\in Syl_2(G)$, then $gHg^{-1}\in Syl_2(G)$ too (and can be a different element if $H$ is not normal).

 

[Mr Davis 97]

$Syl_2(G)$ is not a subgroup of $G$; rather, it is the set of all of the 2-Sylow subgroups of $G$. If $H\in Syl_2(G)$, then $gHg^{-1}\in Syl_2(G)$ too (and can be a different element if $H$ is not normal).

Got it. One more question. Why does that fact that $\rho$ is transitive imply that it is nontrivial?

 

[Infrared]

Let $H_1,H_2\in Syl_2(G)$. Since the action is transitive, there is an element $g\in G$ such that $gH_1g^{-1}=H_2$. We can pick $H_1,H_2$ to be different, and then this means that $\rho(g)$ is not the trivial permutation.