# Understanding a Sylow theory proof

## Main Question or Discussion Point

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?

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fresh_42
Mentor
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
Gold Member
$Syl_2(G)$ 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.
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