# Why does group action by conjugation on sylow subgroups define a homomorphism into the symmetric group?

Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?

• This action is used in the proof that the Sylow $p$-subgroups are conjugate. – Derek Holt Apr 6 '15 at 11:03

## 1 Answer

In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $g\in G$ then $gSg^{-1}$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^{-1}$ is a subgroup of cardinal $|S|=p^k$ that $gSg^{-1}$ is also a $p$-Sylow. This answers your last question.

For the first question. Suppose $X=\{S_1,...,S_n\}$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $n\geq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $g\in G$ then :

$$g.S:=gSg^{-1}\in X$$

It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^{-1}=ghSh^{-1}g^{-1}=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :

$$\rho: G \rightarrow Bij(X)$$

$$g\mapsto [S\mapsto gSg^{-1}]$$

The fact that $\rho(g)$ is one to one and onto comes from the fact that $\rho(g^{-1})=\rho(g)^{-1}$ furthermore from the rules of group action, $\rho$ is actually a group morphism from $G$ to $Bij(X)$ (with the law of composition).

Now the last thing to use is that $X$ is in bijection with $\{1,...,n\}$. Let us set :

$$\psi: \{1,...,n\}\rightarrow X$$

$$i\mapsto S_i$$

Then you have $\psi^{-1}\circ\rho(g)\circ\psi$ is a bijection of $\{1,...,n\}$ i.e. an element of $\mathfrak{S}_n$ (a permutation). So the morphism you are looking for is explicitely :

$$\rho_0:G\rightarrow \mathfrak{S}_n$$

$$g\mapsto \psi^{-1}\circ\rho(g)\circ\psi$$

It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $\psi$) nevertheless, it is canonical up to conjugation in $\mathfrak{S}_n$.

• Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group. – Sundip Apr 6 '15 at 11:03
• Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $n\geq 2$ and $|G|>n!$ then $G$ cannot be simple. – Clément Guérin Apr 6 '15 at 11:12