11-1100-Pgadey-Lect6

Theorem

Every $G$-set is a disjoint union of "transitive $G$-sets"

Theorem

If $X$ is a transitive $G$-set and $x\in X$ then $X\simeq G/Stab(x)$ where the isomorphism an isomorphism of $G$-sets.

Transitive $G$-set

A $G$-set $X$ is transitive is $\forall _{x,y\in X}\exists _{g\in G}\ st.\ gx=y$.

Stabilizer of a point

We write $Stab(x)=\{g\in G:gx=x\}$ for the stabilizer subgroup of $x$.

Proof We define an equivalence relation $x\sim y\iff \exists _{g\in G}gx=y$. This relation is reflexive since $x=ex$ and thus $x\sim x$. This relation is symmetric since $y=gx$ implies $g^{-1}y=x$. This relation is transitive, since if $x=gy$ and $y=hz$ then $x=ghz$. It follows that $X=\coprod _{i\in I}Gx_{i}$ where $Gx_{i}$ denote the orbit of a point $x_{i}$.

We then claim that $Gx_{i}$ is a transitive $G$-set. [Dror: "[This fact] is too easy."]

We show that $Gx$ is isomorphic to $G/Stab(x)$ as a $G$-set.

We produce two morphism $\Psi :Gx\rightarrow G/Stab(x)$ and $\Phi :G/Stab(x)\rightarrow Gx$.

To define $\Psi$ there is only one thing we can do. We have $y\in Gx\Rightarrow y=gx$ and then we define $\Psi (y)=gStab(x)$. We check that this map is well defined. If $y=gx=g'x$ then $g^{-1}g'x=x$ and hence $g^{-1}g\in Stab(x)$. It follows that $gStab(x)=g'Stab(x)$. Thus $\Psi$ is well defined.

To define $\Phi$ we take $gStab(x)\in G/Stab(x)$ and define $\Phi (gStab(x))=gx$. We show that this map is well defined. If $gStab(x)=g'Stab(x)$ then $g^{-1}g'\in Stab(x)$ and hence $g^{-1}g'x=x$. It follows that $gx=g'x$ and hence $\Phi$ is well defined.

We need to check that $\Psi$ and $\Phi$ are mutually inverse and $G$-set morphisms. We quickly check that $\Phi$ is a $G$-set morphism. If $y=gx$ and $g_{1}\in G$ then $g_{1}\Psi (y)=g_{1}(gStab(x))=(g_{1}g)Stab(x)$. Similarly, $\Psi (g_{1}y)=g'Stab(x)=(g_{1}g)Stab(x)$. The last inequality follows since we can take any $g'$ such that $g'y=g_{1}y$. Why not take $g'=g_{1}g$ -- since we know that works.

Theorem (Orbit-Stabilizer)

If $|X|<\infty$ and $X=\coprod _{i\in I}Gx_{i}$ then $|X|=\sum _{i}{\frac {|G|}{Stab(x_{i})}}$.

This is just a rewriting of the theorem above.

$p$-Group

A $p$-group is a group $G$ with $|G|=p^{k}$ for some $k$.

$G=(Z/2Z)^{3},(Z/2Z)\times (Z/4Z),Z/8Z,D_{8},Q=\{\pm 1,\pm i,\pm j,\pm k:i^{2}=j^{2}=k^{2}=-1,ij=k\}$

The last group $Q$ is the famous unit quaternions -- They need a better description here.

Theorem

Any $p$-group has a non-trivial centre.

Let $G$ act on itself by conjugation. Decompose $G=\coprod Gx_{i}$. Then, $|G|=\sum _{|Gx_{i}|=1}1+\sum _{|Gx_{i}|>1}{\frac {|G|}{|Stab(x)|}}$ Observe that $|Gx_{i}||=1$ iff $x_{i}\in Z(G)$. It follows that $|G|=|Z(G)|+\sum _{|Gx_{i}|>1}{\frac {|G|}{|Stab(x)|}}$ The formula above is called "the class formula". We have that $|G|/|Stab(x)|=p^{k}$ for some $1<k$ since $Stab(x)$ is a subgroup. It follows that $|G|\equiv 0\ \mod \ p$ and $\sum _{|Gx_{i}|>1}{\frac {|G|}{|Stab(x_{i})|}}\equiv 0\ \mod \ p$. It follows that $|Z(G)|\equiv 0\ \mod \ p$. Since $e\in Z(G)$ we have $1\leq |Z(G)|$ and thus $p\leq |Z(G)|$.

SYLOW

A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.

Cauchy's Lemma

If $A$ is an abelian group and $p$ divides $|A|$, then there is an element of order $p$ in $A$.

Proof. Pick $x\in A$. If $p$ divides the order of $x$ then we have $x^{np}=e$ for some $n$. It follows that $(x^{n})^{p}=e$. We then have that the order of $x^{n}$ is $p$. If $p$ does not divide the order of $p$, then consider $A/<x>$. Since $A$ is abelian, $<x>$ is a normal subgroup. We have that $p$ divides $|A/<x>|$, and $|A/<x>|<|A|$. We then induct. Let $y<x>$ have order $p$, that is $(y<x>)^{p}=<x>$. We then have that $y^{p}=x^{k}$ for some $k$. We write $|<y>|=np+r$ where $0\leq p<r$. We then have $e=y^{|<y>|}=y^{np+r}=x^{nkp}y^{r}\Rightarrow y^{r}\in <x>$. It follows that $(y<x>)^{r}=<x>$ contradicting the assumption that the order of $y<x>$ is $p$.

Sylow set

If $|G|=p^{k}m$ for $m\not \equiv 0\ \mod \ p$ then $Syl_{p}(G)=\{P\leq G:|P|=p^{k}$.

Sylow I

$Syl_{p}(G)\neq \emptyset$

We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. [Dror: "Stare at the class equation.] Since $|G|\equiv 0\ \mod \ p$ we have either:

• $|G|\equiv 0\ \mod \ p$ and $\sum |G|/|Stab(x_{i})|\equiv 0\ \mod \ p$.
• $|G|\equiv 0\ \not \mod \ p$ and $\sum |G|/|Stab(x_{i})|\not \equiv 0\ \mod \ p$.

If $|Z(G)|\not \equiv 0\ \mod p$ then there exists $x_{i}$ such that $|G|/|Stab(x_{i})|\not \equiv 0\ \mod \ p$. Thus $p^{k}$ divides $|Stab(x_{i})|$. We have that $|Stab(x_{i})|<|G|$ [Why happens here?] We then have that $p^{k}\leq Stab(x_{i})<|G|$ and by induction there is $|P|=p^{k}$ such that $P\leq Stab(x_{i})$. It follows $P\leq Stab(x_{i})\leq G$. We've obtained the Sylow $p$-subgroup.

WIf $|Z(G)|\equiv 0\ \mod p$ then by Cauchy's Lemma, there is $x\in Z(G)$ with $|<x>|=p$. Consider the group $G/<x>$. By the induction hypothesis there is $P'\leq G/<x>$ where $|P'|=p^{k-1}$. Then, there is the canonical projection $\pi :G\rightarrow G/<x>$. By the fourth isomorphism theory $P=\pi ^{-1}(P')\leq G$ and $|\pi ^{-1}(P')|=p(p^{k-1})=p^{k}$.

Sylow 2

Every Sylow $p$-subgroup of $G>isconjugate.Moreover,every<math>p$-subgroup is contained in a Sylow $p$-subgroup.

Sylow 3

Let $n_{p}(G)=|Syl_{p}(G)|$. We have $n_{p}\equiv 0\ \mod \ |G|$ and $n_{p}\equiv 1\ \mod \ p$.

A Nearly Tautological Lemma

If $P\in Syl_{p}(G)$ and $H\leq N(P)$ is a $p$-group, then $H\leq P$.

If $x\in G$ has $|<x>|=p^{k}$ and $x\in N(P)$ then $x\in P$.

[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]

We show the first statement. We have that $|P/P\cap H|=p^{k}$ since $P$ is a $p$-group. We then know that $PH/H\simeq P/P\cap H$ by the second isomorphism theorem. It foolows that $|PH|=p^{k'}$. But since $P$ is maximal, we have $P=PH$ and thus $H\subseteq P$. The first statement implies the second by taking $H=<x>$.

GROUPS OF ORDER 15

If $|G|=15$ then $n_{3}\equiv 0\ \mod \ 15$ and $n_{3}\equiv 1\ \mod \ 3$. These imply $n_{3}=1$. Moreover, $n_{5}\equiv 0\ \mod \ 15$ and $n_{5}\equiv 1\ \mod \ 5$. These imply $n_{5}=1$. Thus we have $P_{3}$ a normal $3$-subgroup. Moreover, we have $P_{5}$ a normal $5$-subgroup. This tells us a lot about the group.