|
|
Line 4: |
Line 4: |
|
|
|
|
|
;Theorem |
|
;Theorem |
|
|
Theorem |
|
: If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets. |
|
: If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets. |
|
|
|
|
Line 45: |
Line 46: |
|
Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that |
|
Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that |
|
<math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math> |
|
<math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math> |
|
The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>. |
|
The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>. |
|
|
|
|
|
==Sylow== |
|
==Sylow== |
Line 58: |
Line 59: |
|
|
|
|
|
; Sylow set |
|
; Sylow set |
|
: If <math>|G| = p^k m</math> for <math>m \not\equiv 0\ \mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>. |
|
: If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>. |
|
|
|
|
|
; Sylow I |
|
; Sylow I |
|
: <math>Syl_p(G) \neq \emptyset</math> |
|
: <math>Syl_p(G) \neq \emptyset</math> |
|
|
|
|
|
We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>|G| \equiv 0\ \mod\ p</math> we have either: |
|
We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: |
|
* <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p</math>. |
|
* <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>. |
|
* <math>|G| \not\equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>. |
|
* <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. |
|
|
|
|
|
If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup. |
|
If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup. |
|
|
|
|
|
WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>. |
|
WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>. |
Line 75: |
Line 76: |
|
|
|
|
|
; Sylow 3 |
|
; Sylow 3 |
|
: Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\ \mod\ |G|</math> and <math>n_p \equiv 1\ \mod\ p</math>. |
|
: Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>. |
|
|
|
|
|
; A Nearly Tautological Lemma |
|
; A Nearly Tautological Lemma |
Line 89: |
Line 90: |
|
==Groups of Order 15== |
|
==Groups of Order 15== |
|
|
|
|
|
If <math>|G| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group. |
|
If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group. |
Theory of Transitive
-sets
- Theorem
- Every
-set is a disjoint union of "transitive
-sets"
- Theorem
Theorem
- If
is a transitive
-set and
then
where the isomorphism an isomorphism of
-sets.
- Transitive
-set
- A
-set
is transitive is
.
- Stabilizer of a point
- We write
for the stabilizer subgroup of $x$.
Proof We define an equivalence relation
. This relation is reflexive since
and thus
. This relation is symmetric since
implies
. This relation is transitive, since if
and
then
. It follows that
where
denote the orbit of a point
.
We then claim that
is a transitive
-set. [Dror: "[This fact] is too easy."]
We show that
is isomorphic to
as a
-set.
We produce two morphism
and
.
To define
there is only one thing we can do. We have
and then we define
. We check that this map is well defined. If
then
and hence
. It follows that
. Thus
is well defined.
To define
we take
and define
. We show that this map is well defined. If
then
and hence
. It follows that
and hence
is well defined.
We need to check that
and
are mutually inverse and
-set morphisms. We quickly check that
is a
-set morphism. If
and
then
. Similarly,
. The last inequality follows since we can take any
such that
. Why not take
-- since we know that works.
- Theorem (Orbit-Stabilizer)
- If
and
then
.
This is just a rewriting of the theorem above.
-Group
- A
-group is a group
with
for some
.
The last group
is the famous unit quaternions -- They need a better description here.
- Theorem
- Any
-group has a non-trivial centre.
Let
act on itself by conjugation. Decompose
. Then,
Observe that
iff
. It follows that
The formula above is called "the class formula". We have that
for some
since
is a subgroup. It follows that
and
. It follows that
. Since
we have
and thus
.
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
is an abelian group and
divides
, then there is an element of order
in
.
Proof. Pick
. If
divides the order of
then we have
for some
. It follows that
. We then have that the order of
is
. If
does not divide the order of
, then consider
. Since
is abelian,
is a normal subgroup. We have that
divides
, and
. We then induct. Let
have order
, that is
. We then have that
for some
. We write
where
. We then have
. It follows that
contradicting the assumption that the order of
is
.
- Sylow set
- If
for
then
.
- Sylow I
![{\displaystyle Syl_{p}(G)\neq \emptyset }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3234b2009030d27ade14ffac845cbcd7c85240eb)
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
we have either:
and
.
and
.
If
then there exists
such that
. Thus
divides
. We have that
[Why happens here?] We then have that
and by induction there is
such that
. It follows
. We've obtained the Sylow
-subgroup.
WIf
then by Cauchy's Lemma, there is
with
. Consider the group
. By the induction hypothesis there is
where
. Then, there is the canonical projection
. By the fourth isomorphism theory
and
.
- Sylow 2
- Every Sylow
-subgroup of
-subgroup is contained in a Sylow
-subgroup.
- Sylow 3
- Let
. We have
and
.
- A Nearly Tautological Lemma
- If
and
is a
-group, then
.
- If
has
and
then
.
[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
since
is a
-group. We then know that
by the second isomorphism theorem. It foolows that
. But since
is maximal, we have
and thus
. The first statement implies the second by taking
.
Groups of Order 15
If
then
and
. These imply
. Moreover,
and
. These imply
. Thus we have
a normal
-subgroup. Moreover, we have
a normal
-subgroup. This tells us a lot about the group.