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;Theorem |
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;Theorem |
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Theorem |
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: 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. |
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: 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. |
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; Stabilizer of a point |
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; Stabilizer of a point |
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: We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$. |
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: We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of <math>x</math>. |
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''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>. |
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''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>. |
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: <math>Syl_p(G) \neq \emptyset</math> |
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: <math>Syl_p(G) \neq \emptyset</math> |
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We proceed by induction on the order of <math>p</math>. 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: |
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We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than <math>|G|</math>. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: |
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* <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>. |
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* <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>. |
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* <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. |
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* <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. |
Theory of Transitive
-sets
- Theorem
- Every
-set is a disjoint union of "transitive
-sets"
- 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
.
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 order of
. Assume the claim holds for all groups of order less than
. [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
is conjugate. Moreover, every
-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.