Simplicity of
.
- Claim
is simple for
.
For
we have that
which is simple.
For
we have that
, and once again
.
For
we have that
which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.
For
we have Dror's Favourite Homomorphism (the map
given by a coloured tetrahedron (link)
[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]
We proceed with some unmotivated computations:
These are the main ingredients of the proof. The first equality says that a product of transpositions of non-disjoint pairs is a 3-cycle, the second equality says that the product of a pair of disjoint transpositions is a product of three cycles. Thus any product of two transpositions can be written a product of three cycles. [ The second equality is amusing with physical objects. ]
- Lemma 1
is generated by three cycles in
. That is,
.
We have that each element of
is the product of an even number of transpositions (braid diagrams, computation with polynomials, etc). But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.
- Lemma 2
- If
contains a 3-cycle then
.
Up to changing notation, we have that
. We show that
for any
. By normality, we have this for
. If
we can write
for
. But then
and thus
Since all 3-cycles are conjugate to
we have that all 3-cycles are in
. It follows by Lemma 1 that
.
- Case I
contains a cycle of length
.
The claim then follows by Lemma 2.
- Case II
- If
contains an with two cycles of length 3.
The claim then follows by Case I.
- Case III
- If
contains ![{\displaystyle \sigma =(123)({\textrm {aproductofdisjoint2-cycles}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/033fd1ddcb14c53faca095abe8b25e3830cf1749)
We have that
. The claim then follows by Lemma 1.
- Case IV
- If every element of
is a product of disjoint 2-cycles.
We have that
But then
.
The claim then follows by Case 1.
[Note: This last case is the only place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]
Throwback to ![{\displaystyle S_{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc1632bc1d95d33ccb5473b9d8cc333c2dd0d13a)
- Claim
contains no normal
such that
.
has an element of order three, therefore
does. We then conjugate to get all the three cycles. Then
is too big.
[ Suppose that
, then
Which implies that
, but since
we have
, a contradiction.]
Group Actions
- A group
acting on a set ![{\displaystyle X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
A left (resp. right) group action of
on
is a binary map
denotes by
satisfying:
(resp.
)
(resp
)
- [The above implies
and
.]
Examples of group actions
acting on itself by conjugation (a right action). ![{\displaystyle (g,g')\mapsto g^{g'}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f32cec63bc405c529581fd235886b4804083dcaa)
- Let
be the set of bijections from
to
, with group structure given by composition. We then have an
-action of
given ![{\displaystyle x\mapsto gx:X\rightarrow X\in S(X)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e2bb18d438dcd010c2e2fea77ad4ecc0323cfcb)
["Where does the shirt come into the business?!" Dror's remark after talking about the symmetry of a student's shirt.]
- If
is a group where
is the underlying set of
and
is the group multiplication. We have an action:
this gives a map
.
is the group of orientation preserving symmetries of the
-dimensional sphere. We have that
as the subgroup of rotations that fix the north and south pole. There is a map
given by looking at the image of the north pole.
- If
which may not be normal, then we have an action of
on
given by
.
- We have
and
.
- Exercise
- Show that
acting on
and
are isomorphic
-sets.
[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space
and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]
- Claim
- Left
-sets form a category.
The objects of the category are actions
. The morphisms, if we have
and
are
-sets, a morphism of
-sets is a function
such that
.
- Isomorphism of
-sets
- An isomorphism of
-sets is a morphism which is bijective.
- Silly fact
- If
and
are
-sets then so is
, the disjoint union of the two.
the next statement combines the silly observation above, with the construction of an action of
on
.
- Claim
- Any
-set
is a disjoint unions of the "transitive
-sets". And If
is a transitive
-set, then
for some
.