# 11-1100-Pgadey-Lect5

## Contents |

## 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

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

- 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

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).
- Let be the set of bijections from to , with group structure given by composition. We then have an -action of given

["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 .