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Simplicity of A_n.

A_n is simple for n \neq 4.

For n = 1 we have that A_n = \{e\} which is simple. For n=2 we have that S_n = \{(12), e\} , and once again A_n = \{e\} . For n = 3 we have that A_n = \{e, (123), (132)\} \simeq Z/3Z which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.

For n = 4 we have Dror's Favourite Homomorphism (the map S_4 \rightarrow S_3 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: (12)(23) = (123) \quad \quad (12)(34) = (123)(234) 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
A_n is generated by three cycles in S_n. That is, A_n = \langle \{ (ijk) \in S_n \} \rangle.

We have that each element of A_n 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 N \triangleleft A_n contains a 3-cycle then N=A_n.

Up to changing notation, we have that (123) \in N. We show that (123)^\sigma \in N for any \sigma \in S_n. By normality, we have this for \sigma \in A_n. If \sigma \not\in A_n we can write \sigma = (12)\sigma' for \sigma \in A_n. But then (123)^{(12)} = (123)^2 and thus  (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N Since all 3-cycles are conjugate to (123) we have that all 3-cycles are in N. It follows by Lemma 1 that N = A_n.

Case I
N contains a cycle of length \geq 4.
 \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N 

The claim then follows by Lemma 2.

Case II
If N contains an with two cycles of length 3.
 \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N 

The claim then follows by Case I.

Case III
If N contains \sigma = (123)(\textrm{a product of disjoint 2-cycles})

We have that \sigma^2 = (132) \in N . The claim then follows by Lemma 1.

Case IV
If every element of N is a product of disjoint 2-cycles.

We have that  \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N But then \tau^{-1}(125)\tau(125)^{-1} = (13452) \in N . 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 S_4

S_4 contains no normal H such that H \simeq S_3 .
S_3  has an element of order three, therefore H  does. We then conjugate to get all the three cycles. Then H  is too big.

[ Suppose that (123) \in H , then

 S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H 

Which implies that |S_3| = 6 < 9 \leq |H| , but since H \simeq S_3 we have |H| = |S_3| , a contradiction.]

Group Actions

A group G acting on a set X

A left (resp. right) group action of G on X is a binary map G \times X \rightarrow X denotes by (g,x) \mapsto gx satisfying:

  • ex = x (resp. xe = x )
  • (g_1g_2)x = g_1(g_2x) (resp (xg_1)g_2 )
  • [The above implies ex = x and gy = x \Rightarrow g^{-1}y=x .]

Examples of group actions

  • G acting on itself by conjugation (a right action). (g,g') \mapsto g^{g'}
  • Let S(X) be the set of bijections from X to X , with group structure given by composition. We then have an S(X) -action of X given x \mapsto gx : X \rightarrow X \in S(X)

["Where does the shirt come into the business?!" Dror's remark after talking about the symmetry of a student's shirt.]

  • If G = (\mathcal{G}, \cdot) is a group where \mathcal{S} is the underlying set of G and \cdot is the group multiplication. We have an action: (g,s) = g \cdot s this gives a map G \rightarrow S(\mathcal{G}) .
  • SO(n) is the group of orientation preserving symmetries of the (n-1) -dimensional sphere. We have that SO(2) \leq SO(3) as the subgroup of rotations that fix the north and south pole. There is a map SO(3)/SO(2) \rightarrow S^2 given by looking at the image of the north pole.
  • If H \leq G which may not be normal, then we have an action of G on G/H given by g(xH) = (gx)H .
  • We have S_{n-1} \leq S_n and |S_n / S_{n-1}| = n!/(n-1)! = n .
Show that S_n acting on \{1, 2, \dots, n\} and S_n / S_{n-1} are isomorphic S_n -sets.

[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space T and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]

Left G -sets form a category.

The objects of the category are actions G \times X \rightarrow X . The morphisms, if we have X and Y are G -sets, a morphism of G -sets is a function \gamma : X \rightarrow Y such that \gamma(gx) = g(\gamma(x)) .

Isomorphism of G -sets
An isomorphism of G -sets is a morphism which is bijective.
Silly fact
If X_1 and X_2 are G -sets then so is X_1 \coprod X_2 , the disjoint union of the two.

the next statement combines the silly observation above, with the construction of an action of G on G/H .

Any G -set X is a disjoint unions of the "transitive G -sets". And If Y is a transitive G -set, then Y \simeq G/H for some H \leq G .