11-1100-Pgadey-Lect5

Simplicity of $A_{n}$.

Claim

$A_{n}$ 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 $\sigma =(123)({\textrm {aproductofdisjoint2-cycles}})$

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 $S_{4}$

Claim

$S_{4}$ 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. $xe=x$)
• $(g_{1}g_{2})x=g_{1}(g_{2}x)$ (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 $x\mapsto gx:X\rightarrow X\in S(X)$

@@color:green ; //[Where does the shirt come into the business?! ]// @@

• 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{n-1} \leq S_n } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |S_n / S_{n-1}| = n!/(n-1)! = n } .

Exercise

Show that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_n } acting on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{1, 2, \dots, n\} } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_n / S_{n-1} } are isomorphic Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T } and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]

Claim

Left Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets form a category.

The objects of the category are actions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G \times X \rightarrow X } . The morphisms, if we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y } are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets, a morphism of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets is a function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma : X \rightarrow Y } such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma(gx) = g(\gamma(x)) } .

Isomorphism of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets

An isomorphism of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets is a morphism which is bijective.

Silly fact

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_1 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_2 } are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets then so is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G/H } .

Claim

Any Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X } is a disjoint unions of the ``transitive Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -sets. And If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y } is a transitive Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G } -set, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y \simeq G/H } for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H \leq G } .