MTH3003 Lecture 13

We wrapped up last time by understanding alternating group $A_n$ as the even permutations inside $S_n$ and seeing that $A_n$ is normal in $S_n$. In this lecture we finish the structural story of $A_n$ (simplicity for $n\ge 5$) and then introduce group actions, which repackage arbitrary groups as permutation groups via homomorphisms into $\mathrm{Sym}(X)$.

The Alternating Group $A_n$

Definition and Basic Properties

Alternating group

The alternating group of degree $n$, written $A_n$, is the subgroup of $S_n$ consisting of all even permutations:

$$A_n=\{g\in S_n:\sigma (g)=1\},$$

where $\sigma :S_n\to \{1,-1\}$ is the signature homomorphism.

Key points:

• We already know $\sigma :S_n\to \{1,-1\}$ is a surjective group homomorphism with kernel equal to the even permutations.
• By definition, $A_n=\ker (\sigma )$, so $A_n\le S_n$.

Using the First Isomorphism Theorem:

• $\mathrm{Im}(\sigma )=\{1,-1\}$, so $|\mathrm{Im}(\sigma )|=2$.
• $\ker (\sigma )=A_n$.

Hence

$$|S_n/A_n|=|\mathrm{Im}(\sigma )|=2\Rightarrow \boxed{|A_n|=\frac{n!}{2}}.$$

Listing Elements of Small $A_n$

We can classify elements of $A_n$ by cycle decomposition shape plus signature.

• In $S_3$ the possible cycle shapes are:
  • $\varnothing$ (the identity),
  • $(2)$ (a transposition),
  • $(3)$ (a $3$-cycle).
• Their signatures:
  • $\varnothing$: signature $+1$, so in $A_3$;
  • $(2)$: signature $-1$, so not in $A_3$;
  • $(3)$: signature $+1$, so in $A_3$.

Thus

$$A_3=\{e,\neg \neg (1\neg \neg 2\neg \neg 3),\neg \neg (1\neg \neg 3\neg \neg 2)\},$$

which is cyclic of order $3$, so we may also write $A_3\cong C_3$.

For $S_4$ the possible cycle shapes are:

• $\varnothing$,
• $(2)$,
• $(2,2)$,
• $(3)$,
• $(4)$.

The shapes with signature $+1$ are:

• $\varnothing$,
• $(2,2)$,
• $(3)$.

So $A_4$ consists of:

• the identity,
• all products of two disjoint transpositions,
• all $3$-cycles.

(You will later list all elements of $A_5$ in this way on a problem sheet.)

$A_n$ Is Normal in $S_n$

Normality of $A_n$

For all $n\ge 2$ we have $\boxed{A_n⊴S_n}$.

Proof idea:

• Take any $g\in S_n$ and any $h\in A_n$.
• Because $\sigma$ is a homomorphism and $\sigma (h)=1$, $\sigma (g^{-1}hg)=\sigma (g^{-1})\sigma (h)\sigma (g)=\sigma (g^{-1})\sigma (g)=\sigma (g^{-1}g)=\sigma (e)=1$.
• Therefore, $g^{-1}hg$ is even, so $g^{-1}hg\in A_n$.
• Thus, $A_n$ is closed under conjugation by elements of $S_n$, hence is normal.

Simplicity of $A_n$ for $n\ge 5$

Recall the notion of a simple group: a nontrivial group with no normal subgroups other than $\{e\}$ and itself.

• $A_1$ and $A_2$ are both the trivial group, hence simple.
• As above, $A_3\cong C_3$ is simple (it has no nontrivial proper subgroups).
• $A_4$ is not simple, because it has a normal Klein four subgroup $K_4⊴A_4$.

The deep result:

Simplicity of $A_n$

For all $n\ge 5$ the alternating group $A_n$ is simple, i.e.

$$\boxed{A_n\text{ has no nontrivial proper normal subgroups when }n\ge 5.}$$

We only saw a sketch of the proof, since the full argument is quite long. The strategy has three main steps.

Step 1: Every Element of $A_n$ is a Product of 3-cycles

Sketch:

1. Every element $g\in A_n$ can be expressed as a product of $2$-cycles (transpositions).
2. Because $\sigma (g)=1$, this product uses an even number of transpositions, say $2k$.
3. Use induction on $k$:
   • For the induction step, write $g=(a_1\neg \neg b_1)(a_2\neg \neg b_2)h$, where $h$ is a product of $2(k-1)$ transpositions and therefore, by induction, a product of $3$-cycles.
   • Consider the size of $\{a_1,b_1\}\cap \{a_2,b_2\}$ (0, 1, or 2).
   • In each case, manipulate $(a_1\neg \neg b_1)(a_2\neg \neg b_2)$ into a product of $3$-cycles or the identity.

Conclusion: $A_n$ is generated by $3$-cycles.

Step 2: Normal Subgroups Containing a 3-cycle Contain All 3-cycles

Let $H⊴A_n$, and suppose $H$ contains some $3$-cycle, say $(a_1\neg \neg a_2\neg \neg a_3)$.

Because $n\ge 5$, we can choose elements

• $b\notin \{a_1,a_2,a_3\}$,
• $z\notin \{a_1,a_2,a_3,b\}$.

Since $A_n$ contains all $3$-cycles, in particular $(a_1\neg \neg z\neg \neg b)\in A_n$.

Using normality of $H$:

• $(a_1\neg \neg z\neg \neg b{)}^{-1}(a_1\neg \neg a_2\neg \neg a_3)(a_1\neg \neg z\neg \neg b)\in H$,
• this conjugate is another $3$-cycle, for example $(b\neg \neg a_2\neg \neg a_3)$ (which is the same as $(a_2\neg \neg a_3\neg \neg b)$).

By varying the choice of elements, we can show that every $3$-cycle lies in $H$.

Thus any normal subgroup $H⊴A_n$ which contains one $3$-cycle must contain all $3$-cycles, hence equals $A_n$ (since $A_n$ is generated by $3$-cycles).

Step 3: Any Nontrivial Normal Subgroup Contains a 3-cycle

Let $H⊴A_n$ with $H\ne \{e\}$.

• Pick $h\in H$ with $h\ne e$ and write it as a product of disjoint cycles:

$$h=c_1{c}_2\cdots c_m.$$

Now analyse the cycle structure of $h$:

1. If some $c_i$ is a $3$-cycle and all other cycles are $2$-cycles, then $h^2\in H$ is itself a $3$-cycle.
2. If the longest cycle has length at least $4$, we can find a $3$-cycle $\rho$ such that the commutator-like element

$$\rho h{\rho }^{-1}{h}^{-1}\in H$$

becomes a $3$-cycle.

1. If that fails, we can find $\rho$ making $\rho h{\rho }^{-1}{h}^{-1}$ a $5$-cycle, and then apply the previous argument to this $5$-cycle to produce a $3$-cycle inside $H$.

Either way, $H$ must contain a $3$-cycle.

Combining with Step 2, any nontrivial normal subgroup $H$ is all of $A_n$.

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Group Actions and $G$-sets

We now change gears and introduce group actions, which encode how groups act as symmetries of sets.

Definition of a Group Action

Group action and $G$-set

Let $G$ be a group and $X$ a set. An action of $G$ on $X$ is a function

$$\lambda :G\times X\to X,(g,x)\mapsto \lambda (g)x$$

such that, for all $g,f\in G$ and all $x\in X$:

1. $\lambda (e_G)x=x$;
2. $\lambda (fg)x=\lambda (f)(\lambda (g)x)$. If there exists such an action, we say $G$ acts on $X$, and we call $X$ a $G$-set.

Intuition:

• For each $g\in G$, the action tells you how $g$ moves points of $X$.
• Axioms say: the identity does nothing, and composition in $G$ corresponds to composing the moves.

Basic Examples

1. Natural action of $S_n$ on $\{1,\dots ,n\}$ Let $X=\{1,2,\dots ,n\}$ and $G=S_n$. Define

$$\lambda (g)x:=gx$$

in the usual permutation sense. For example, if $g=(1\neg \neg 2\neg \neg 3)$ then $\lambda (g)3=1$.

1. Trivial action For any group $G$ and any set $X$, define

$$\lambda (g)x:=x$$

for all $g\in G$, $x\in X$. Every element acts as the identity permutation on $X$.

Actions as Homomorphisms into $\mathrm{Sym}(X)$

Let $X$ be a $G$-set with action $\lambda$.

Action as a permutation representation

For each $g\in G$, the map $\lambda (g):X\to X$ is a bijection, so $\lambda (g)\in \mathrm{Sym}(X)$. Moreover,

$$\boxed{\lambda :G\to \mathrm{Sym}(X)}$$

is a group homomorphism, called the permutation representation associated to the action.

Proof sketch:

• Fix $g\in G$. Using the axioms, one checks:
  • Injective: if $\lambda (g)x=\lambda (g)y$, then applying $\lambda (g^{-1})$ and using the action property shows $x=y$.
  • Surjective: for any $y\in X$, set $x:=\lambda (g^{-1})y$. Then $\lambda (g)x=y$.
• Thus $\lambda (g)$ is a permutation of $X$, so $\lambda (g)\in \mathrm{Sym}(X)$.
• The map $g\mapsto \lambda (g)$ is a homomorphism because for all $f,g\in G$ and all $x$,

$$\lambda (fg)x=\lambda (f)(\lambda (g)x),$$

which is exactly the condition $\lambda (fg)=\lambda (f)\lambda (g)$ in $\mathrm{Sym}(X)$.

Consequence: every action gives us a homomorphism

$$\lambda :G\to \mathrm{Sym}(X),$$

so we can talk about its kernel and image.

Kernel and image of an action

The kernel of the action is

$$\ker (\lambda )=\{g\in G:\lambda (g)x=x\text{ for all }x\in X\}.$$

The image $\mathrm{Im}(\lambda )$ is a subgroup of $\mathrm{Sym}(X)$ that describes all permutations of $X$ realised by elements of $G$.

Common notation: we often drop the symbol $\lambda$ and simply write $gx$ for $\lambda (g)x$.

Why Group Actions Matter

The slogan:

• A group action describes how $G$ behaves as a collection of symmetries of some set $X$.
• Via the associated homomorphism $G\to \mathrm{Sym}(X)$, every group can be represented as a permutation group.

This is the bridge between abstract group theory and explicit permutations.

Key Examples of Actions on the Group Itself

1. Regular (left multiplication) action Take $X=G$ and define

$$gx:=gx$$

in the usual group multiplication sense. This is an action; each $g$ permutes $G$ by left multiplication.

1. Conjugation action Take again $X=G$, but now define

$$gx:=gx{g}^{-1}$$

for all $g,x\in G$. This is the (left) conjugation action.

These two actions are fundamental: the regular action encodes the group’s structure in terms of orbits, and the conjugation action is closely tied to normal subgroups and class equations.

A More Geometric Example: $D_8$ Acting on a Cube

Consider the dihedral group $D_8$, the symmetries of a square with vertices labelled $1,2,3,4$.

• Let $\rho =(1\neg \neg 2\neg \neg 3\neg \neg 4)$ be a rotation, and $\sigma =(1\neg \neg 4)(2\neg \neg 3)$ a reflection.
• Then

$$D_8=\{e,\rho ,{\rho }^2,{\rho }^3,\sigma ,\sigma \rho ,\sigma {\rho }^2,\sigma {\rho }^3\}.$$

Now imagine the square is one face of a cube whose vertices are labelled $1,\dots$. We can define an action of $D_8$ on the set of cube vertices $\{1,\dots \}$ by:

• $\lambda (\rho )=(1\neg \neg 2\neg \neg 3\neg \neg 4)(5\neg \neg 6\neg \neg 7\neg \neg 8)$,
• $\lambda (\sigma )=(1\neg \neg 4)(2\neg \neg 3)(5\neg \neg 8)(6\neg \neg 7)$,
• $\lambda ({\sigma }^i{\rho }^j)=\lambda (\sigma {)}^i\lambda (\rho {)}^j$.

One checks that the group action axioms hold; this formalises how the 2D symmetries of the square extend to natural symmetries of the cube.

A Non-example

Not every formula that looks conjugation-like gives a valid action.

Consider the rule

$$\lambda (g)x=g^{-1}xg$$

for $g,x\in G$ and $X=G$.

• For a fixed $g$, this is indeed conjugation by $g^{-1}$, hence a permutation of $G$.
• However, when you try to check the action condition

$$\lambda (gf)x\stackrel{?}{=}\lambda (g)(\lambda (f)x),$$

you get

$$\lambda (gf)x=(gf{)}^{-1}x(gf)=f^{-1}{g}^{-1}xgf,$$

whereas

$$\lambda (g)(\lambda (f)x)=g^{-1}(f^{-1}xf)g=g^{-1}{f}^{-1}xfg.$$

These two expressions coincide only in special situations (e.g. when $G$ is Abelian, or $x$ lies in the centre), so this is not an action in general.

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Pre-Lecture Notes from University Notes

• Defined the alternating group $A_n$ as $\ker (\sigma )$ inside $S_n$ and used the First Isomorphism Theorem to obtain $|A_n|=n!/2$.
• Classified elements of small $A_n$ by cycle shape: $A_3=\{e,(1\neg \neg 2\neg \neg 3),(1\neg \neg 3\neg \neg 2)\}\cong C_3$; described which cycle shapes in $S_4$ give elements of $A_4$.
• Proved $A_n⊴S_n$ using the fact that $A_n$ is the kernel of the signature homomorphism; noted that $A_1,A_2,A_3$ are simple, while $A_4$ is not.
• Sketched the three-part proof that $A_n$ is simple for $n\ge 5$: express elements as products of $3$-cycles, show any normal subgroup with one $3$-cycle has all of them, and deduce any nontrivial normal subgroup must contain a $3$-cycle.
• Introduced group actions and G-sets, with definition via axioms; emphasised that each action induces a homomorphism $G\to \mathrm{Sym}(X)$ with well-defined kernel and image.
• Worked through examples: natural action of $S_n$ on $\{1,\dots ,n\}$, trivial action, actions of a group on itself by left multiplication and by conjugation, and $D_8$ acting on the vertices of a cube; contrasted with a conjugation-like map that fails the action axioms.
• Next time: use group actions to define orbits and stabilisers, and derive the orbit–stabiliser theorem and class equation, tying back to normal subgroups and simplicity.