NoteNextra-origin/content/Math4302/Math4302_L22.md

Math4302 Modern Algebra (Lecture 22)

Groups

Group acting on a set

Let X be a $G$-set, recall that the orbit of x\in X is \{g\cdot x|g\in G\}.

The orbit-stabilizer theorem

For any x\in X, ,$G_x={g\in G|g\cdot x=x}\leq G$.

Let (G:G_x) denote the index of G_x in G, then (G:G_x)=\frac{|G|}{|G_x|}, which equals to the number of left cosets of G_x in G.

Proof

Define \alpha:gG_x\mapsto g\cdot x.

\alpha is well-defined and injective.

gG_x=g'G_x\iff g^{-1}g'\in G_x\iff (g^{-1}g')\cdot x=x\iff g^{-1}\cdot(g'\cdot x)=x\iff g'\cdot x=g\cdot x

\alpha is surjective, therefore \alpha is a bijection.

Example

Number of elements in the orbit of x is 1 if and only if g\cdot x=x for all g\in G.

if and only if G_x=G.

Theorem for orbit with prime power groups

Suppose X is a $G$-set, and |G|=p^n for some prime p. Let X_G be the set of all elements in X whose orbit has size 1. (Recall the orbit divides X into disjoint partitions.) Then |X|\equiv |X_G|\mod p.

Examples

Let G=D_4 acting on \{1,2,3,4\}=X.

X_G=\emptyset since there is no element whose orbit has size 1.

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Let G=\mathbb{Z}_{11} acting on a set with |X|=20 if the action is not trivial, then what is |X_G|?

Using the theorem we have |X_G|\equiv 20\mod 11=9. Therefore |X_G|=9 or 20, but the action is not trivial, |X_G|=9.

An instance for such X=\mathbb{Z}_{11}\sqcup\{x_1,x_2,\ldots,x_9\}, where \mathbb{Z}_{11} acts on \{x_1,x_2,\ldots,x_9\} trivially. and \mathbb{Z}_{11} acts on x_1 with addition.

Proof

If x\in X such that |Gx|\geq 2, then \frac{|G|}{|G_x|}=|Gx|\geq 2.

So |G|=|G_x||Gx|\implies |Gx| divides |G|.

So |Gx|=p^m for some m\geq 1.

Note that X is the union of subset of elements with orbit of size 1, and distinct orbits of sizes \geq 2. (each of them has size positive power of p)

So p|(|X|-|X_G|).

this implies that |X_G|\equiv |X_G|\mod p.

Corollary: Cauchy's theorem

If p is prime and p|(|G|), then G has a subgroup of order p.

This does not hold when p is not prime.

Consider A_4 with order 12, and A_4 has no subgroup of order 6.

Proof

It is enough to show, there is a\in G which has order p: \{e,a,a^2,\ldots,a^{p-1}\}\leq G.

Let X=\{(g_1,g_2,\ldots,g_p)|g_i\in G,g_1g_2g_3\ldots g_p=e\}.

Then |X|=|G|^{p-1} since g_p is determined uniquely by g_p=(g_1,g_2,\ldots,g_{p-1})^{-1}.

Therefore we can define \mathbb{Z}_p acts on X by shifting.

i\in \mathbb{Z}_p i\cdot (g_1,g_2,\ldots,g_p)=(g_{i+1},g_2,\ldots,g_p,g_1,\ldots,g_i).

X is a $\mathbb{Z}_p$-set.

• 0\cdot (g_1,g_2,\ldots,g_p)=(g_1,g_2,\ldots,g_p).
• j\cdot (i\cdot (g_1,g_2,\ldots,g_p))=(i+j)\cdot (g_1,g_2,\ldots,g_p).

By the previous theorem, |X|\equiv |X_G|\mod p.

Since p divides |G|^{p-1}, p also divides |X_G|. Therefore (e,e,e,\ldots,e)\in X_G. Therefore |X_G|\geq 1.

So |X_G|\geq 2, we have (a,a,\ldots,a)\in X_G, a\neq e, but a^p=e, so ord(a)=p.