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 $1$ .

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 $ord(a)=p$ .