Math4302 Modern Algebra (Lecture 22)

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.

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 $G_x=G$.

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.

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 $|X_G|\equiv |X_G|\mod p$.

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 $ord(a)=p$.