# Algebra, group action

• Dec 2nd 2006, 02:30 PM
PvtBillPilgrim
Algebra, group action
Studying for a final and had the following practice problem, which I can't answer. Could anyone help?

Let G be a finite group acting on a finite set. Prove that <a>=<b> for two elements a,b in G then I(a)=I(b), where for any element g in G, I(g)={s in S: gs=s}.

Thank you.
• Dec 2nd 2006, 03:19 PM
ThePerfectHacker
Quote:

Originally Posted by PvtBillPilgrim
Studying for a final and had the following practice problem, which I can't answer. Could anyone help?

Let G be a finite group acting on a finite set. Prove that <a>=<b> for two elements a,b in G then I(a)=I(b), where for any element g in G, I(g)={s in S: gs=s}.

Thank you.

I presume you are talking about G-sets.
The notation I will use is, $\displaystyle G$ for the group.
And $\displaystyle X$ for the non-empty set.
The binary operation $\displaystyle *$ will simply be juxtaposition of two elements,
$\displaystyle *:G\times X\to X$
The standard way to show,
$\displaystyle X_a=X_b$
It to show,
$\displaystyle X_a\subseteq X_b \mbox { and }X_b\subseteq X_a$
Now,
$\displaystyle X_a=\{x \in X| ax=x\}$
$\displaystyle X_b=\{x \in X|bx=x\}$

Let,
$\displaystyle x\in X_a$
Then,
$\displaystyle x\in X_b$
Because,
$\displaystyle b\in <a>, b=a^c$
Thus,
$\displaystyle bx=a^cx=a^{c-1}x=...=ax=x$ implies $\displaystyle x\in X_b$
Similarly,
$\displaystyle a\in <b>,a=b^d$
Thus,
$\displaystyle ax=b^dx=b^{d-1}x=...=bx=x$ implies $\displaystyle x\in X_a$.
Thus,
$\displaystyle X_a\subseteq X_b\mbox{ and }X_b\subseteq X_a\to X_a= X_b$

(The facts that the sets are finites and groups are finite was not necessary).