# 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, $G$ for the group.
And $X$ for the non-empty set.
The binary operation $*$ will simply be juxtaposition of two elements,
$*:G\times X\to X$
The standard way to show,
$X_a=X_b$
It to show,
$X_a\subseteq X_b \mbox { and }X_b\subseteq X_a$
Now,
$X_a=\{x \in X| ax=x\}$
$X_b=\{x \in X|bx=x\}$

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

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