# $g^{k}S=S \Rightarrow g$ has finite order

Let $G$ be a group (not necessarily finite), $g \in G$, $g \ne 1$. Suppose that $S \subseteq G$, $S$ is finite, $1 \in S$, and $gS=S$. It follows that $g^{k}S=S$ for all $k \in \mathbb{N}$. Does it also follow that the order of $g$ is finite?

My attempt: Since $S$ is finite, let $|S|=n$. Since $1 \in S$, $g^{k} \in g^{k}S = S$ for all $k \in \mathbb{N}$. Consider $A=\{g,g^{2},\ldots , g^{n}, g^{n+1}\}$. Since for all $k \in \mathbb{N}$ we have that $g^{k} \in S$, it follows that $A \subseteq S$. Since $A$ and $S$ are finite, $|A| \le |S|=n$. Thus, by the Pigeonhole principle, $g^{s}=g^{t}$ for some $s \ne t$. WLOG, suppose $s < t$. Then $g^{t-s}=1$, so the order of $g$ is finite.

Two questions: Is the above proof correct and is there an easier way to do this by letting $G$ act on itself and considering permutation representations?

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It is fine. But as soon as you proved that the group generated by $g$ is finite, there was no real need for the part that started with "thus, by the Pigeonhole $\dots$". – André Nicolas Jun 14 '12 at 0:08
Good point, that simplifies things a bit. – Jackson Jun 14 '12 at 0:18

Perm rep reprising of your argument: Do the same thing as before, but let "x" be in S, instead of 1. Conclude $g^s x = g^t x$, so that $g^{t-s}$ stabilizes $x$. However, in a regular permutation representation, only the identity stabilizes any element, so $g^{t-s}=1$.