About the set of periods of a group subset

Question

Let $G$ be a group and $A\subseteq G$. Put $$ S(A):=\{g\in G: gA=A\}. $$

It can be shown that $S(A)$ is a subgroup of $G$, and $S(A)=A \ \iff \ A\leq G$.

Now, is it true that:

(1) $|S(A)|\leq |A|$?

(2) if $|S(A)|=|A|<\aleph_0$, then $A\leq G$ or $|A|=1$?

Answer 1

The elements $g_ia$ for a fixed $a\in A$ and distinct $g_i\in S(A)$ are all distinct elements in $A$. So clearly $|S(A)| \leq |A|$.

(2) is false, because for groups $G$ and $H$ form the group $G \times H$ and let $A=\{(g,a): g\in G\}$ where $a$ is a fixed non-identity element of $H$. Then $S(A)=G \times \{e\}$. Even though $|S(A)|=|A|$ but $A$ is not a subgroup.

Answer 2

You may want to consider the set-up in the following way.

Let $X=\{gA:g\in G\}$

Then, we have that $G$ acts transitively on $X$ and so by the orbit stabiliser theorem, $|Stab_G(X)|\times |X|=|G|$

$|S(A)|=|G|/|X|$

If $X$ partitions $G$ into different classes, we have $|G|/|X|=A$ but otherwise, $|G|/|X|<|A|$ as an element of $G$ may be in two different elements of $X$.

So yes, $|S(A)|\leqslant |A|$.

As for the second part, the clue is in the comment above; $|S(A)|=|A|$ if elements of $X$ partitions $G$ so cosets of a subgroup also satisfy this as well.