In a finite group, if $e \notin A$, does $gA = A \implies g = e$?

Question

Let $G$ be a finite group, and let $e \notin A$ be a subset not containing the identity. Suppose $g \cdot A = A$. Does this imply that $g = e$?

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I am able to show that this is true in the case that $G = \mathbb{Z} / p \mathbb{Z}$, but am not sure how to proceed in the general case.

Answer 1

Not in general, no. Consider the group $\Bbb Z_{6}$ with standard addition, and let $A$ be the subset of odd numbers (more correctly odd congruence classes). Then $2+A=A$, yet $2\neq 0$.

This is a case of a more general example: if $A$ is a right coset of a subgroup $H\subseteq G$, then $gA=A$ for any $g\in H$. If $A\neq H$, and $H$ is non-trivial, then you have a counterexample to your statement.

Answer 2

No. E.g., let $G = \Bbb{Z}/4\Bbb{Z}$ be the cyclic group of order four and take $A = \{[1], [3]\}$ then $[2] + A = A$, but $[2] \neq [0]$. In general, if $G$ is any group and $H$ is any proper subgroup, a coset $Hx$ where $x \notin H$ does not contain $e$ but is fixed by left multiplication by elements of $H$.

Answer 3

Arthur has given a general recipe to come up with examples: given a group $G$ and a nontrivial proper subgroup $H$, if we let $A=Hx$ with $x\notin H$, then $e\notin A$, but for any $h\in H$ we have $hA = hHx = Hx = A$, even when $h\neq e$. You can also take a union of right cosets.

This works even when $G$ is infinite.

In fact, all examples are of this type.

Let $A$ be a subset of $G$. Let $H=\{g\in G\mid gA=A\}$. Then $H$ is nonempty, as $e\in H$. If $h_1,h_2\in H$, then $(h_1h_2)A = h_1(h_2A) = h_1A = A$, so $H$ is closed under products. And if $gA=A$, then $A=g^{-1}gA = g^{-1}A$, so $H$ is also closed under inverses. Thus, $H$ is a subgroup.

Pick $a\in A$. Then $a\in Ha\subseteq A$, hence $$A = \bigcup_{a\in A}\{a\} \subseteq \bigcup_{a\in A} Ha\subseteq A,$$ so $A$ is a union of right cosets of $H$.