Prove or disprove: $A$ is a subgroup of $G$ if and only if $AA=A$.

Question

I have a question about groups. I need to prove or disprove:

Let $G$ be a group, and $A$ non-empty subset of $G$. $A$ is a subgroup of $G$ if and only if $AA=A$,
where $AA=$ $\{a*a'|a,a' \in A\}$.

If $A$ is a subgroup then of course $AA=A$.
However, I couldn't prove the other direction. I know $A$ is closed under multiplication, but I think something must be wrong with the inverse. However, I couldn't prove it.
Any help will be appreciated!

Answer 1

$G=\mathbb{Z}$ and $A=\mathbb{N}\cup\{0\}$ is a counterexample.

The statement is true if $G$ is finite though. (because then the inverse of $g$ is a power of $g$, so if a subset is closed under multiplication then it has to be closed under inverses)

Answer 2

The other dorection is wrong: the group of integers, subset of natural numbers. It is true for finite groups.

Answer 3

Note that, for any group $G$ and for any finite nonempty subset $A\subseteq G$, $A$ is a subgroup of $G$ if and only if $A\cdot A=A$. A proof can be done using a similar argument to the proof of the proposition below.

If $A$ is an infinite subset of $G$, then it is possible that $A\cdot A=A$, yet $A$ is not a subgroup of $G$ (nonetheless, the condition $A\cdot A=A$ guarantees that $A$ is a subsemigroup of $G$). A different counterexample is $G:=\mathbb{R}$ where $A:=\mathbb{R}_{>0}$.

Alternatively, for any nonempty subset $A\subseteq G$ whose elements all have finite orders, $A$ is a subgroup of $G$ if and only if $A\cdot A=A$. A proof is, again, similar to the proof of the proposition below.

Proposition. We say that a group $G$ is neat if, for any nonempty subset $A$ of $G$, $A$ is a subgroup of $G$ if and only if $A\cdot A=A$. Then, a group $G$ is neat if and only if every cyclic subgroup of $G$ is finite.

If $G$ has an infinite cyclic subgroup $\langle g\rangle$ for some $g\in G$, then take $A:=\{g^k\,|\,k\in\mathbb{Z}_{\geq 0}\}$. Consequently, $A\cdot A=A$, but $A$ is not a subgroup of $G$. That is, $G$ is not neat.

Conversely, suppose that every cyclic subgroup of $G$ is finite. Suppose that $A\subseteq G$ is a nonempty subset such that $A\cdot A=A$. Pick $a\in A$. We can see that $a^k\in A\cdot A$ for all positive integer $k$. Since the subgroup $\langle a\rangle$ of $G$ is cylic, it is finite. Suppose $n$ is the order of the subgroup $\langle a\rangle$. If $n=1$, then $a=1_G=a^{-1}$. For $n>1$, note that $a^n=1_G$ and $a^{n-1}=a^{-1}$. Thus, $A$ contains the identity element of $G$, and for any $a\in A$, we have $a^{-1}\in A$. Since $A\cdot A=A$, we conclude that $A$ is a subgroup of $G$.

Examples. Here are some neat groups:

• finite groups,
• direct limits of finite groups (such as the Prüfer groups and the general linear group $\text{GL}_\infty(\mathbb{F})$ where $\mathbb{F}$ is a finite field), and
• weak products (direct sums) of neat groups (such as $\mathbb{Q}/\mathbb{Z}$).