Nonempty characteristic subsemigroup of abelian group implies subgroup

Statement
Suppose $$G$$ is a abelian group and $$M$$ is a nonempty subsemigroup  of $$G$$ such that $$\sigma(M) = M$$ for any automorphism $$\sigma$$ of $$G$$. Then, $$M$$ is a subgroup, and hence a characteristic subgroup, of $$G$$.

Similar facts

 * Nonempty finite subsemigroup of group is subgroup

Opposite facts

 * Characteristic submonoid of group not implies subgroup
 * Characteristic submonoid of nilpotent group not implies subgroup

Facts used

 * 1) uses::inverse map is automorphism iff abelian

Proof
By Fact (1), $$M$$ is closed under the inverse map. Since it is non-empty, we can pick an element $$g \in M$$, obtain that $$g^{-1} \in M$$, and then get that their product, the identity element, is in $$M$$. Combining, we get that $$M$$ is closed under all the group operations (multiplication, inverse map, and identity) hence is a subgroup. Further, by our assumptions, it must be a characteristic subgroup.