Corollary 3

Let $f : G \to G'$ be a homomorphism of groups with operators and X a subset of G. The image under f of the stable subgroup of G generated by X is the stable subgroup of G' generated by $f(X)$.

Let $X' = f(X)$. Then $\hat{X}' = f(\hat{X})$ and ${X'}^{-1} = f(X^{-1})$. Hence

The corollary then follows from § 1, no. 4, Proposition 1.

Example (2). Let G be a group and x an element of G. The subgroup generated by $\{x\}$ (called more simply the subgroup generated by x) is the set of $x^n, n \in \mathbf{Z}$. The stable subset (under the law on G) generated by $\{x\}$ is the set of $x^n$ where $n \in \mathbf{N}^*$. These two sets are in general distinct.

Thus, in the additive group $\mathbf{Z}$, the subgroup generated by an element x is the set $x.\mathbf{Z}$ of $xn, n \in \mathbf{Z}$, and the stable subset generated by x is the set of $xn, n \in \mathbf{N}^*$. These two sets are always distinct if $x \neq 0$.

The union of a right directed family of stable subgroups of G is obviously a stable subgroup. It follows that, if P is a subset of G and H a stable subgroup of G not meeting P, the set of stable subgroups of G containing H and not meeting P, ordered by inclusion, is inductive (Set Theory, III, § 2, no. 4). Applying Zorn's Lemma (Set Theory, III, § 2, no. 4), we obtain the following result: