Exponent semigroup

Definition
Suppose $$G$$ is a group. The exponent semigroup of $$G$$, denoted $$\mathcal{E}(G)$$ is the following submonoid of the multiplicative monoid of integers:

$$\mathcal{E}(G) := \{ n \in \mathbb{Z} \mid (xy)^n = x^ny^n \ \forall \ x,y \in G \}$$

In other words, it is the set of $$n$$ for which the $$n^{th}$$ power map is an endomorphism (and hence a defining ingredient::universal power endomorphism). For each such $$n$$, we say that $$G$$ is a defining ingredient::n-abelian group.

Facts

 * $$\mathcal{E}(G)$$ is a multiplicative submonoid of $$\mathbb{Z}$$ containing zero. In other words, it contains 0 and 1, and is closed under multiplication.
 * If $$G$$ has finite exponent $$n$$, then $$\mathcal{E}(G)$$ contains all multiples of $$n$$. Moreover, $$a \in \mathcal{E}(G) \iff a + n \in \mathcal{E}(G)$$ (periodicity).
 * $$\mathcal{E}(G)$$ is closed under reflection about $$1/2$$, i.e., $$a \in \mathcal{E}(G) \iff 1 - a \in \mathcal{E}(G)$$. This follows from n-abelian iff (1-n)-abelian.
 * $$G$$ is an abelian group if and only if its exponent semigroup is all of $$\mathbb{Z}$$. For one direction, see abelian implies universal power map is endomorphism. For the other direction, note that the element 2 is in $$\mathcal{E}(G)$$ iff $$G$$ is abelian (square map is endomorphism iff abelian). Alternately, we can use that -1 is in $$\mathcal{E}(G)$$ iff $$G$$ is abelian (inverse map is automorphism iff abelian).
 * Characterization of exponent semigroup of a finite p-group