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 universal power endomorphism). For each such $n$, we say that $G$ is a 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