TITLE: Why does exponentiating group elements by integers always make sense?
QUESTION [6 upvotes]: Let $G$ be a group.
For any $g\in G$, we can define a mapping $\mathbb Z\to G$ via $n\mapsto g^n$.
This map is a homomorphism between the additive group $\mathbb Z$ and $G$.
On the one hand, it's obvious why we can always do this. We just define $g^n$ as "multiply $g$ with itself $n$ times" (for $n>0$, and then straightforwardly generalise to all $n\in\mathbb Z$).
Thus, $\mathbb Z$ comes into play because we are "counting" the number of times $g$ is operating on itself.
On the other hand, this means that for any group $g\in G$, there is a homomorphism $\mathbb Z\to G$ sending $n$ to $g^n$.
Is there any reason why this group in particular, $\mathbb Z\equiv(\mathbb Z,+)$, ends up being homomorphic to subgroups of every group? Are there groups other than $\mathbb Z$ sharing this property?

REPLY [2 votes]: As you explain, given any group $G$ and any element $g\in G$, there is a (unique) homomorphism $\mathbb Z\to G$ which takes $1$ to $g$. You ask whether other groups have similar properties. As mentioned in a comment on another answer, the relevant concept is that of a free group. The infinite cyclic group $\mathbb Z$ is a free group on one generator. Now, for any set $S$ there is a free group over $S$, i.e. a free group $F$ with generators (identified with) the elements of $S$, and any (set) map from $S$ to $G$ can be uniquely extended to a homomorphism $F\to G$.