Should infinite order elements not be called zero order elements? For a unital ring $R$, there is a homomorphism $f:\mathbb{Z}\rightarrow R$ and the kernel is an ideal of the form $n\mathbb{Z}$ for unique $n\in \mathbb{N}$, which we call the characteristic of the ring. 
In a similar kind of way, for a group $G$, and $g\in G$ there is a homomorphism $f:\mathbb{Z}\rightarrow G$ sending $n$ to $g^n$. The kernel of this homomorphism is again of the form $n\mathbb{Z}$ for unique $n\in \mathbb{N}$ which we call the order of $g\in G$, except for when $n=0$ and then we say $g$ has infinite order. 
Wouldn't it be better to say $g$ has order zero in this case, for consistency? 
 A: The nomenclature for an infinite order element of a group is not in terms of the kernel of the relevant homomorphism, but rather the order of the cyclic subgroup generated by the element. For rings, they went with the nonnegative generator of the kernel of the homomorphism $\mathbb Z\to R$, though this is not explicit in the usual definition (which is inconsistent, saying the characteristic is the minimal positive integer such that $n1=0$, then giving $0$ as an exception when no such integer exists).
One could argue that the subgroup concept makes more sense for general groups because it is given in terms of the order of a group, whereas the kernel definition is additive which is appropriate for the additive group of a ring.
A: I think it's harmless to call such elements of order zero; probably it's been done occasionally. If for some reason it's practical (e.g., you talk of elements of order $n$ and $n$ can be zero), just say it at the beginning.
Actually it's "almost" done somewhere: the characteristic of a unital ring $R$ is by definition the order of $1$ in $(R,+)$... and one says "characteristic zero", and (almost?) never "infinite characteristic".
A: Here's a situation where this convention is useful. Suppose we have a group homomorphism $f : G \to H$ and let $a$ be in $G$. What is the order of $f(a)$ in $H$? In general, $f$ need not preserve order (just consider the constant map $f(a) = 1$), but what's true is that $\operatorname{ord}_H(f(a))$ must be a divisor of $\operatorname{ord}_G(a)$. If we set $\operatorname{ord}(a) = 0$ for elements of "infinite" order, then, keeping in mind that everything divides 0 while 0 divides only 0, this divisibility relation shows that homomorphic images of elements of infinite order can have arbitrary order, while no element of finite order can map to an element of infinite order.
