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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?

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    $\begingroup$ For me $0\in \mathbb{N}$ $\endgroup$ – Joshua Tilley Mar 18 at 14:03
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    $\begingroup$ @MoisheKohan Whether $0$ is a natural number is not a question of language, but of the whims of the particular mathematician writing, and often depends more on which choice is more convenient in the given context (unlike whether $0$ is a positive number, which does depend on language). $\endgroup$ – Tobias Kildetoft Mar 18 at 14:06
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    $\begingroup$ I have voted to close this as "opinion-based". $\endgroup$ – Geoffrey Trang Mar 18 at 14:10
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    $\begingroup$ What you say makes sense, but it is only one view. The other view is that the order of $g\in G$ is the order of the cyclic subgroup $\langle g\rangle$. I see no reason why we should adopt either view over the other. $\endgroup$ – user1729 Mar 18 at 14:28
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    $\begingroup$ @Moishe As someone who was UK-educated and has only ever worked in UK maths departments, I find that the reality of this tradition is closer to what Tobias describes (possibly because when you write maths in English you do not expect your readers to be native English speakers and so be aware of such traditions). $\endgroup$ – user1729 Mar 18 at 14:34
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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.

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    $\begingroup$ Maybe for rings it has more to do with the kernel of $\mathbb Z\to R$. $\endgroup$ – rschwieb Mar 18 at 15:37
  • $\begingroup$ @rschwieb You're probably right, and I just realized that's in the question. I edited my answer, $\endgroup$ – Matt Samuel Mar 18 at 15:44
  • $\begingroup$ In the case of rings, we can also define the characteristic to be the order of the subring generated by the homomorphism though, so no real asymmetry. $\endgroup$ – Joshua Tilley Mar 18 at 17:33
  • $\begingroup$ @JoshuaTilley If we defined it that way the characteristic would be infinity instead of $0$. $\endgroup$ – Matt Samuel Mar 18 at 17:41
  • $\begingroup$ Yes, it would be a different convention, but a more consistent one $\endgroup$ – Joshua Tilley Mar 18 at 20:41
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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".

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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.

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    $\begingroup$ Nice observation! $\endgroup$ – Joshua Tilley Mar 19 at 10:52

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