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I have no problem seeing how having an inverse implies cancellation property, but I've problems proving (or even seeing why) the converse must be also true. Can you please suggest a way to approach this?

wvxvw
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2 Answers2

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No. Cancellation property does not imply existence of inverse element. Consider the additive monoid $(\Bbb{N},+)$. (We wiil assume that $\Bbb{N}$ contains 0.) Then $(\Bbb{N},+)$ satisfies associativity and cancellation property but all elements except 0 have no inverse element.

Hanul Jeon
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  • So is it accurate to say that cancellation works because there is an inverse, but the inverse element is not nessesarily part of the set and that's what invertability is? – Pineapple Fish Jan 08 '19 at 11:22
  • @BenjaminThoburn I do not think your explanation is correct in full detail. Every non-zero element of N has no inverse, and so we do not know how to talk about what the inverse of 1 is, for example. In my example, it should be −1, but it may not be obvious for more complex structure. – Hanul Jeon Jan 08 '19 at 11:34
  • However, I guess every monoid with cancellation property can be extended to a group. It really holds if we assume the monoid is abelian. I do not know my guess really holds and I have no time to check it. – Hanul Jeon Jan 08 '19 at 11:36
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However, if you require the set to be finite, then you do get invertibility (this is a common way to show that $\mathbb{Z} / p \mathbb{Z}$ is a field).

Jon Hillery
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