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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ Jan 8, 2019 at 11:22
  • $\begingroup$ @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. $\endgroup$
    – Hanul Jeon
    Jan 8, 2019 at 11:34
  • $\begingroup$ 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. $\endgroup$
    – Hanul Jeon
    Jan 8, 2019 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).

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