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In textbook which I am studying,

LEMMA 1
A commutative monoid $S$ can be embedded in a group if and only if it admits cancellation by all elements: $ac=bc$ implies $a=b,$ for all $a,b,c \in S.$

I think reverse of this lemma 1 is only right when $S$ is a finite set.

How do you think about it?

Thomas Andrews
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HB Y
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    Do you have an infinite counterexample? – Randall Jul 13 '21 at 01:57
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    What do you mean by "reverse"? The Lemma is an "if and only if" statement, not merely an implication. And it holds regardless of whether $S$ is finite or infinite. – Arturo Magidin Jul 13 '21 at 01:58
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    If $S$ is finite and cancellative, then in fact, it must actually already be an abelian group. – Geoffrey Trang Jul 13 '21 at 02:30
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    It’s pretty easy to define the abelian group - the same way we define $\mathbb Z$ from $\mathbb N.$ – Thomas Andrews Jul 13 '21 at 02:33
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    The right-to-left direction of this wouldn't be true without the assumption that $S$ is commutative. Is that what you missed? – Rob Arthan Jul 13 '21 at 03:02
  • Thank you very much for many helpful comments. I misunderstood what 'be embedded in' is.
    Tomas Andrews comments it's pretty easy. but I think it's not as easy as defining Z from N because we cannot define '-x' freely.
    We need several steps for embedding like equivalence class and new symbol.
    Do I misunderstand one more?     – HB Y Jul 14 '21 at 01:16

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