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Why do integer mod integer sets with the operation of subtraction not form groups?

For example, integers mod 3 is {0,1,2}, which has an identity (0) and inverses (self inverses). And subtraction is an operation because any arguments into the operation outputs something still within integers mod 3. I suspect I am missing something as to why that is not a group.

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    $\begingroup$ Hint $ $ Subtraction is associative $\iff$ subtraction = addition, e.g. in the integers mod $2$ or any ring of characteristic $2$. See my answer for the simple proof. $\endgroup$ – Bill Dubuque Oct 2 at 23:03
  • $\begingroup$ thanks, that makes sense for integers mod 2 $\endgroup$ – tau Oct 2 at 23:08
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Have you checked associativity?

For example, is $(2-1)-1=2-(1-1)$?

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  • $\begingroup$ thanks! i was thinking about that because i heard another explanation and got stuck thinking about how there are inverses. completely forgot about associativity though. $\endgroup$ – tau Oct 2 at 21:48
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Subtraction is associative $\!\iff\! (x\!-\!y)\!-\!z = \overbrace{x\!-\!(y\!-\!z)}^{\textstyle x-y+z}\! \iff\! -z =z\iff\! \overbrace{x - y = x+ y}^{\!\!\!\!\!\text{subtraction}\,{\bf =}\, \text{addition}}$

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