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Lemma:Suppose a finite set $ G$ is closed under an associative product and that both cancellation laws hold in $G$,then $G$ must be a group.

Using the result, prove that the nonzero integers modulo $p$, $p$ a prime number, form a group under multiplication $\mod p$.

However, I wanted to solve it in a general way and by using the above lemma. If we solve it in a " general" way, my solution goes like this:

First we prove the closure property . If $a,b\in G$, then $a.b\in G$. Now, if $a,b$ are non-zero integers modulo p and $a,b\in G $ , where $G=\{0,1,...,p-1\}$ and then $a.b (\mod p)\in G$, is true. Hence, closure property is satisfied. Next, we go for associative property, which is always true as multiplication ($\mod p$) is always associative . If $a,b,c\in G $ then $(a.(b.c))(mod p)=((a.b).c)(mod p)$. Hence, associative property is satisfied. Next, we go for the identity property i.e $\exists e\in G$ such that $\forall a\in G$, $a.e=e.a=a$. If $e=1$ , then $\forall a\in G$, $a.1(\mod p)=1.a(\mod p)=a$. Thus, identity property is satisfied. Next, we prove the inverse property i.e for every $a\in G $ , $\exists a^{-1}\in G $ such that $a.a^{-1}=a^{-1}.a=e$. Hence , for every $a\in G $, $\exists a^{-1}\in G$, such that $a.a^{-1}\equiv 1 (mod p)$, which is true (by Bezout's Lemma) and if $a^{-1}\equiv a_1(\mod p)$ then, $a.a^{-1}\equiv a.a_1\equiv 1 (\mod p)$, and $a_1\in G$. Thus, inverse property is satisfied. Hence , $G$ is a group under multiplication modulo $p$.

Is the above proof correct? Is the above proof valid? If not, then why it isn't true? Also,I am not getting how to use the lemma to prove that $G$ is a group under this binary operation. How to prove it? I am not quite getting it...

Arthur
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    https://math.stackexchange.com/questions/1564859/let-s-be-a-non-empty-set-with-an-associative-cancellative-operation-and-for-e might help you, also https://math.stackexchange.com/questions/1554361/confused-about-this-exercise-question-if-a-set-with-a-certain-binary-operation and https://math.stackexchange.com/questions/142572/if-cancellation-laws-hold-then-a-finite-semi-group-is-a-group and https://math.stackexchange.com/questions/203023/a-finite-cancellative-semigroup-is-a-group and probably others. – Gerry Myerson Oct 13 '22 at 06:19
  • @GerryMyerson but is my proof valid? Also, how to draw the same conclusion using the lemma stated above in my original post ?...Thank you!... – Arthur Oct 13 '22 at 07:30
  • OK, all the links are actually about proving the lemma, not about using it. Sorry. But you're trying to prove the nonzero integers mod $p$ form a group, so to use the lemma, you just have to show the set is closed under multiplication, that multiplication is associative, and that cancellation works. How much of that can you do? What stops you from completing the work? – Gerry Myerson Oct 13 '22 at 11:50
  • @GerryMyerson to show that $G$ is closed under multiplication modulo $p$ is obvious as : if $a,b\in G $, then if $ab\equiv x (\mod p)$, then $x\in G$. Multiplication under modulo p is also associative as $(a.b).c\equiv x_1(\mod p)$ and $a.(b.c)\equiv x(\mod p)$ . But how to show that that cancellation works? Are the previous steps correct? Also, is the proof in the "general way" in the original post, correct?... – Arthur Oct 13 '22 at 14:01
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    If $ab=ac$, then $p$ divides $ab-ac$, that is, $p$ divides $(b-c)a$, but $p$ is prime, so $p$ divides $b-c$ or $a$ (or both). If $p$ divides $a$, then $a$ is zero, but zero is not in our set, so $p$ divides $b-c$, so $b=c$, proving left cancellation. Right cancellation, same ideas. – Gerry Myerson Oct 14 '22 at 03:36
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    @GerryMyerson Thanks a lot!...I do get it now...Thank you!... – Arthur Oct 14 '22 at 08:11

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