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Let $G$ be a set with an associative operation defined on it.

Assuming $G$ is finite and both cancellation laws hold, show that $G$ is a group.

Is knowing that $G$ is associative under an operation and finite enough to say that it is closed? Also how can I build an identity from this?

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marked as duplicate by BCLC, Tyrone, user99914, Sil, José Carlos Santos Aug 16 '18 at 21:34

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  • $\begingroup$ In case you were unsure if finiteness really is necessary: $(\Bbb N,+)$ is an associative (even commutative) algebraic structure and satisfies the cancellation law, but there is no identity. Also closure should be assumed, otherwise it doesn't make sense to consider $a\cdot b$ for all $a,b\in G$. You could probably glean this from the fact that the cancellation laws hold if you wanted to go that route. $\endgroup$ – Cameron Williams Oct 16 '15 at 0:58
  • $\begingroup$ Simply being a binary operation makes it clised, I think. Doesn't closed simply mean every a*b is a member of the set? $\endgroup$ – fleablood Oct 16 '15 at 2:01
  • $\begingroup$ Thank you for the input, this definitely helps my understanding $\endgroup$ – Burgundy Oct 16 '15 at 15:54
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Hint: show that cancellation means that the maps $L_g: G \to G$ and $R_g: G\to G$ defined by $L_g(a) = g\ast a$ and $R_g(a) = a\ast g$ are injective, and since $G$ is finite, they are surjective, as well.

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I think this is the answer, but I'm not sure what "cancelation rules" mean.

We are given a binary associative operation. So we need to show an identity and inverses.

I think "cancelation" means for every c and every a and c*a and a*c there are c' and c_ such that c'ca = a and acc_ =a. We need to show c'c = cc_ = b'b = bb_ for all c and b.

If we can do this c'c_ would be an identity.

Then we'd have to show for inverses that c' = c_ for all c.

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  • $\begingroup$ Why did you answer this question if you don't know? $\endgroup$ – Matt Samuel Oct 16 '15 at 4:45
  • $\begingroup$ I gave an outline. The OP can file in the rest. $\endgroup$ – fleablood Oct 17 '15 at 4:08

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