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Let $*$ be a binary operation defined on a non empty, finite set $G$ such that it follows associative, commutative and cancellation law.Show that $G$ under the operation $*$ is abelian.
Now for $G$ to be abelian, it is missing identity and inverse existence. But above statement doesn't seem to contribute anything for proving above. Also there is the bit about the group $G$ being finite, the importance of which I fail to recognize.

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    Yup, an abelian group.2012-09-10

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A start: Let $a$ be any element of our set.

Consider the sequence that goes $a$, $a\ast a$, $a\ast(a\ast a)$, and so on like that forever. Let's switch to more standard juxtaposition notation for the product, and exponential notation. So we are looking at the sequence $a,a^2,a^3,\cdots$.

Because our set $G$ is finite, there are positive integers $i$ and $j$, with $i\lt j$, such that $a^i=a^j$. In fact, if $G$ has $n$ elements, there will be such $i$, $j$ with $j\le n+1$.

Thus for any $x$, $a^i x=a^j x$. By cancellation we have $a^{j-i}x=x$. We have established the existence of an identity element. It remains to show the existence of inverses.

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    OK - I misread, sorry, I thought he was trying to prove it was abelian.2012-09-10
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inverse existence

is $a \in G$ we know there is $0 such that $a ^ k$ is neutral, then the inverse of $a$ is $a^{k-1}$ because $aa^{k-1} = a^{k-1}a = a^k$.