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I've been proposed by a very interesting theoretical problem in my abstract algebra class. It seems like one of those open problems with a concise and simple wording, but the proof and answer could probably take up more than 20 pages to explain.

Given any set (nonempty and finite), is it possible to impose a binary operation on the set such that it can turn into

1) A group

2) An abelian group

3) A cyclic group

I think answering (3) will answer (2) immediately. Anyways, I thought about it and I think for (3), I just have to take an element, keep multiplying by itself to create a cyclic.

As for (1), I am not sure. I think if you have some ingenuity, you could. But there could be some crazy sets out there that doesn't have this property

EDIT: Okay, so the bijective map and the inverse part I got. So I take an element from the set S and map it to T. That is I get $\phi(a * b) = \phi(a) *' \phi(b)$.

I don't understand how you got the + operation and I don't understand what you mean by "evaluation by last residues"

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The problem is a lot simpler than you're making it out to be. It should be noted that answering (3) answers all of the other ones immediately (why?).

Hint: What does $|S|=n$ mean for $S$ a set? It means there's a bijective map between BLANK and BLANK. Do you see a way to turn the second blank into any of the above types of groups?

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Given a finite set $S$ of cardinality $n$, there's obviously a bijection $\phi$ from $S$ to $T =\{1,2,...,n\}$. Therefore every element of $S$ is equal to $\phi^{-1}(a)$ for some $a \in \{1,2,...,n\}$. Note $T$ is a cyclic group with respect to addition (or formally, addition and then evaluation to the least residue). Therefore for any elements $\phi^{-1}(a)$ and $\phi^{-1}(b)$ in $S$ you can just define $\phi^{-1}(a)\ast\phi^{-1}(b)$ to be $\phi^{-1}(a+b)$.

Then $S$ will be a cyclic, abelian group.

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    @Hawk This means that $a_k + a_l = a_{k+l \bmod n}$ where $a_i$ is the $i$-th element of $S$. The modulus operation takes care of the "wrap around": $a_1 + a_{n-1} = a_{n\bmod n} = a_0$.2015-09-11