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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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up vote 1 down vote accepted

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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How do we assume it even maps from S to some set T? How do we know T even exists? – Hawk Oct 21 '12 at 3:49
The problem states that your starting set, $S$, is non-empty and finite. So $S$ contains some finite numbers of elements, say $n\neq 0$. Then $S$ is obviously in bijection with the set $\{1,\ldots,n\}$, which we call $T$ for ease. – Bey Oct 21 '12 at 4:18
you seem to be getting caught up in the formalism. The set $S$ is finite, so you can think of its elements as the "first," "second," "third," etc.You can make $S$ into a cyclic group in the sense that you can define the group operation of, say, the "second" and "fifth" elements to be your "seventh" element. Your "i"th and "j"th elements operated together will be your "i+j"th element (of course you must reduce to the least residue). If $|S| = n$, there is no difference between $S$ and the group cyclic group $Z_{n}$ apart from the names you give the elements and what you call the operation. – D_S Oct 21 '12 at 5:26
Could you explain what this means (or formally, addition and then evaluation to the least residue) – Hawk Oct 22 '12 at 19:06
@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": $a_1 + a_{n-1} = a_{n\bmod n} = a_0$. – AlexR Sep 11 '15 at 17:13

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