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A group of prime order is cyclic and so unique.

But is this the only possibility?

How do you characterize finite group of order $n$ for which it is unique up to isomorphism?

As pointed out by friend earlier, if $n=pq$ is composite, then $C(p) \times C(q)$ is not isomorphic to $C(n)$ and we are done.

But Here is a theorem:

Let $n$ be a positive integer. Then the cyclic group $C(n)$ of order $n$ is the only group of order $n$ if and only if one has $(n, \phi(n)) = 1,$ where $\phi$ denotes the Euler phi function.

I'm just trying to understand it.

Brian Moehring
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patamon
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    There is a partial answer for semiprime group orders. – Eric Towers Dec 13 '19 at 22:56
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     – patamon Dec 13 '19 at 22:57
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    here is a paper which classifies all $n$ for which all groups of order $n$ are cyclic. worth remarking: your friend's comment is not correct. All groups of order $15=3\times 5$ are cyclic, for example. – lulu Dec 13 '19 at 23:06
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    Note that your statement "if $n=pq$ is composite, then $C(p)\times C(q)$ is not isomorphic to $C(n)$" is false. In fact, if $p \neq q$ are primes, then we always have $C(p)\times C(q) \cong C(pq)$ – Brian Moehring Dec 13 '19 at 23:08

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