# A Finite Group of Nonprime Order which is Unique up to Cyclic Group

Maybe I have to wait until I learn and study more, but I just became curious. I know that every finite group of prime order is cyclic, and hence unique up to isomorphism. I have 2 questions about this.

(1) [Converse] If I know that a finite group of nonprime order $n$ is always cyclic, can we conclude that $n$ is a prime or 1? In other words, if $n$ is not a prime or 1, can we always find a group of order $n$ which is not cyclic?

(2) If I know that a finite group of nonprime order $n$ is always unique up to isomorphism, can we conclude that every finite group of order $n$ is cyclic as well?

(1) No, the smallest counterexample is $n = 15$; up to isomorphism the only group of order $15$ is $Z_3 \times Z_5 \cong Z_{15}$ but $15$ is not prime.

Much more generally, if $$n = p_1 \cdots p_k$$ for distinct primes $p_1, \ldots, p_k$ such that $p_i \, \nmid \, (p_j - 1)$ for all $1 \leq i, j \leq k$, then there is only one group of order $n$ up to isomorphism, namely $Z_n \cong Z_{p_1} \times \cdots \times Z_{p_k}$. In fact, such numbers account for all counterexamples; see http://groupprops.subwiki.org/wiki/Classification_of_cyclicity-forcing_numbers .

The list of such numbers is given as OEIS A050384 (with $1$ not excluded as in the question, NB we can regard it as the empty product of primes):

$$1, 15, 33, 35, 51, 65, 69, 77, 85, 87, \ldots.$$

With $1$ included, this is precisely the list of positive integers $n$ that are coprime with their cototient $n - \phi(n)$, where $\phi$ is Euler's totient function.

(2) If I understand you correctly---namely, if you're asking that if for some $n$ there is exactly one group of order $n$ up to isomorphism, is it cyclic)---the answer is yes, as for any $n$, there is a cyclic group $Z_n$ of order $n$.

• Your answer was perfect, from the counterexample to all the references! Thanks a lot. One question: $\not\vert$ here means not divisible, right? It seems a bit awkward in TeX. – Taxxi Jan 19 '15 at 3:36
• @TaxxiDriver You're welcome, and thanks, I'm glad you found it useful. Yes, that's what it means; I've replaced it with $\nmid$ (\nmid), which I'd temporarily forgotten about but which looks much better. – Travis Jan 19 '15 at 3:45
• Much better now :) – Taxxi Jan 19 '15 at 3:51
• Nice answer. Here's a paper from the Monthly that gives an elementary proof of your claim. – André 3000 Jan 19 '15 at 4:53
• @SpamIAm Thanks! I was curious about it, too. – Taxxi Jan 19 '15 at 9:43