Nonprime N Where All Groups of Order N Are Isomorphic

Is there nonprime n where all groups of order n are isomorphic?

Yes—there are plenty of composite (non‑prime) integers $n$ for which every group of order $n$ is forced to be the same (indeed, cyclic). The smallest example is

$$ n = 15 = 3\cdot5 . $$

Because $\operatorname{Aut}(C_{5})\cong C_{4}$ has no element of order 3, no non‑trivial semidirect product $C_{5}\rtimes C_{3}$ can be formed, so the only group of order 15 is the cyclic one $C_{15}$.

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The complete criterion

Let

$$ n=p_{1}p_{2}\cdots p_{k}\quad (p_{1}<p_{2}<\dots<p_{k}) $$

be the prime–factorisation of $n$.

All groups of order $n$ are isomorphic iff

1. $n$ is square‑free (no $p_{i}^{2}\mid n$), and

2. for every pair of primes with $p_i < p_j$ one has

   $$ p_i\nmid (p_j-1). $$

Why these two conditions are necessary.

• If a prime square $p^{2}$ divides $n$, the abelian groups $C_{p^{2}}$ and $C_{p}\times C_{p}$ give two non‑isomorphic examples.
• If $p_i\mid(p_j-1)$ for some $i<j$, there exists a non‑trivial action $C_{p_i}\to\operatorname{Aut}(C_{p_j})$, producing the non‑abelian semidirect product $C_{p_j}\rtimes C_{p_i}$ alongside the cyclic one.

Why they are sufficient. With $n$ square‑free, every Sylow subgroup is cyclic. Condition (2) forces each Sylow subgroup to be normal (Sylow counts give $n_{p_i}\equiv1\pmod{p_i}$ and $n_{p_i}\mid n/p_i$; if $p_i$ divided some $p_j-1$ we could have a non‑normal Sylow). A direct product of mutually normal cyclic Sylow subgroups is cyclic, so every group of that order is $C_n$.

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Infinitely many such $n$

Choose any set of distinct primes with no divisibility relations of the form $p_i\mid(p_j-1)$. Products of two primes are easiest:

$n=pq$	Is every group cyclic?	Reason
$15=3\cdot5$	Yes	$3\nmid 4$
$33=3\cdot11$	Yes	$3\nmid10$
$35=5\cdot7$	Yes	$5\nmid6$
$39=3\cdot13$	Yes	$3\nmid12$

Whenever $p<q$ and $p\nmid(q-1)$, there is only the cyclic group of order $pq$; thus there are infinitely many composite $n$ with the desired property.

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Summary

So the answer is yes: the composite integers satisfying the two conditions above (beginning with 15) have exactly one group structure—namely the cyclic group $C_n$.