For which $n$ is there only one group of order $n$?

hbghlyj

例如$\gcd(21,ϕ(21))=3$,所以21阶群不唯一.
Cyclic numbers是$\gcd(n,ϕ(n))=1$的正整数解$n$.
1, 2, 3, 5, 7, 11, 13, 15, 17, 19, 23, 29, 31, 33, 35, 37, 41, 43, 47, 51, 53, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 123, 127, 131, 133, 137, 139, 141, 143, 145, 149, 151, 157, 159, 161, 163, 167, 173...
对于这些$n$, $n$阶群只有$C_n$.
由Tibor Szele于1947年证明, 根据MathOverflow

hbghlyj

Three-group numbers Jørn B. Olsson
设$n=p_1^{a_1} p_2^{a_2} \ldots$使得
若$a_i \geq 2$则存在2个不同构的$p_i^{a_i}$阶群$C_{p_i^{a_i}}$和$C_{p_i^{a_i-1}}\times C_{p_i}$, 矛盾. 故$a_1=a_2=\dots=1$.
若$p_j\mid p_i-1$则存在2个不同构的$p_ip_j$阶群$C_{p_i}\rtimes C_{p_j}$和$C_{p_ip_j}$, 矛盾. 故$\gcd(n,\phi(n))=1$.

---

后半段证明不懂metacyclic

Conversely if $n$ is a product of distinct prime numbers with no relations, an abelian group of order $n$ is cyclic. It is known that a group $G$ of order $n$ is metacyclic. If $G$ is not abelian then $G / G^{\prime}$ and $G^{\prime}$ are nontrivial and there has to exist a relation between a prime divisor of $\left|G: G^{\prime}\right|$ and a prime divisor of $\left|G^{\prime}\right|$. Otherwise $G^{\prime}$ would be contained in the center of $G$ and thus be a direct factor of $G$. This is clearly not possible.