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I just learned the theorem that for every prime number $p,$ there is just one group of order $p$ down to isomorphism (the cyclic group $\mathbb{Z}/p\mathbb{Z}$).

But interestingly, the converse of this statement is not true, as there is just a single group of order $15$. This lead me to wonder: Is there a nice partial converse?

Consider a composite natural number $m \in \mathbb{N}$. I want to find the conditions for there to be multiple groups of order $m$. Since there is a cyclic group of every order, this is equivalent to finding a non-cyclic group of order $m.$ I have two conditions so far that imply multiple groups:

  1. $m$ is even ($>2$)

Then $m=2k$ for some $k>1$. As the dihedral group $D_k$ has order $2k=m$ and is not cyclic, there are multiple groups of order m.

  1. $p^2 \mid m$ for some prime $p$

Then let $m = p^2k$ for some $k\in \mathbb{N}$. Then the group $(\mathbb{Z}/k\mathbb{Z}) \times (\mathbb{Z}/p^2\mathbb{Z})$ and the group $(\mathbb{Z}/k\mathbb{Z}) \times (\mathbb{Z}/p\mathbb{Z}) \times (\mathbb{Z}/p\mathbb{Z})$ both have order $p^2k = m$ and are not isomorphic.


But these two conditions definitely don't cover all cases. Is there a nice condition for a composite number m such that there are multiple groups of order $m$?

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There is a characterization, but I wouldn't call it simple. If we define $f(n)$ to be the number of isomorphism classes of groups of order $n$, we see that $f(n) =1 $ if and only if $\gcd(n,\phi(n)) = 1$, where $\phi$ is the Euler totient function. More details about this can be found here (https://mathoverflow.net/questions/148731/for-which-n-is-there-only-one-group-of-order-n).

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  • $\begingroup$ Thanks! Very cool that the same formula works for primes as well. $\endgroup$ Dec 14, 2020 at 17:05

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