Pairwise isomorphic groups of the same order

Question

I know those cases exist for groups of order $n,n$ is prime but what if n is not prime?

Can we find such $n=p_1 p_2 ...$, so every 2 groups of order n are pairwise isomorphic?

The first isomorphism theorem might play a role here but I can't find a way to begin.

Answer 1

For every finite number $n$, the following conditions are equivalent:

1. There is only one group of order $n$ up to isomorphism (necessarily the cyclic group $\mathbb{Z}/n\mathbb{Z}$).

2. $n$ is a product of distinct primes $n = p_1p_2\dots p_k$ such that for all $i$ and $j$, $p_i$ does not divide $p_j-1$.

For example, this is true whenever $n$ is a prime number. The smallest non-prime example is $15$. Indeed, $15 = 3\times 5$, and $3$ does not divide $4 = 5-1$.

See here for more information.