Let $v$ be a positive integer. Show that if any group of order $v$ is cyclic then $v$ is not divisible by the square of a prime.
This was originally an iff proof but I've proved the other direction. I have a feeling this direction should be rather straight forward but I can't seem to get it. I was trying a proof by contradiction.
This is obvious in the case where $v$ is prime, so let's only look at composite $v$. Take two numbers $u$ and $w$ with $uw=v$. Then $\mathbb{Z}_u\times\mathbb{Z}_w$ is of order $v$ and therefore isomorphic to $\mathbb{Z}_v$.
From this we can conclude that any $u$ and $w$ we find are coprime (because $\mathbb{Z}_m\times\mathbb{Z}_n\cong\mathbb{Z}_{mn}$ iff $m$ and $n$ are coprime), and from that we get that every prime factor in $v$ appears no more than once.
