If $p$ is a prime number, then $\Bbb Z_{pn} \cong \Bbb Z_{p} \times \Bbb Z_{n}$ for a random integer $n$

Question

I don't think so, because $\gcd(p,n)$ is not $1$ for a random $n$, but I have no idea how to prove it.

Answer 1

$\mathbb{Z_8}\ncong \mathbb{Z_2} \times \mathbb{Z_4}$ as $\mathbb{Z_8}$ has an element of order $8$ but $\mathbb{Z_2} \times \mathbb{Z_4}$ does not.

Answer 2

If you have p=2, n=2. You will get a counterexample. And hence it can't hold for any n, but it could hold for some, like p=2, n=3.

Answer 3

You're right.Take $n=p$. Then the group $\mathbf{Z}_{p^{2}}$ is cyclic, $\mathbf{Z}_{p} \times \mathbf{Z}_{p}$ is not as every element vanishes when you add it to itself p times!

Answer 4

You're correct. Take $n=p$. Then $\mathbb{Z}/{p^2}$ is not isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$ because the latter isn't even cyclic: Suppose $\mathbb{Z}_p\times\mathbb{Z}_p=\langle (m,n) \rangle$. Then $(m,n)^p=(m^p,n^p)=(1,1)$, a contradiction since $(m,n)$ should have order $p^2$.