 Solution

For $p\equiv 1(\mathrm{mod}20)$, put $\theta =\sqrt{-p}$. Since $-p\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$. Modulo five,The two roots are distinct, so the Dedekind factorization theorem givesIn particular $P\ne \overline{P}$.

We prove that $[P]$ has order of a group element greater than $n$. Suppose instead that $P^k$ is a principal ideal for some $1\le k\le n$. Since $P^k\subseteq R$, its generator is in $R$: write $P^k=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Taking ideal norms gives the positive field normBut $5^k\le 5^n<p$, so $b=0$. Thus $a^2=5^k$, forcing $k$ to be even and $a=\pm 5^{k/2}$. This would givecontrary to unique factorization of ideals in a number field and $P\ne \overline{P}$. Consequently no such $k$ exists. The finiteness of the ideal class group ensures that $[P]$ has finite order of a group element, soFor every positive integer $n$, the allowed infinitude of prime numbers in the progression $1(\mathrm{mod}20)$ supplies one larger than $5^n$. Applying the construction proves unbounded element orders in imaginary quadratic ideal class groups. In particular, this establishes elements of arbitrarily large order of a group element, rather than merely unbounded class numbers.