Prove for $n \in \mathbb N$ and prime $p\mid n$, $\Phi_{np}(t) = \Phi_n(t^p)$

Question

I am asked to prove the result for cyclotomic polynomials that if a prime $p$ divides $n$, then $\Phi_{np}(t) = \Phi_n(t^p)$

By partitioning the set of factors of $n$ into sets:

$\{d\in \mathbb N \mid d\mid x, \; p \nmid d\} = X,\; pX, \dots, p^kX$ where $p^k \mid x, \; p^{k+1} \nmid x$

And by using an induction argument, I have reduced the equation to:

$$\Phi_{np}(t)\prod_{d\in X}\Phi_{d}(t)\Phi_{pd}(t) = \Phi_n(t^p)\prod_{d \in X} \Phi_d(t^p)$$

This then leads me to believe $$\prod_{d\in X}\Phi_{d}(t)\Phi_{pd}(t) = \prod_{d \in X} \Phi_d(t^p)$$

And more specifically:

$$\Phi_d(t)\Phi_{pd}(t) = \Phi_d(t^p)\;\; \forall d\in \mathbb N, p\nmid d$$

However, I am unsure how to prove either of these last two results. I am also unsure if there is perhaps a better way to show the main result I am trying to achieve, since even with this inductive argument I'm not too sure what the base case should be.

Any help would be appreciated, thank you.

Answer 1

Your last claim is true:

$\Phi_d(t)\Phi_{pd}(t) = \Phi_d(t^p)\;\; \mbox{when} \;\; p\nmid d$

This follows directly from $\Phi_n(x^m)= \displaystyle\prod_{d\mid m} \Phi_{dn}(x)$ when $\gcd(m,n)=1$. See a proof here.

Just set $n=d$ and $m=p$.