Disclaimer: I'm a relative beginner in this area. I'm trying to prove that if one has a commutative ring $R$ and a prime number $p$, then there is an exact sequence of the form
$$\DeclareMathOperator\THH{THH}
0 \rightarrow \pi\_n(\THH(R)\_{h\mathbb{T}})/p\pi\_n(\THH(R)\_{h\mathbb{T}}) \rightarrow \pi\_n(\THH(R; \mathbb{F}\_p)\_{h\mathbb{T}}) \rightarrow \pi\_{n-1}(\THH(R)\_{h\mathbb{T}})[p] \rightarrow 0.
$$
My basic intuition is that arguing formally should go quite far. It would suffice to show that
$$
\THH(R)\_{h\mathbb{T}} \xrightarrow{\times p} \THH(R)\_{h\mathbb{T}} \rightarrow \THH(R; \mathbb{F}\_p)\_{h\mathbb{T}}
$$
is a fibration, for one could then take the long exact sequence in homotopy groups and then do some manipulations to quickly obtain the result. I'm not really sure how to show that one gets a fibration out of this however, or if one should take an entirely different approach. It does resemble the standard exact sequence one gets in the Bockstein spectral sequence, except $\THH(R)\_{h\mathbb{T}}$ is far from needing to be torsion-free and that wouldn't help give a fibration anyway. Thanks in advance.



An exact sequence involving THH
Let's extract a clear question (about spectra in general) from your question, and then answer it. Let $E$ be any spectrum.


There is the degree $p$ map $p:S\to S$ from the sphere spectrum to itself. Smashing with $E$, this gives a map $E\to E$; we also call this map $p$. It multiplies elements of $\pi\_n(E)$ by $p$. Denote its spectrum cofiber by $E/pE$. The cofibration sequence $E\to E\to E/pE$ is also a fibration sequence, because we are talking about spectra. From the exact sequence
$$
\dots \to \pi\_n(E)\to \pi\_n(E)\to \pi\_n(E/pE)\to \pi\_{n-1}(E)\to \pi\_{n-1}(E)\to \dots
$$
you get what I think you want: an exact sequence
$$
0\to coker(p)\to \pi\_n(E/pE)\to ker(p)\to 0
$$
where the group on the left is
$$
coker (p:\pi\_n(E)\to \pi\_n(E))=\pi\_n\otimes \mathbb F\_p
$$
(it could also be called $\pi\_n(E)/p$) and the group on the right is
$$
ker (p:\pi\_{n-1}(E)\to \pi\_{n-1}(E))=Tor(\pi\_{n-1},\mathbb F\_p)
$$
(I think you are also calling it $\pi\_{n-1}(E)[p]$.)


$\pi\_n(E/pE)$ is called the $n$th mod $p$ homotopy group of $E$.


Curiously, it can have elements of order $4$ if $p=2$.


