In Krause and Nikolaus' "Lectures on topological Hochschild homology and cyclotomic spectra" (which can be found [here](https://www.uni-muenster.de/IVV5WS/WebHop/user/nikolaus/papers.html)), there is a lemma I'm trying to understand, but one line in the proof eludes me. The lemma and proof are as follows:



> 
> **Lemma A.4.** Let $X\to Y$ be a $G$-equivariant map of connective spectra where $G$ is a discrete group. Assume that $G$ acts trivially on $\Bbb{F}\_p$-homology of $X$ and $Y$ and that the induced map $$X\_{hG}\to Y\_{hG}$$ is a $p$-adic equivalence. Then also the initial map $X\to Y$ is a $p$-adic equivalence.
> 
> 
> *Proof*: Let $Z$ be the cofibre of the map $X\to Y.$ Then it also carries a $G$-action and the homotopy orbits $Z\_{hG}$ are $p$-adically trivial. We want to show that $Z$ is $p$-adically trivial, i.e. that the $\Bbb{F}\_p$-homology of $Z$ vanishes. Assume that it has a lowest non-trivial homology group $H\_n(Z;\Bbb{F}\_p).$ We get that $$H\_n(Z\_{hG},\Bbb{F}\_p) = H\_n(Z,\Bbb{F}\_p)/G.$$ The action of $G$ on $H\_n(Z,\Bbb{F}\_p)$ is not necessarily trivial, but still $2$-stage nilpotent by the long exact sequence. Thus the coinvariants are non-trivial which gives a contradiction.
> 
> 
> 


**Question 1:** I'm assuming that $H\_n(Z,\Bbb{F}\_p)/G$ refers to the coinvariants of the $G$ action on $H\_n(Z,\Bbb{F}\_p),$ although I'm not 100% sure about this. It's the only thing that seems to make sense in the context, but I'm not familiar with this notation being used to mean coinvariants. Is this indeed what is meant? (Also, where can I find a proof of the stated isomorphism? I think I have a spectral sequence argument that proves this, but it would be nice to find a reference as well.)


**Question 2:** Aside from the coinvariant notation, I follow the argument until the penultimate sentence. I have never heard of the notion of a "$2$-stage nilpotent group action," and I did not find any explanation when I did a search. I only find references to nilpotent groups, which does not seem to be what is meant here -- all abelian groups, trivial or not, are nilpotent, so this doesn't seem like it would be particularly useful in the argument. What does the "$2$-stage nilpotence" here mean, and why does it prove the result?


Although I could not figure out the meaning of $2$-stage nilpotence, I played around with long exact sequences in homology associated to $X\to Y\to Z$ to try to show that $H\_n(Z,\Bbb{F}\_p)\_G$ is nontrivial, and I think I've found a proof (which may or may not be what Krause and Nikolaus had in mind); I've included it for completeness.


**Proof:** Consider the long exact sequence in homology
$$
\cdots\xrightarrow{\partial\_{n+1}}H\_n(X,\Bbb{F}\_p)\xrightarrow{f\_n}H\_n(Y,\Bbb{F}\_p)\xrightarrow{g\_n}H\_n(Z,\Bbb{F}\_p)\xrightarrow{\partial\_n}H\_{n-1}(X,\Bbb{F}\_p)\xrightarrow{f\_{n-1}}H\_{n-1}(Y,\Bbb{F}\_p)\to 0.
$$
Set $K := \ker f\_{n-1} = \operatorname{im}\partial\_n.$ Observe that $K$ fits into an exact sequence
$$
0\to K\xrightarrow{i\_{n-1}}H\_{n-1}(X,\Bbb{F}\_p)\xrightarrow{f\_{n-1}}H\_{n-1}(Y,\Bbb{F}\_p)\to 0
$$
and that the map $\partial\_n$ factors as $H\_n(Z,\Bbb{F}\_p)\twoheadrightarrow K\xrightarrow{i\_{n-1}} H\_{n-1}(X,\Bbb{F}\_p).$ Then there are two cases to consider:


1. $K = 0.$ In this case, it follows that $\partial\_n = 0,$ and so we get a short exact sequence
$$
0\to K'\xrightarrow{i\_n} H\_n(Y,\Bbb{F}\_p)\xrightarrow{g\_n}H\_n(Z,\Bbb{F}\_p)\to 0,
$$
where $K' =\ker g\_n\subseteq H\_n(Y,\Bbb{F}\_p).$ Taking coinvariants, we get an exact sequence
$$
(K')\_G\xrightarrow{(i\_n)\_G} H\_n(Y,\Bbb{F}\_p)\_G\xrightarrow{(g\_n)\_G}H\_n(Z,\Bbb{F}\_p)\_G\to 0.
$$
However, the $G$-action on $H\_n(Y,\Bbb{F}\_p)$ is assumed to be trivial, which implies that the $G$-action on $K'$ is also trivial and that the map $(i\_n)\_G = i\_n.$ Thus,
$$
\ker\left((g\_n)\_G\right) = \operatorname{im}\left((i\_n)\_G\right) = \operatorname{im}i\_n = \ker g\_n,
$$
so the $G$-action on $H\_n(Z,\Bbb{F}\_p)$ is trivial. This implies that $H\_n(Z,\Bbb{F}\_p)\_G = H\_n(Z,\Bbb{F}\_p)\neq 0,$ and we get the desired contradiction.
2. $K\neq 0.$ Because $H\_n(Z,\Bbb{F}\_p)$ surjects onto $K$ and taking coinvariants is right exact, it suffices to prove that $K\_G\neq 0.$ Taking coinvariants of the short exact sequence
$$
0\to K\xrightarrow{i\_{n-1}} H\_{n-1}(X,\Bbb{F}\_p)\xrightarrow{f\_{n-1}}H\_{n-1}(Y,\Bbb{F}\_p)\to 0
$$
gives the exact sequence
$$
K\_G\xrightarrow{(i\_{n-1})\_G} H\_{n-1}(X,\Bbb{F}\_p)\_G\xrightarrow{(f\_{n-1})\_G}H\_{n-1}(Y,\Bbb{F}\_p)\_G\to 0,
$$
which simplifies to
$$
K\_G\xrightarrow{(i\_{n-1})\_G} H\_{n-1}(X,\Bbb{F}\_p)\xrightarrow{f\_{n-1}}H\_{n-1}(Y,\Bbb{F}\_p)\to 0
$$
because the action of $G$ is trivial on both $H\_{n-1}(X,\Bbb{F}\_p)$ and $H\_{n-1}(Y,\Bbb{F}\_p).$ But now $K\_G$ surjects onto $\operatorname{im}\left((i\_{n-1})\_G\right),$ and we have
$$
\operatorname{im}\left((i\_{n-1})\_G\right) = \ker((f\_{n-1})\_G)= \ker f\_{n-1}= K\neq 0,
$$
so we are finished.



$p$-adic equivalence of spectra with $G$-action
I don't know if "$2$-stage nilpotent" is a standard term for this, but I'm sure that what the author means is this: a group $A$ that $G$ is acting on has a subgroup $B$ such that the action of $G$ fixes every element of $B$ and such that also the resulting action of $G$ on $A/B$ is trivial. This is true for $A=H\_n(Z;F\_p)$, with $B$ being the image of $H\_n(Y;F\_p)$. 
That implies that the group of coinvariants $A\_G$ is nontrivial if $A$ is nontrivial, because either (1) $B\neq A$, so that $A/B$ is nontrivial and $A\_G$ maps onto the nontrivial group $(A/B)\_G$, or (2) $B=A$ so that $A\_G=A$ is nontrivial.
When $G$ acts on a spectrum $Z$ and $H\_j(Z)=0$ for $j<n$ then there is a spectral sequence for $H\_\ast(Z\_{hG})$ with $E^2\_{i,j}=H\_i(G;H\_j(Z))$. The corner group $E^2\_{0,n}=E^\infty\_{0,n}=H\_0(G;H\_n(Z))=H\_n(Z)\_G$ is $H\_n(Z\_{hG})$.


