Stable Homotopy groups of $X$ with coefficients in $G$ I have been reading the following lectures notes https://klein.mit.edu/~hrm/papers/cobordism.pdf on Thom's Cobordism theorem and the proof of Theorem $12.4$ in page $33$ the author introduces the concept of stable homotopy groups of $X$ with coefficients in an abelian group $G$. First he "shows" that for each abelian group $G$ there exists a spectrum $M_G$ with trivial singular homology in all dimensions, except zero, where it's $G$.
Then we define the homotopy groups $\pi_n(X;G):=[S^n,M_G\wedge X]$.
The author claims that we have an exact sequence of the form
$$...\rightarrow \pi_n(X;\mathbb{Z}_{p^{k}-1})\rightarrow \pi_n(X,\mathbb{Z}_{p^k})\rightarrow \pi_n(X,\mathbb{Z}_p)\rightarrow ...$$
but I don't see why this follows from the short exact sequence of the coefficients. I would see that this would be the case if $M_G$ were Eilenberg-Maclne spectra, since these represent singular homology, but In this more general context I don't see why this follows.
Moreover the author also claims that there exists a Whitehead theorem and that $\pi_n(X,\mathbb{Q})\cong \pi_n(X)\otimes \mathbb{Q}$. This also I don't know why is true.
Therefore I would like to know if anyone knows of a refrence where this theory of homotopy groups with coefficients is done, and where I could find these statements proved ? Or any insight regarding why they would be true.
Thanks in advance.
 A: Any cofiber sequence $X\to Y\to Z$ gives rise to a long exact sequence in homotopy groups. Smashing with a fixed spectrum preserves cofiber sequences, so this gives rise to the long exact sequence you write. These are standard facts and can be found in Stefan Schwede's book project on symmetric spectra or his lecture notes on equivariant spectra (use the case of the trivial group).
The claim that $\pi_\ast(-;\mathbb{Q})\cong\pi_\ast(-)\otimes \mathbb{Q}$ can be seen as follows. First note that $H_\ast(-;\mathbb{Q}\cong \pi_\ast(-)\otimes\mathbb{Q}$, since both sides form a homology theory on spectra, because $\mathbb{Q}$ is flat, and they have have the same value on the sphere spectrum, namely $\mathbb{Q}$ in degree zero and $0$ in higher degrees. Next, note that $H\mathbb{Q}\simeq M\mathbb{Q}$, since the cofiber of the Hurewicz map $\mathbb{S}\to H\mathbb{Z}$ has torsion homotopy groups, hence it vanishes after smashing with $H\mathbb{Q}$, so that $H\mathbb{Z}\wedge H\mathbb{Q}\simeq H\mathbb{Q}$, which is the defining property of a Moore spectrum.
