I'm following some notes on group cohomology (namely these ones by Peter J. Webb: http://www-users.math.umn.edu/~webb/oldteaching/Year2010-11/8246CohomologyNotes.pdf) and I'm stuck with a proposition I'm not understanding. The setting is the following: I have a presentation of $G$ $$1 \rightarrow R \rightarrow F \rightarrow G \rightarrow 1$$ and the author proves that it induces an exact sequence of $\mathbb{Z}G$-modules $$0 \rightarrow \dfrac{R}{R'} \rightarrow \dfrac{IF}{IR\cdot IF} \rightarrow IG \rightarrow 0$$ Here the author states that the middle term is $\mathbb{Z}G \otimes_{\mathbb{Z}G} IF = (\mathbb{Z}G)^{\mathrm{rk}F}$.
I can't understand the reasoning here. I know that being $F$ free the augmentation ideal $\mathbb{Z}F$ is free of rank $\mathrm{rk}F$, but the most I can prove is that that term is $\mathbb{Z}\otimes_{\mathbb{Z}R}IF$. I also noted that if I can prove that the middle term is $\mathbb{Z}G \otimes_{\mathbb{Z}F} IF$ then I can easily conclude that it is isomorphic to $(\mathbb{Z}G)^{\mathrm{rk}F}$ as a $\mathbb{Z}G$-module, but I don't know how to put it all together.
I think I've figured it out. We have the following isomorphism: $$\dfrac{IF}{IR\cdot IF} \cong \dfrac{IF}{IR\cdot \mathbb{Z}F\cdot IF} \cong IF \otimes_{\mathbb{Z}F}\dfrac{\mathbb{Z}F}{IR\cdot \mathbb{Z}F}$$
The canonical map $\mathbb{Z}F \rightarrow \mathbb{Z}G$ fits in the following short exact sequence:
$$0 \rightarrow IR\cdot \mathbb{Z}F \rightarrow \mathbb{Z}F \rightarrow \mathbb{Z}G \rightarrow 0$$
so this means that $\dfrac{IF}{IR\cdot IF}$ is actually isomorphic to $IF \otimes_{\mathbb{Z}F} \mathbb{Z}G$. Since $IF$ is free of rank $\mathrm{rk}F$ the thesis follows.