Why does $\underline{cd}\:G\leq silp\:G\leq\underline{cd}+1$ hold?

Question

I read the about text in a book but don't understand how or why this inequality is "straightforward" can anybody explain this to me?

Answer 1

First, you can show that in the definition of $\underline{cd}$, you can use ${\mathbb Z}G$-projective modules in the second slot instead of ${\mathbb Z}G$-free modules.

The first inequality then follows from the fact that the second sup is taken over all ${\mathbb Z}G$ modules rather than just ${\mathbb Z}$-free modules.

For the second inequality, take an arbitrary ${\mathbb Z}G$ module $A$ and a free ${\mathbb Z}G$ module $F$ mapping onto it, with kernel $K$:

$$0 \to K \to F \to A \to 0$$

Since $F$ is ${\mathbb Z}$-free and $K$ is a subgroup of $F$ as an abelian group, it's also ${\mathbb Z}$-free. The inequality then follows by looking at the long exact sequence for Ext.