$\mathbb{Z}$ as $\mathbb{Z}G$ - module.

Question

Let G group. Recently i read that $\mathbb{Z}$ can be considered as $\mathbb{Z}G$- module with the trivial action, such that : $$\forall x\in \mathbb{Z}G,\ z\in \mathbb{Z}:\quad x\cdot z:=z$$ But if that is true then for every $x,x'\in \mathbb{Z}G$ and $z\in \mathbb{Z}$ we have that $$(x+x')\cdot z=x\cdot z+x'\cdot z\Rightarrow z=z+z.$$ Where is the problem about the definition of this action ?

Answer 1

The correct definition is to define $g\cdot x =x$ for all $g \in G$ and $x \in \mathbb{Z}$. Then, for an arbitrary element of $\mathbb{Z}G$, we let it act by the obvious linear extension. So, given $w=\sum_{g\in G} a_gg$ with $a_g \in \mathbb{Z}$ we would define $$ w\cdot x = \left( \sum_{g\in G} a_gg\right) \cdot x = \sum_{g \in G} a_g (g\cdot x) = \sum_{g \in G} a_gx. $$ So, in your case, for two group elements $g_1, g_2$ you would have $$ (g_1+g_2) \cdot x = g_1 \cdot x + g_2 \cdot x = x+x. $$ This is not $x$ because $g_1+g_2$ is not an element of $G$, but a linear combination in $\mathbb{Z}G$.

This is exactly analogous to defining a linear map on vector space $V$ by defining it solely on a basis and then just saying "we extend it to all of $V$ by linearity." Here, $G$ is the "basis" of $\mathbb{Z}G$ and $\mathbb{Z}$ is the "scalars."