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I am learning some group cohomology from Romyar Sharifi's online notes. (Everything I have written here can be found on page 5-6)

Let $G$ be a finite multiplicative group. Consider the group ring $\mathbb{Z}[G]$. The augmentation ideal is defined as follows:

Consider the ring homomorphism $\epsilon:\mathbb{Z}[G]\rightarrow\mathbb{Z}$ $$\epsilon:\sum_{g\in G}a_gg\mapsto\sum_{g\in G}a_g$$

The augmentation ideal is the kernel of $\epsilon$.

$\text{}$ An abelian group $A$ (written additively) is called a $G$-module if there is an action of $G$ on $A$ such that

(1) $1\cdot a=a$ for all $a\in A$.

(2) $g\cdot (a+b)=g\cdot a+g\cdot b$ for all $g\in G$ and $a,b\in A$.

(3) $g\cdot(h\cdot a)=(gh)\cdot a$ for all $g,h\in G$ and $a\in A$

The action of $\mathbb{Z}[G]$ on $A$ is defined as $$(\sum_{g\in G}a_gg)\cdot a=\sum_{g\in G}a_g(g\cdot a)$$

The group of $G$-coinvariants is defined as

$$A_G=\frac{G}{I_GA}$$

$\text{}$ I have doubt about the definition of $I_GA$. Is it defined like this: $$I_GA=\{\sum_{g\in G}(g\cdot a_g-a_g) | a_g\in A \}$$

I have one more doubt: It is mentioned that $A_G$ is the largest quotient of $A$ fixed by $G$. Does $G$ act on $A/I_GA$ (if so, how is the action defined)?

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    Yes, $G$ acts on the quotient since it acts on $A$ and $I_GA$ is a $G$-submodule. – 2017-01-18
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    For $I_GA$ to be a submodule, we must have $g\cdot (I_GA)\subseteq I_GA$ for all $g\in G$. But I can't understand how that is happening. – 2017-01-18
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    Is my definition of $I_GA$ correct ? – 2017-01-18
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    Yes, your definition is correct, and it should be fairly straightforward to show that it is invariant under the action of $G$. – 2017-01-18
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    Let $h\in G$ be arbitrary. Then $h\cdot(\sum_{g\in G}[ga_g-a_g])=\sum_{g\in G} [(hg)\cdot a_g-h\cdot a_g]$. What after this ? If $G$ is abelian, then the proof is done. If not, then how to proceed? – 2017-01-18

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