Let $k$ be a commutative ring with unit and let $(A,\varepsilon)$ be a (not necessarily commutative) augmented, finitely generated $k$-algebra with augmentation ideal $I$. 


If $\mu(A)$ denotes the minimal number of algebra generators of $A$ and $\mu(I)$ the minimal number of generators of $I$ as ideal, we have $\mu(I) \le \mu(A)$. 


(For, if $A$ is generated as algebra by $x\_1,...,x\_n$, then it's also generated by $y\_i := x\_i - \varepsilon(x\_i)\cdot 1\in I$. If $y$ is any element in $I$, write $y = a + f\langle y\_1,...,y\_n\rangle$ where $a \in k$ and $f$ is a polynomial in the (noncommuting) vaiables $y\_1,..,y\_n$ having no constant term. Hence $0=\varepsilon(y)=\varepsilon(a)=a$, i.e. $y = f\langle y\_1,...,y\_n\rangle$ is in the ideal generated by $(y\_1,...,y\_n)$. This shows $I=(y\_1,...,y\_n)$ and hence $\mu(I) \le n$)


I wonder what can be said in the case of group algebras: 


**Question:** Does for finitely generated groups the equality 
$\mu(\mathbb{Z}[G]) = \mu(I)$ hold ? 


Note: If $\mu(G)$ denotes the minimal number of generators of the group $G$, we have for finite $G$: 
$$\mu(I) \le \mu(\mathbb{Z}[G]) \le \mu(G).$$
So the equality would follow from $\mu(I)=\mu(G)$ but I don't know if the latter is true. 



Minimal number of algebra generators of a group ring
The paper [The presentation rank of a direct product of finite groups](http://www.sciencedirect.com/science/article/pii/0021869374900623) by Cossey, Gruenberg and Kovacs shows that the difference between the minimal number of generators of $G$ and $I$ can be as big as you like for finite groups.


