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Let $G$ be a group, define the augmentation map $\varepsilon:\mathbb{Z}G\rightarrow\mathbb{Z}$ such that $\varepsilon(g)=1$ for all $g\in G$. The kernel $I$ of $\varepsilon$ is called the augmentation ideal of $\mathbb{Z}G$.

What I want to prove is this: if $S$ is a set of elements of $G$ such that the elements $s-1\;(s\in S)$ generate $I$ as a left ideal then $S$ generates $G$.

Here what I have done: every element of $I$ is of the form $\sum m_gg(s_g-1)$ where $m_g\in\mathbb{Z}$ and $s_g\in S$. So if $g\in G$ then $g-1\in I$ so $g-1=\sum m_{g_i}g_i(s_i-1)$ from this I want to deduce that $g_1=g, g_0=1$ and $g_i=g_{i+1}s_i^{\pm 1}$.

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    @Aaron:$I$don't know if my approach is right, but it was the only thing that I thought.2011-08-23

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