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Let $G$ be a group and $I_G$ be the augmentation ideal of the group ring $\mathbb{Z}G$, i.e. $I_G$ consists of formal linear combinations $\sum n_i g_i$ ($n_i\in\mathbb{Z}$, $g_i\in G$) such that $\sum n_i=0$. Is there a characterization of the groups $G$ for which $\bigcap_k I_G^k=\{0\}$?

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up vote 6 down vote accepted

I'll restrict to the case that $G$ is finite.


This is true if and only if $G$ is solvable.

Let $J$ be a two-sided ideal in $\mathbb{Z}[G]$. We call $J$ idempotent if $J^2=J$. Note that $\bigcap I_G^k$ is idempotent. We call an idempotent ideal nontrivial if it is neither $\{ 0 \}$ nor $\mathbb{Z}[G]$.

The following is a theorem of Roggenkamp: $\mathbb{Z}[G]$ contains nontrivial idempotent ideals if and only if $G$ is non-solvable.

In one direction, this answers your question immediately. If $G$ is solvable, then $\mathbb{Z}[G]$ contains no nontrivial idempotent ideals, so $\bigcap I_G^k$ must be trivial and, as it does not contain the identity, it must be $\{ 0 \}$.

In the other direction, look at the specific nontrivial idempotent ideal $J$ which Roggenkamp constructs. (In the first Proposition, on the first page of his paper.) It clearly obeys $I_G \supseteq J$. So, for every $k$, $I_G^k \supseteq J^k = J$ so $\bigcap I_G^k \supseteq J$ and the intersection is not $\{ 0 \}$.


In the particular case that $G$ is perfect (equals its own commutator subgroup), Roggenkamp claims that $I_G$ itself is idempotent. Unfortunately, Roggenkamp claims this without citation. Fortunately, I think I have reconstructed a proof. For any $x$ and $y$ in $G$, note that $$\left[ (x-1)(y-1) - (y-1)(x-1) \right] x^{-1} y^{-1} = xyx^{-1} y^{-1} -1$$ is in $I_G^2$. By definition, the commutator subgroup of $G$ is all elements of the form $(x_1 y_1 x_1^{-1} y_1^{-1}) (x_2 y_2 x_2^{-1} y_2^{-1}) \cdots (x_r y_r x_r^{-1} y_r^{-1})$. So, for any $g$ in the commutator subgroup,

$$\begin{multline} g-1 = \left( x_1 y_1 x_1^{-1} y_1^{-1} -1 \right) (x_2 y_2 x_2^{-1} y_2^{-1}) (x_3 y_3 x_3^{-1} y_3^{-1}) \cdots (x_r y_r x_r^{-1} y_r^{-1}) + \\ \left( x_2 y_2 x_2^{-1} y_2^{-1} -1 \right) (x_3 y_3 x_3^{-1} y_3^{-1}) \cdots (x_r y_r x_r^{-1} y_r^{-1}) + \cdots + \left( x_r y_r x_r^{-1} y_r^{-1} -1 \right) \end{multline}$$

is in $I_G^2$.

So, if $G$ is its own commutator subgroup, then $I_G^2 = I_G$, as claimed.

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Another way to prove that $I_G=I_G^2$ when $G$ is its own commutator is to notice that $g\mapsto(g-1)+I_G^2$ induces a group isomorphism from the abelianization of $G$ to $I_G/I_G^2$. –  Keenan Kidwell Jun 2 '11 at 15:47
    
Beautiful! Do you know what happens with $\cap_k I_G^k$ in $\mathbb{Q}G$? (I believe it's not $0$ for any nontrivial finite group, but it is $0$ for free groups) –  user8268 Jun 2 '11 at 16:46
1  
Certainly not zero for any nontrivial finite group. Let $V$ be a nontrivial $\mathbb{Q}$-irrep of $G$. Let $\pi_V$ be the projection onto $V$. Then $\pi_V^2 = \pi_V$, and $\pi_V \in I_G$, so $\pi_V$ is in every power of $I_G$. –  David Speyer Jun 2 '11 at 17:38
    
If you have mathscinet access, check out ams.org/mathscinet/search/… and start chasing references –  David Speyer Jun 2 '11 at 17:46

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