Re: exercise on group rings

On Jul 7, 8:12 am, Ignacio <ignat.sor...@xxxxxxxxx> wrote:
On Jul 7, 8:36 am, Jack Schmidt <Jack.Schmidt.SciM...@xxxxxxxxx>
wrote:



On Jul 6, 8:50 pm, Ignacio <ignat.sor...@xxxxxxxxx> wrote:

Could you please help me with the following exercise:

Let G be a group and ZG its group algebra over integers. Also, let IG
be the augmentation ideal, i.e. the kernel of the homomorphism eps: ZG
--> G, mapping every element g to 1.

We need to prove that if IG is finitely generated as a left ideal,
then G is finitely generated.

Thank you!

eps:ZG -> Z

Suppose IG is generated by elements b_1,...,b_n in ZG. Let X be the
union of the supports of the b_i, that is the set of all g in G such
that the coefficient of g in b_i is nonzero for some i. Then IG is
also generated by the finite set {1-g : g in X} since each b_i is in
the Z-linear span of X. For any arbitrary g in G, express 1-g as a ZG-
linear combination of the 1-h for h in X, and show that the support of
the ZG-coefficients can be chosen to come from the subgroup generated
by X, and in particular that g itself is in the subgroup generated by
X. Since g is arbitrary, G = <X> is finitely generated.

Thank you, Jack! This really helps. It was a little tricky to "show
that the support of ZG-coefficients can be chosen to come from the
subgroup generated by X", so I did a workaround. Let H = <X>. Then as
you showed, 1-g = sum a_i (1-h_i), where h_i belongs to X. Consider
the permutation module Z[G/H]. The right-hand side of 1-g stabilizes
element H of this module. Therefore, 1-g itself stabilizes H. This
means that g belongs to H=<X> for all g in G.

The idea to consider supports was fruitful indeed. Thank you again!

Thanks for posting your much improved solution. The permutation
module made the argument much shorter and clearer.

.

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