Editing Group ring (section)

==Some basic properties==
Using 1 to denote the multiplicative identity of the ring ''R'', and denoting the group unit by 1<sub>''G''</sub>, the ring ''R''[''G''] contains a subring isomorphic to ''R'', and its group of invertible elements contains a subgroup isomorphic to ''G''. For considering the [[indicator function]] of {1<sub>''G''</sub>}, which is the vector ''f'' defined by
:<math>f(g)= 1\cdot 1_G + \sum_{g\not= 1_G}0 \cdot g= \mathbf{1}_{\{1_G\}}(g)=\begin{cases}
1 & g = 1_G \\
0 & g \ne 1_G
\end{cases},</math>
the set of all scalar multiples of ''f'' is a subring of ''R''[''G''] isomorphic to ''R''. And if we map each element ''s'' of ''G'' to the indicator function of {''s''}, which is the vector ''f'' defined by
:<math>f(g)= 1\cdot s + \sum_{g\not= s}0 \cdot g= \mathbf{1}_{\{s\}}(g)=\begin{cases}
1 & g = s \\
0 & g \ne s
\end{cases}</math>
the resulting mapping is an injective group homomorphism (with respect to multiplication, not addition, in ''R''[''G'']).

If ''R'' and ''G'' are both commutative (i.e., ''R'' is commutative and ''G'' is an [[abelian group]]), ''R''[''G''] is commutative.

If ''H'' is a [[subgroup]] of ''G'', then ''R''[''H''] is a [[subring]] of ''R''[''G''].  Similarly, if ''S'' is a subring of ''R'', ''S''[''G''] is a subring of ''R''[''G''].

If ''G'' is a finite group of order greater than 1, then ''R''[''G''] always has [[zero divisors]]. For example, consider an element ''g'' of ''G'' of order {{math|1= {{!}}''g''{{!}} = ''m'' > 1}}. Then 1 − ''g'' is a zero divisor:
<math display="block">
(1 - g)(1 + g+\cdots+g^{m-1}) = 1 - g^m = 1 - 1 =0.
</math>

For example, consider the group ring '''Z'''[''S''<sub>3</sub>] and the element of order 3 ''g'' = (123). In this case,
<math display="block">
(1 - (123))(1 + (123)+ (132)) = 1 - (123)^3 = 1 - 1 =0.
</math>
A related result: If the group ring <math> K[G] </math> is [[Prime ring|prime]], then ''G'' has no nonidentity finite normal subgroup (in particular, ''G'' must be infinite).

Proof: Considering the [[contrapositive]], suppose <math> H </math> is a nonidentity finite normal subgroup of <math> G </math>. Take <math> a = \sum_{h \in H} h </math>. Since <math> hH = H </math> for any <math> h \in H </math>, we know <math> ha = a </math>, therefore <math> a^2 = \sum_{h \in H} h a = |H|a </math>. Taking <math> b = |H|\,1 - a </math>, we have <math> ab = 0 </math>. By normality of <math> H </math>, <math> a </math> commutes with a basis of <math> K[G] </math>, and therefore
:<math> aK[G]b=K[G]ab=0 </math>.
And we see that <math> a,b </math> are not zero, which shows <math> K[G] </math> is not prime. This shows the original statement.