Let $\mathbb{C}_2$ denote the group of order $2$ and let $\mathbb{Z}_4$ denote the ring of integers modulo $4$. The group ring $R = \mathbb{Z}_4\mathbb{C}_2$ is commutative.

My problem is how to generate all the elements of this ring and the operations involved. I am a learner try to teach myself something to do with group rings.


Let's denote by $0,1,2,3$ the elements of $\mathbb{Z}_4$ and by $\mathbf{1}$ (the identity) and $\mathbf{u}$ the elements of $C_2$. An element of the group ring is of the form $$ a\mathbf{1}+b\mathbf{u} $$ where $a,b\in\mathbb{Z}_4$. So we have sixteen elements in all. The zero element is $0\mathbf{1}+0\mathbf{u}$, the identity is $1\mathbf{1}+0\mathbf{u}$.

Addition is performed in the obvious way: $$ (a\mathbf{1}+b\mathbf{u})+(c\mathbf{1}+d\mathbf{u})= (a+c)\mathbf{1}+(b+d)\mathbf{u} $$ Multiplication is performed with the usual rules: $$ (a\mathbf{1}+b\mathbf{u})(c\mathbf{1}+d\mathbf{u})= ac\mathbf{1}\mathbf{1}+ad\mathbf{1}\mathbf{u}+b\mathbf{u}\mathbf{1}+ bc\mathbf{u}\mathbf{u}= (ac+bd)\mathbf{1}+(ad+bc)\mathbf{u} $$ because $\mathbf{u}\mathbf{u}=\mathbf{1}$ in $C_2$.

Note that we haven't used any special property of $\mathbb{Z}_4$, just that it is a ring. By definition, the elements of the group (in this case $C_2$) commute with the elements of the ring (here $\mathbb{Z}_4$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.