The trivial units of a group ring form a group.
A group ring is a construction that combines a group and a ring. Given a group $G$ and a ring $R$, the group ring of $G$ over $R$, denoted as $R[G]$, is the set of all finite formal linear combinations of elements of $G$ with coefficients from $R$. The elements of $R[G]$ are typically expressed as sums of the form $\sum {r_ig_i}$ where $r_i$ are elements of $R$ and $g_i$ are elements of $G$.
The trivial units of $R[G]$ are the elements that can be expressed as a product of a unit in R and an element in $G$. This set is denoted as $U(R)G$, where $U(R)$ represents the group of units in the ring $R$.
To show that $U(R)G$ forms a group, we need to verify the group axioms:
Closure: For any two trivial units $u = rg$ and $v = sh$ in $U(R)$, their product $uv$ is also a trivial unit. This is true because the product is given by $(rg)(sh)$ = $(rs)(gh)$, where $rs$ and $gh$ are units in $R$and elements in $G$, respectively.
Associativity: The product of trivial units is associative, following the associativity of the ring $R$ and the group $G$.
Identity: The identity element in $U(R)G$ is the trivial unit $1$, where $1 = 1_G$ is the identity element of the group $G$.
Inverses: For each trivial unit $u = rg$ in $U(R)G$, its inverse is given by $u^{-1} = (r^{-1})(g^{-1})$, where $r^{-1}$ is the inverse of $r$ in $R$ and $g^{-1}$ is the inverse of $g$ in $G$. Since both $r^{-1}$ and $g^{-1}$ exist in their respective groups, the inverse $u^{-1}$ is also a trivial unit.
Therefore, the set of trivial units $U(R)G$ forms a group under multiplication.
But $U(R)G$ is a set, it is not a ring. It is not a group ring. It is a group. It is isomorphic to $U(R)\times G$ as a group.
The isomorphism can be defined as follows: For any element $u = rg$ in $U(R)G$, where $r$ is a unit in $R$ and $g$ is an element in $G$, we can map $u$ to the pair $(r, g)$ in $U(R) × G$ This mapping is a bijective homomorphism, and under this isomorphism, the group operations in $U(R)G$ and $U(R) × G$ correspond.
However, for a generic group ring $R[G]$, I wonder what can be said about the group ring $T[G]$, where $T$ is the subring of $R$ generated by the group $U(R)$ (formed by closing $U(R)$under addition).
Thanks.
