Prove that $G$ embeds as proper subgroup into group of units of the group ring $\mathbb{Z}(G)$ I need to prove that a group $G$ embeds as a proper subgroup into the group of units of the group ring $\mathbb{Z}(G)$.
The group ring $\mathbb{Z}(G)$ consists of the set of formal sums $z_{1}g_{1} + z_{2}g_{2} + \cdots + z_{k}g_{k}$ where the $z_{i}$ are integers, and the $g_{i}$ are elements of the group $G$. To start out, I have no idea what the group of units of $\mathbb{Z}(G)$ is, so if someone could please tell me how to figure that out, I would appreciate it very much.
Next, I need to show that $G$ embeds as a proper subgroup of these units. So, there must exist an injective map under which $G$ is an image, and while $G \subset U(\mathbb{Z}(G))$ (the group of units of $\mathbb{Z}(G)$) there must exist at least one element of $U(\mathbb{Z}(G))$ that is not in $G$ in order to make that conclusion proper. As to what that map is, I don't know. 
This is pretty much all I was able to gather from this problem. As to what the map is and how to find it,  I do not know. Therefore, I would very much appreciate some extremely patient person (who is willing to entertain lots and lots of follow-up questions as I figure out what I need to do) pointing me in the right direction of what I am supposed to be doing here.
Thank you. 
Also, I don't know anything about modules or categories, so please don't mention them in your answers.

If you write an element  $x\in ZG, x=\sum x_i1_{g_i}$, you can define $f:G\rightarrow ZG$ by $f(g)=1_g$.. You have $1_g.1_{g^{-1}}=1$. This shows that $1_g$ is a unit.
