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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.

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    Do you know what a unit is? If so, this question should be fairly straightforward. – 2017-02-27
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    @Qudit it's an element with a multiplicative inverse, right? – 2017-02-27
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    @Qudit if it were just $\mathbb{Z}$, then $U(\mathbb{Z}) = \{ -1, 1 \}$, but it's not just $\mathbb{Z}$, it's $\mathbb{Z}(G)$, and I'm having trouble visualizing multiplicative inverses in $\mathbb{Z}(G)$. And somehow $G$ is supposed to embed as a proper subgroup of $U(\mathbb{Z}(G))$ here... – 2017-02-27
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    I find very unfortunate that notation $R[G]$ for group rings (or monoid rings, or semigroup rings, whatever...). Why the hell the (a lot more) intuitive notation $R[X^G]$ is not used? sigh... – 2017-02-27

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It's typical to view $G$ as a subset in $\mathbb{Z}[G]$ by identifying $g\in G$ with the formal sum $1g\in\mathbb{Z}[G]$. These are invertible since under the multiplication in $\mathbb{Z}[G]$, $1g\cdot 1g^{-1}=(1\cdot 1)gg^{-1}=1e$, which is the identity of $\mathbb{Z}[G]$. In this case, the group $G$ is proper in the group of units since the elements $-1g$ are also units in $\mathbb{Z}[G]$, for instance.

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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.