Linear combination of group elements as a ring? 
I'm having trouble understanding this example. Why is $\mathbb{Z}$G considered a ring? How would this be shown? Any advice would be helpful.
 A: The group ring $\mathbb{Z}G$ is one way to turn a group into a ring (although it's a little weird).  Throughout this discussion, I am thinking of $G$ as a finite group (so that everything makes sense).  If you want to work where $G$ is infinite, then the sums are all finite sums, in other words, all but finitely many of the coefficients of elements of $G$ are zero.
Set theoretically, you can think of $\mathbb{Z}G$ as $\mathbb{Z}^{|G|}$.  In other words, for each element of $G$, you have a copy of $\mathbb{Z}$.  So, you can think of this as a vector of length $|G|$.  When we write $a_gg$, we mean that in the $g$'th coordinate, the value is $a_g$.  Then, we add elements just as we would vectors, so elements in the same dimension, i.e., $a_gg+b_gg=(a_g+b_g)g$ because they are both in the same coordinate.
Observe that the addition does not use the group structure.  Multiplication, on the other hand, is based on the group structure.  In particular, if we want to multiply two elements $(a_gg)(b_hh)=(a_gb_h)(gh)$, so that we multiply the coefficients and change the coordinate from $g$ and $h$, to the group element $gh$.  Then, we extend this by linearity.
It might be helpful to consider the example $\mathbb{Z}S_3$.  Then, an element of this group ring has 6 entries, one for each group element: 
$$
(n_e,n_{(12)},n_{(13)},n_{(23)},n_{(123)},n_{(321)})
$$
For "simplicity" we write this as
$$
n_ee+n_{(12)}(12)+n_{(13)}(13)+n_{(23)}(23)+n_{(123)}(123)+n_{(321)}(321).
$$
Think of it as writing the sum in terms of basis elements (the basis elements are the group elements).  Addition is just done component-wise (we add corresponding dimensions in the vector or the other representation).
If you want to compute $(2(13))\cdot(-3(23))=-6[(13)(23)]=-6(321)$ so, in terms of the vectors,
$$
(0,0,2,0,0,0)\cdot(0,0,0,-3,0,0)=(0,0,0,0,0,-6).
$$
The value is determined by multiplying the integers and the location is determined by the group elements.  Finally, if we'd like to multiply the following,
$$
(2(13)-3(23))(-1e+1(13)+2(23))
$$
you use the distributive law to get
$$
-2[(13)e]+2[(13)(13)]+4[(13)(23)]+3[(23)e]-3[(23)(13)]-6[(23)(23)]
$$
Simplifying, the result is
$$
-4e-2(13)+3(23)+4(123)-3(321).
$$
A: As g and h are essentially permutations, which may be represented equivalently as permutation matrices, those sums are simply linear combinations of those permutation matrices. As ordinary matrix addition and multiplication obtains, the addition is the straightforward $a_g\mathbf{g}+b_g\mathbf{g}=(a_g+b_g)\mathbf{g}$ where $\mathbf{g}$ is now the permutation matrix representing that particular group operation.
A multiplication, for example
$$(\dots + a\mathbf{g_1}+b\mathbf{g_2} + \dots)(\dots + c\mathbf{g_1}+d\mathbf{g_2} + \dots)$$
is now a sum of matrix products
$$\dots + ac\mathbf{g_1}\mathbf{g_1}+bc\mathbf{g_2}\mathbf{g_1}+ad\mathbf{g_1}\mathbf{g_2}+bd\mathbf{g_2}\mathbf{g_2} + \dots$$
where each of the products is just another permutation matrix directly equivalent to the composition of the two corresponding group operations. Maybe $\mathbf{g_1}\mathbf{g_2}=\mathbf{g_2}\mathbf{g_1}=\mathbf{I}$ and $\mathbf{g_1}\mathbf{g_1}=\mathbf{g_3}$?
