Question: If after subtracting two generating series, we get an expression such as:

$1−(1/x^2)$

does this mean that the resulting series has coefficient 0 for all (x, x^2, x^3, x^4,....) or is it not even considered a valid one because the x^2 is in the denominator?

Answer: 
I think you are confusing $1$ (the identity) and $\mathbf{0}$ here. It is a problem with using $1\_G$ as the identity too often over $e\_G$ (there are arguments for both). The identity in the first group is $(0,0,0)$ and in the second $(0,0)$. Certainly an isomorphism would have to send $(0,0,0)$ to $(0,0)$ or vise versa. But this hasn't told you anything about an isomorphism between them because *every* homomorphism between them has this problem. 

Perhaps what you are thinking is trying to map a generator '1' in one to another. But neither of these groups are cyclic. The element $(1,1,1)$ does not generate the first group and $(1,1)$ does not generate the second. It is almost always the case that $(1,1,\ldots,1)$ does not generate $C\_{n\_1} \times \cdots \times C\_{n\_k}$, where $C\_{n\_i}$ is the cyclic group of order $n\_i$. [There are some special cases when it does. You learn about these in the Fundamental Theorem of Finitely Generated Abelian Groups.] But here's a hint:

**HINT.** An isomorphism preserves the order of elements, i.e. if $\phi$ is an isomorphism of groups, then $g$ and $\phi(g)$ have the same order. So isomorphic finite groups have the same number of elements of a given order. Can you find an element of $\mathbb{Z}\_4 \times \mathbb{Z}\_2$ of 'higher' order that cannot be found in $\mathbb{Z}\_2 \times \mathbb{Z}\_2 \times \mathbb{Z}\_2$?

Alternatively, notice that if $g \in \mathbb{Z}\_2 \times \mathbb{Z}\_2 \times \mathbb{Z}\_2$, then $g+g= (0,0,0)$ for all $g$. But what about the element $\phi(g+g)$ as $g$ varies over all $g \in \mathbb{Z}\_2 \times \mathbb{Z}\_2 \times \mathbb{Z}\_2$? Can you use the properties of an homomorphism $\phi$ to show that if $\phi$ were an isomorphism that this cannot hold for *all* possible $\phi(g)$? That is, can you find $h \in \mathbb{Z}\_4 \times \mathbb{Z}\_2$ so that $h+h \neq (0,0)$? If so, can there be a $g \in \mathbb{Z}\_2 \times \mathbb{Z}\_2 \times \mathbb{Z}\_2$ with $\phi(g)=h$?
One technique which can sometimes be useful is to pick a positive integer $k$ then, for each element, calculate its $k$th power. Then find how many elements arose how many times among the $k$th powers.

Here, $k=2$ is suitable: find the square of each element. In $\mathbb{Z}\_2\times\mathbb{Z}\_2\times\mathbb{Z}\_2$, the identity turns up all 8 times, and this is enough to distinguish this group from any other 8-element group. But the technique can also be useful to distinguish between groups both of which have several elements of order 4 with various elements for their squares.